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CUBO A Mathematical JournalVol.16, No01, (71110). June 2014
Trisectors like Bisectors with equilaterals instead of Points
Spiridon A. Kuruklis
Eurobank,
Group IT Security and Risk Management Division
Athens, GREECE
skuruklis@gmail.com
ABSTRACT
It is established that among all Morley triangles ofABCthe only equilaterals are the
ones determined by the intersections of the proximal to each side ofABC trisectors
of either interior, or exterior, or one interior and two exterior angles. It is showed
that these are in fact equilaterals, with uniform proofs. It is then observed that the
intersections of the interior trisectors with the sides of the interior Morley equilateral
form three equilaterals. These along with Paschs axiom are utilized in showing that
Morleys theorem does not hold if the trisectors of one exterior and two interior angles
are used in its statement.
RESUMEN
Se establece que entre todos los triangulos de Morley de ABC, los unicos equilateros
son theones determinados por las intersecciones del proximal a cada lado de los trisec-
tores ABCde angulos interior, o exterior, o uno interior y dos exteriores. Se muestra
que estos estan en triangulos equilateros de facto con demostraciones uniformes. Luego,
se observa que las intersecciones de trisectores interiores con los lados de un equilatero
Morley interior forman tres triangulos equilateros. Junto con el axioma de Pasch, se
utilizan para probar que el Teorema de Morley no se satisface si se usan los trisectores
de un angulo exterior y dos interiores.
Keywords and Phrases: Angle trisection, Morleys theorem, Morley trisector theorem, Morley
triangle, Morley interior equilateral, Morley central equilateral, Morley exterior equilateral, Paschs
axiom, Morleys magic, Morleys miracle, Morleys mystery.
2010 AMS Mathematics Subject Classification: 51M04
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1 Introduction
The systematic study of the angle trisectors in a triangle starts after 1899, when Frank Morley, a
Cambridge mathematician, who had just been recently appointed professor at Haverford College,
U.S.A. while investigating certain geometrical properties using abstract algebraic methods, made
the following astonishing observation, known since then as Morleys theorem.
In any triangle the trisectors of its angles, proximal to the three sides respectively, meet at the
vertices of an equilateral.
Fig.1
A Morley triangle ofABC is formed by the three
points of intersection of pairs of angle trisectors con-
nected by each triangle side. Obviously for a particular
side there are four possibilities for pairing trisectors since
there are four of them that the side connects. Thus Mor-
leys theorem claims that a Morley triangle ofABC is
equilateral, if it is formed by the intersections of trisectors
proximal to the three sides ofABCrespectively.
It should be noted that Morleys theorem, as it is
stated, is subject to interpretation as the term angle
could mean either interior or exterior angle, or even a
combination of both for the different instances of the term
in the statement.
According to the angle meaning, Morleys theorem
gives the following Morley equilaterals ofABC. The
intersections of the proximal trisectors of the interior angles form the interior Morley equilateral
ofABC. Also the intersections of the proximal trisectors of the exterior angles form the central
Morley equilateral ofABC. In addition the intersections of the proximal trisectors of one interiorand two exterior angles form an exterior Morley equilateral ofABC, and thus there are three
exterior Morley equilaterals ofABC. Fig.1 depicts the above Morley equilaterals. Proofs that
the above Morley triangles are in fact equilaterals are given in Part 3 of this work.
Fig.2
But so an obvious question, that several authors have raised,
begs for an answer. In aABC are there other Morley equilaterals
besides the interior, the central and the three exterior Morley equi-
laterals?
Apparently the requirement of Morleys theorem is satisfied by
three more Morley triangles formed by combinations of proximal tri-
sectors of an exterior and two interior angles. One of them is por-trayed in Fig.2. Some experimentation using computer generated
graphs for these triangles has tempted the belief that Morleys theo-
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Trisectors like Bisectors with equilaterals instead of Points . . . 73
rem holds for them as well [14]. But in Part 5, it will be proved that these are not equilaterals.
After the examination of all Morley triangles it will be shown that the equilateral ones are
exactly the interior, the central and the three exterior Morley equilaterals.
This enables the establishment of an analogy between the structures of the angle bisectorsand the angle trisectors in a triangle. Namely, the structure of trisectors resembles the structure
of bisectors with the inner and the exterior Morley equilaterals ofABC corresponding to the
incenter and the excenters ofABCrespectively, while the central Morley equilateral corresponds
to the triangle with vertices the excenters ofABC.
Morleys theorem is considered among the most surprising discoveries in mathematics as it
went curiously unnoticed across the ages. Ancient Greeks studied the triangle geometry in depth
and they could find it. But curiously they did not and it was overlooked during the following two
thousand years.
Angle trisectors exist regardless of how they can be constructed. If the structure of angletrisectors maintains the regularity which characterizes the triangle geometry then theorems must
exist for expressing it.
The first observation about this regularity may have forgotten. Morley didnt publish it until
25 years later by providing a sketchy proof, when the theorem had become already famous. But
Morley, excited by his discovery, travelled back to England to mention it to his expert friends. In
turn mathematical gossip spread it over the world and several journals proposed it for a proof.
Obviously, the simplicity of the theorem statement creates the expectation of an equally simple
proof. This simplicity challenges the mathematical talent.
The vast majority of publications on Morleys theorem has treated only the trisectors of the
interior angles and gave proofs for the interior Morley equilateral. In the preface of the firstpublication on the subject, by Taylor and Marr [12], it is recognized that the Morleys work
on vector analysis, from which the above theorem follows, holds for both interior and exterior
trisectors. The papers treatment of the theorem with only the interior trisectors is explained
as Morleys work never published and it was only the particular case of internal trisectors that
reached the authors. The very respectable given effort has produced proofs of many kinds,
exploiting a variety of features. Trigonometric, analytic and algebraic proofs supplement the
proofs of a purely geometric kind. Site Cut the Knot[13] presents 27 different proofs of Morleys
theorem from many more available. Notably, Roger Penrose [9] used a tiling technique, Edsger
Dijkstraapplied the rule of sines three times and then the monotonicity of the functiony = sin(x)
in the first quadrant [3], Alain Connes offered a proof in Algebraic Geometry [1], John Conway
showed it in plane geometry like a jigsaw puzzle solution [2], whileRichard Guyproved that it is aconsequence of his Lighthouse theorem [5]. However, a geometric, concise and logically transparent
proof is still desirable.
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Richard Guy notes: There are a few hints that there is more than one Morley triangle, but
Hosberger [p. 98] asks the reader to show that Morleys theorem holds also in the case of the
trisection of the exterior angles of a triangle [5]. Rose [10] and Spickerman [11] have proposed
proofs, using different methods, for the central Morley equilateral. In Parts 3 and 4 proofs for the
exterior Morley equilaterals will be offered.
The most popular technique for proving Morleys theorem is encountered asindirect, backwards
or reverse construction methodand fits in the following scheme.
Given a triangle assume that its angles are trisected and equal to 3, 3and3, respectively,
where+ += 60o. In order to show that one of its Morley triangles is equilateral, start with
an equilateralA B C and construct aABCwith angles 3, 3 and 3, so thatA B C is
the appropriate (interior, central or exterior) Morley triangle ofABC. ThusABC would be
similar to the given triangle and so would be their corresponding Morley triangles.
Proofs of the above method most often construct ABCby erecting B AC , C BA and
A CB with proper choice of the angles formed on the sides ofA B C . However repeated
requests have been recorded in geometry discussion forums for an explanation of the particular,seemingly arbitrary, choice of angles made at the beginning of these proofs. Of course the rea-
soning of the choice is not necessary for their validity. But the readers unfulfilling understanding
may have encouraged the mathematical folklore the use of words mystery, magic or mira-
cle for referring to Morleys theorem. This is not justifiable as there is nothing mathematically
extraordinary related to the theorem.
The presented proofs for showing that the interior, the central and the exterior Morley tri-
angles are equilaterals use the classicalAnalysis and Synthesis method. They exploit the inherent
symmetries of the problem and characterized by their uniform structure, logical transparency,
remarkable shortness and the distinct aesthetics of the Euclidean geometry. The Synthesis partfollows the previous method scheme. But it is empowered by two simple observations, supplying
necessary and sufficient conditions for a point to be the incenter or one of the excenters of a given
triangle. Even though they are almost trivial have a subtlety that enables to confront the messy
complexity of the triangle trisectors by enforcing clean simplicity and create proofs by harnessing
the power of the triangle angle bisector theorem. In addition these proofs reveal fundamental
properties of the Morley equilaterals stated as Corollaries. Besides their extensive use for showing
Morley triangles as not equilaterals, their fertility is demonstrated by proving the following: (1)
The two sides extensions of the inner Morley equilateral meet the corresponding inner trisectors at
two points which with the two sides common vertex form an equilateral. (2) The sides of Morley
equilaterals are collinear or parallel. (3) In any triangle the exterior trisectors of its angles, proxi-
mal to the three sides respectively, meet at the vertices of an equilateral, if and only if, the interiortrisectors of an angle and the exterior trisectors of the other two angles, proximal the three sides
respectively, meet at the vertices of an equilateral.
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In short, this work advocates that for Morleys observation a natural theoretical setting is
Euclidean geometry.
2 Notation and Counting of all Morley triangles
In a Morley triangle ofABC each vertex is the intersection of two trisectors, each of which is
either proximal or distal to a side ofABC. Hence a vertex is called proximal, distal or mix with
respect to the triangle side it belongs in the case the trisectors are both proximal, both distal, or
one proximal and one distal to the side, respectively.
Fig.3
So we may denote a proximal, dis-
tal or mix vertex with respect to a side
by using as superscripts p, d or * to
the letter of the corresponding angle
ofABCopposite to the side.Thus Ap, Ad and A denote the
proximal, distal and mix vertex of a
Morley triangle with respect toBC re-
spectively. In Fig.3 the notations for
all intersections of the interior trisec-
tors of ABC are showed. Notice
that a Morley triangle may have either
proximal vertices, or distal vertices, or
exactly two mix vertices.
Specifically, ApBpCp denotes
the inner Morley triangle of proximal
vertices, which is the inner Morley tri-
angle determined by the intersections
of proximal to each side trisectors. Also there is just one Morley triangle with distal vertices which
is denoted by AdBdCd. In addition there are three Morley triangles with one vertex proximal
and two vertices mix. They are denoted by ApBC, BpCA and CpAB
Moreover there are three more Morley triangles with one vertex distal and two vertices mix.
They are denoted by AdBC, BdCA and CdAB. Notice that a proximal or a distal
vertex is uniquely determined but a mix vertex is not as there are two such denoted by the same
letter. However in a Morley triangle with a pair of mix vertices, given its proximal or distal vertex,
the mix vertices are uniquely specified due to the choice restrictions in pairing trisectors for thesecond and then for the third vertex. Hence there are 8 interior Morley triangles formed by the
trisectors of the interior angles ofABC.
Similarly the trisectors of the exterior angles ofABC form Morley triangles. These are
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denoted by ApABpBC
pC , for the Morley triangle of proximal vertices, A
dAB
dBC
dC, for the Morley
triangle of distal vertices, ApAB
BC
C, BpBC
CA
A and CpCA
AB
B, for the Morley triangles with
one proximal and two mix vertices, AdAB
BC
C, BdBC
CA
A andCdCA
AB
B for the Morley trian-
gles of one distal and two mix vertices. In this notation we use subscripts in order to distinguish
a vertex determined by the interior trisectors from the vertex of the same type determined by theexterior trisectors. Hence, in general, there are 8 Morley triangles formed by the trisectors of the
exterior angles ofABC. Their vertices are in the exterior ofABC and due to their rather
central location with respect to ABCare calledcentral Morley triangles.
There are two more possibilities for the formation of a Morley triangle. One is by combining
the trisectors of an interior angle with the trisectors of the other two exterior angles of ABC.
Another is by combining the trisectors of two interior angles with the trisectors of the third exterior
angle ofABC.
The Morley triangles formed by combining the trisectors of the interior Awith the trisectors
of the exterior Band Care denoted byApAbpAc
pA, for the Morley triangle of proximal vertices,
AdAb
dAc
dA, for the Morley triangle of distal vertices, C
p
Ca
Cb
C, ap
Cb
CC
C and bp
CC
Ca
C,for the Morley triangles with one proximal and two mix vertices, and AdAb
Ac
A, bdAc
AA
A,
cdAA
Ab
A for the Morley triangles with one distal and two mix vertices. The use of a small letter
is for denoting the intersection of an interior and an exterior trisector ofABC. The vertices of
these 8 triangles formed by the trisectors of the interior A with the trisectors of the exterior B
and Care in the exterior ofABCand thus they are called exterior Morley triangles relative to
A.
Similarly are denoted the Morley triangles relative to B, which are formed by combining the
trisectors of the interior angle B with the trisectors of the exterior C and A, and also the
ones relative to C formed by combining the trisectors of the interior C with the trisectors of
the exterior A and B. Hence, in general, there are 24 exterior Morley triangles determined by
the intersections of trisectors of an interior and two exterior angles ofABC.
The Morley triangles formed by combining the trisectors of the interior B and C with
the trisectors of the exterior A are denoted by ApbpAcpA, for the Morley triangle of proximal
vertices,AdbdAcdA, for the Morley triangle of distal vertices, A
pbAc
A, bpAc
AA, cpAA
bA,
for the Morley triangles of one proximal vertex and two mix, AdbAc
A, bdAc
AA, cdAA
bA,
for the Morley triangles of one distal vertex and two mix. It should be remarked that in this
notation the same symbol for the intersection of an interior with an exterior trisector may refer to
two different points, an ambiguity which is clarified in a Morley triangle since one of its vertices
specifies its type and so the vertex that the symbolism refers. Hence, there are 8 Morley triangles
relative to the exterior A, which obviously have one vertex inside and two outside ABC. In
general, there are 24 Morley triangles ofABC determined by the intersections of trisectors ofone exterior and two interior angles ofABC.
Conclude that in total there are, in general, 64 Morley triangles ofABC.
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3 Uniform Proofs for all Morley Equilaterals
Fig.4
In this part we will prove that five Morley triangles are
equilaterals. The proofs are uniform and utilize two basic
observations for determining the incenter or an excenterofABCusing only one of its bisectors.
Observe that the incenter I is lying on a unique arc
passing through two vertices and I. In Fig.4 the unique
arc passing throughA, B and I is depicted. Obviously
AIB=1800 12ABC= 12BAC= 90o + 12ACB.
ThusI may be characterized as the intersection in the
interior ofABCof a bisector with the arc of size 90o +12ACB passing through A and B. Clearly an analogous
result holds for the other two pairs of vertices ofABC.
We refer to this as the Incenter Lemma.
IfIC is the excenter relative to Cthen
AICB= 90o
12ACB, BICC=
12BACand CICA=
12CBA.
Thus IC is determined by the intersection in the exterior ofABCof a bisector, either of the
interiorCor the exterior Aor Bwith the arc of size90o 12ACBpassing through A and B,
or with the arc passing through B and C of size 12BAC, or with the arc of size 12CBApassing
through C and A. Evidently analogous results hold for the other two excenters IA and IB. We
refer to this as the Excenter Lemma.
Theorem 1. In any triangle the interior trisectors of its angles, proximal to the sides, meet atthe vertices of an equilateral.
Proof.
Analysis: Let ABC be a triangle with A =3, B= 3 and C= 3, where ++= 60o.Suppose that ApBpCp is equilateral, whereAp,Bp and Cp are the intersections of the trisectorsproximal to the sides BC, CA and AB respec-tively. The aim of this step is to calculate theangles formed by the sides ofApBpCp and the
trisectors ofABC. See Fig.5.Let Cd be the intersection ofABp and BAp. SinceACp andBCp are angle bisectors inACdB, Cp
is the incenter.
Fig.5
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LetP and Q be the orthogonal projections ofCp on ACd andBCd respectively. ThusCpP=
CpQ and CdP = CdQ. But so CpPBp = CpQBp as right triangles having two pairs of
sides equal. HenceBpP = ApQ. ThenCdAp = CdBp and so ApCdBpis isosceles. Now from
ACdB we have ACdB = 180o (2+2) = 60o +2.Therefore CdBpAp = CdApBp =
12 [180o
(60o + 2)] = 60o. Consequently
CpBpA= CpApB= 180o 60o (60o ) =60o += +.
LetAd be the intersection ofBCp andCBp. Also letBd be the intersection ofCAp andACp.
Then from BAdC and CBdA find similarly
ApCpB= ApBpC= + and BpApC= BpCpA= +.
Fig.6
Synthesis: Suppose that a triangle is
given and assume that its angles are trisected
and equal to 3, 3 and 3, respectively,
where ++= 60o. Then around an equi-
lateral ApBpCp will constructABCwithangles 3, 3 and 3 so that Ap, Bp and Cp
will be the intersections of the proximal to the
sides interior trisectors.
On the side BpCp erect BpACp with
adjacent angles + = +60o and + = +
60o.
Similarly, erect CpBAp andApCBp on
the sides CpAp and ApBp respectively with
corresponding angles as shown in Fig.6, which
were found in the Analysis step.Let Cd be the intersection ABp and BAp.
Notice that ApCdBp is isosceles as two of
its angles are180o 60o + =60o .
Thus
CdAp =CdBp (1) and ApCdBp =180o 2(60o ) =60o + 2 (2)
Since ApBpCp has been taken equilateral, CpAp =CpBp. Combine this with (1) and infer
that Cp is on the ApBp bisector and so on the ACdB bisector. Moreover from (2) ACpB =
3600(++60o++)=180o(+) = 90o+ 12
(60o+2)=90o+ 12ApCdBp =90o+ 1
2ACdB.
Hence, by the Incenter Lemma, Cp is the incenter ofACdB.
Similarly it is shown that Ap andBp are the incenters ofBAdCand CBdA, respectively,whereAd is the intersection ofBCp andCBp, whileBd is the intersection ofCAp andACp. Thus
CpAB= CpABp = CABp and so ABp, ACp are trisectors ofA.
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Also the choice of angles in the construction ofBpACp implies CpABp = . Hence
A = 3. Likewise infer thatBCp, BAp are trisectors ofB with B = 3 and CAp, CBp are
trisectors ofC with C= 3.
Corollary 1. a) The angles between the trisectors ofABC and the sides of its inner
Morley equilateralApBpCp are: ApBpC = ApCpB = +, BpCpA = BpApC = +,
CpApB= CpBpA= +.
b) The heights of the equilateralApBpCp are: ApAd, BpBd andCpCd .
Theorem 2. In any triangle the exterior trisectors of its angles, proximal to the sides, meet at
the vertices of an equilateral.
Proof.
Analysis: Let ABCbe a triangle with A= 3, B= 3and C= 3, where ++=60o. Let ApA, B
pB and C
pC be the intersections of the exterior trisectors proximal to the sides BC,
CA and AB respectively. Suppose ApABpBC
pC is equilateral. The aim of this step is to calculate
the angles formed by the sides ofApABpBC
pC and the exterior trisectors ofABC.
Fig.7a (< 30o) Fig.7b (> 30o) Fig.7c (= 30o)
Let P andQ be the orthogonal projections ofCpC on ABpB and BA
pA respectively.
Notice that ABpB and BApA may intersect each other or be parallel since
PAB + QBA= 2( +) +2( + ) =120o + 2.
IfABpB and BApA intersect each other let C
dC be their intersection. Next consider all possible
cases.
If< 30o thenCdC and C
pC are at the same side ofAB. In AC
dCB, AC
pC and BC
pC are interior angle bisectors
and soCpC is the incenter, while it is calculated ACdCB= 60
o 2. Fig.7a.
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If> 30o then
CdC and CpC are on different sides ofAB. In AC
dCB, AC
pC and BC
pC are exterior angle bisectors
and so CpC is the excenter relative to CdC, while it is calculated AC
dCB= 2 60
o. Fig.7b.
Hence in both the above cases ( =30o)it holds CpCP = CpCQ. Thus C
pCPB
pB = C
pCQA
pA,
as right triangles having two pairs of sides equal. Consequently CpCBpBP = C
pCA
pAQ and so
ApACdCB
pB is isosceles. Thus:
If> 30o then
CpCBpBP= C
pCA
pAQ=
12 [180
oA
pAC
dCB
pB] 60
o = 12 [180o (60o 2)] 60o =.
If< 30o then
CpCBpBP = C
pCA
pAQ= 180
o60o 1
2[180oA
pAC
dCB
pB] = 180
o60o 1
2[180o (260o)] = .
Deduce that for =30o it holds CpCBpBA= C
pCA
pAB= .
If =30o then + = 30o andABpB//BApA. Fig.7c.
Notice ACp
C
B= 180o (30o + ) (30o + ) and so ACp
C
B= 90o.
LetM be the midpoint ofAB. Since ACpCBis right triangle, CpCM= MA = MB. ThenC
pCM=
MA gives CpCAM = MCpCA. But AC
pC is the PAB bisector and thus C
pCAM = C
pCAP.
Hence MCpCA= CpCAP and so B
pBA//C
pCM//A
pAB.
SinceMA= MB, CpCM bisects ApAB
pB and so C
pCM A
pAB
pB, as A
pAB
pBC
pC is equilateral.
Then ABpB, BApA A
pAB
pB. Therefore AB
pBA
pA = 90
o and given that CpCBpBA
pA = 60
o infer
CpCBpBA= 30
o. Similarly infer CpCApAB= 30
o.
Deduce that for = 30o it holds CpCBpBA= C
pCA
pAB= .
Conclude that for any value ofit holds CpCBpBA= C
pCA
pAB= .
Then from CpCAB
pB and C
pCBA
pA deduce B
pBC
pCA = and A
pAC
pCB = respectively.
Also from BApAC and CBpBA infer B
pBA
pAC= and A
pAB
pBC= .
Synthesis: Let a triangle be given in which its angles are equal to 3, 3and 3 respectively,
where + + = 60o. Then around an equilateral, which is denoted by ApABpBC
pC, will
construct a ABCwith angles 3, 3 and3 so that ApA, BpB and C
pC will be the meeting points
of the exterior angle trisectors proximal to the sides ofABC. On the side BpBCpC erect B
pBAC
pC
with adjacent angles and which were calculated in the Analysis step. Similarly, erect CpCBApA
and ApACBpB on the sides C
pCA
pA and A
pAB
pB with corresponding angles as they are depicted in
Fig.8. Hence ABChas been determined. So it remains to be proved that the resulting ABC
has angles3,3and 3respectively and the erected sides are the trisectors of its exterior angles.Let P andQ be the orthogonal projections ofCpC on the extensions ofAB
pB and BA
pA respec-
tively. Next consider all cases regarding ABpB andBApA.
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Fig.8a (< 30o) Fig.8b (> 30o) Fig.8c (= 30o)
Assume =30o. Set s = (30o )/|30o |and let CdC be the meeting point ofABpB and
ApAB. The choice of angles in the erection ofBpBAC
pC and C
pCBA
pA implies:
CpC is lying on the ApAC
dCB
pB bisector (1) and AC
pCB= 90
o + 12sACdCB (2)
To verify (1) notice that ApACdCB
pB is isosceles, as two of its angles are by construction
either 60o + ( < 30o) or 120o ( > 30o). But ApABpBC
pC is assumed equilateral and so
CpCApA = C
pCB
pB. ThusC
dCC
pC bisects side A
pAB
pB of the isosceles A
pAC
dCB
pB and so C
pC is lying on
the ApACdCB
pB bisector.
To verify (2) notice that in the isosceles ApACdCB
pB either AC
dCB = 180
o 2(60o +)
60o2 (< 30o)or ACdCB=180o2(60o+) =260o (> 30o). Thus ACdCB= s(60
o2).
Hence ACpCB= + 60o + = 60o + (60o ) = 90o + 1
2s(60o 2) =90o +sACdCB.
Therefore from (1) and (2), by the Incenter Lemma (< 30o, s= 1) or the Excenter Lemma
(> 30o, s= 1), CpC is the incenter or the excenter ofACdCB respectively.
Thus ACpC and BCpC are bisectors (interior or exterior) in AC
dCB. So, using C
pCAB
pB and
CpCBApA, deduce C
pCAB = C
pCAP = + and C
pCBA = C
pCBQ = + . Consequently
CpCAB= + and CpCBA= + while AC
pC and BC
pC bisect the angles formed by AB and
the extensions ofABpB andBApA respectively.
Assume = 30o . Then+= 30o. NoticeABpB , BApA A
pAB
pB and soA
pAB//B
pBA. Also
ACpCB= +60o+= 90o and so ACpCBis right triangle. LetMbe the midpoint ofAB. Hence
CpCM=MA=MA. But C
pCM = MA implies C
pCAM = MC
pCA. Since C
pCAM = C
pCAP,
CpCAM= ACpCM. Consequently C
pCM//AP. Thus A
pAB//C
pCM//B
pBAand since MA= MA,
CpCM passes through the midpoint of ApAB
pB. As a result C
pCM is a height of the equilateral
ApABpBC
pC and so A
pAC
pCM= B
pBC
pCM= 30
o. Therefore
Cp
C
AB= MCp
C
A= MCp
C
Bp
B
= Bp
B
Cp
C
A= 30o + = + .
Similarly it is shown CpCBA= 30o + = + .
Also ApAB//CpCM//B
pBA implies C
pCAP=AC
pCMand QBC
pC = BC
pCM. So
CpCAP= CpCABand C
pCBQ= C
pCBA.
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Conclude for any it holds CpCAB = + and CpCBA = + , while AC
pC and BC
pC
bisect the angles formed by AB and the extensions ofABpB and BApA respectively.
The rest cases are treated similarly. Considering BCpC and CBpB it shown that A
pABC= +
and ApACB= + while BApAand CA
pAbisect the angles formed by BC and the extensions ofBC
pC
andCBpB respectively, and eventually consideringCApA and AC
pCit is proved that B
pBCA= +
and BpBAC= + while CBpB and AB
pB bisect the angles formed by ACand the extensions of
ACpC and CApA respectively. Conclude that AB
pB bisects the angle between AC and the extension
ofACpC, whileACpC bisects the angle between AB and the extension ofAB
pB. ThusAB
pB andAC
pC
are trisectors of the exterior A.
Also CpCAB= BpBAC= + . Hence A=
12 [360
o 6( +)] =180o 3( +) =3.
Similarly it is shown that BCpC, BApA are trisectors of the exterior B with B = 3 and CA
pA,
CBpB are trisectors of the exterior Cwith C= 3.
Corollary 2. a) The angles between the exterior trisectors ofABCand the sides of its cen-
tral Morley equilateralApABpBC
pC are: A
pAB
pBC = A
pAC
pCB = , B
pBC
pCA = B
pBA
pAC = ,
CpCApAB= C
pCB
pBA= .
b) The heights of the equilateralApABpBC
pC are: A
pAA
dA, B
pBB
dB andC
pCC
dC.
Theorem 3. In any triangle the interior trisectors of an angle and the exterior trisectors of
the other two angles, proximal the three sides respectively, meet at the vertices of an equilateral.
Proof.
Analysis: LetABCbe a triangle with A= 3, B= 3and C= 3, where ++= 60o.
Let CpC be the intersection of the exterior trisectors of B and C, proximal to AB, while apCand bpC are the intersections of the interior with the exterior trisectors proximal to BC and CA
respectively. Suppose that apCCpCb
pC is equilateral. The aim of this step is to calculate the angles
between the sides ofapCCpCb
pC and the interior trisectors ofC and also the exterior trisectors
ofAand B.
Let P and Q be the orthogonal projections of CpC on AbpC and Ba
pC, respectively. It was
observed in the course of the Analysis Step of Theorem 2 that the trisectors AbpC and BapC inter-
sect each other iff = 30o. Recall that if = 30o CpC is the incenter ( < 30o) or the excenter
(> 30o)ofBCdCCwhile for = 30o AbpC//Ba
pC. But it was shown that in either case it holds
CpCP = CpCQ and hence C
pCPb
pC =C
pCQa
pC, as right triangles having two pairs of sides equal.
This implies
Ab
p
CC
p
C =
Ba
p
CC
p
C (1). Consequently:
If = 30o then CdC is determined. Hence (1) implies bpCa
pCC
dC = a
pCb
pCC
dC. Thus
apCC
dCb
pC is isosceles. However from AC
dCB it is calculated that
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Fig.9a (< 30o) Fig.9b (> 30o) Fig.9c (= 30o)
ACdCB= 60o 2 (< 30o)or ACdCB= 2 60
o (> 30o).
Then we have respectively.
For < 30o, CdC is on the other side ofapCbpC from A, B andAbpCa
pC = Ba
pCb
pC = 180
o
12 [180
o (60o 2)] = 120o .
For > 30o, CdC is on the same side ofapCb
pC with A,B and
AbpCapC = Ba
pCb
pC =
12
[180o (2 60o)] = 120o .
In either case CpCapCB= C
pCb
pCA= 60
o= + .
If = 30o then AbpC//BapC. Also C
pCP, C
pCQ are collinear and + = 30
o. Thus
apCCpCb
pC = 180
o (30o + ) (30o + ) = 60o and so by (1) apCC
pCQ = b
pCC
pCP =
12
(180o apCC
pCb
pC) =
12
(180o 60o) =60o. Hence CpCapCB= C
pCb
pCA= 30
o =+ .
In conclusion for any it holds AbpCCpC = Ba
pCC
pC = 60
o= + .
Finally from BapCC and CbpCA we find Ba
pCC = and Ab
pCC = respectively,
and so CpCapCC = and C
pCb
pCC = . Yet from b
pCAC
pC and C
pCBa
pC calculate that
bpCCpCA= ( + )
+ and apCCpCB= ( + )
+.
Synthesis: Suppose that a triangle is given with angles equal to 3, 3 and 3, respectively,
where + + = 60o. Then from an equilateral, which we denote apCCpCb
pC, will construct
a ABC with angles 3, 3 and 3 so that the sides of the erected triangles are the proper
angle trisectors of the resulting ABC. On the side apCbpC erect a
pCCb
pC with adjacent angles
+ = 60o + and + = 60o + so that CpC is inside apCCb
pC. Next on the sideb
pCC
pC erect
bpCACpC with adjacent angles + and (+ )+. Finally on the side CpCapC erect apCBCpCwith adjacent angles + and ( + )+. ThusABChas been determined. See Fig.10 for the
corresponding value of. So it remains to be proved that the resulting ABChas angles 3, 3
and 3, respectively and the erected sides CapC, CbpC are trisectors ofC, while Ab
pC, AC
pC, and
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BCpC, BapC are trisectors of the exterior angles Aand B respectively.
Fig.10a (< 30o) Fig.10b (> 30o) Fig.10c (= 30o)
Notice that if either < 30o or > 30o then AbpC and BapC intersect each other, while for
= 30o Abp
C//Bap
C.
First we deal with the erected sides CapC andCbpC and prove that they are trisectors ofC.
We also show that C= 3. See Fig.11.
Fig.11
Let bdC be the intersection of ACpC and Ca
pC.
Notice the choice of angles in the construction of
apCCbpC and a
pCBC
pC yields Ca
pCC
pC = and
apCCpCb
dC = 180
o ( + )+ 60o = . Hence
apCb
dCC
pC is isosceles. Thus Ab
dCC = 2 and
AbpCC= 12AbdCC.
Since a
p
Cbd
CC
p
C is isosceles and from the as-sumption apCC
pCb
pC is equilateral, infer that b
dCb
pC
bisectsapCCpC and sob
dCb
pCis the a
pCb
dCC
pCbisector.
HencebdC is lying on the exterior bisector ofAbdCC.
Thus, by the Excenter Lemma, bpC is the excenter of
AbdCCrelative to C. But so CbpC is the ACa
pC bisector.
Similarly show that CapC is the BCbpC bisector.
ThereforeCbpC and CapC are trisectors ofC. Also
apCCbpC = 180
o CbpCa
pC Ca
pCb
pC = 180
o + + =.
Conclude C= 3.
Next we deal with the erected sides ACpC, AbpC and BC
pC, Ba
pC and prove that are trisectors
ofAand B. We also show that A= 3and B= 3.
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Assume =30o. Set s = (30o )/|30o |and let CdC be the meeting point ofAbpC and
BapC. The choice of angles in the erection ofCpCAb
pCandC
pCBa
pC implies:
CpC is lying on the ACdCB bisector (1) and AC
pCB= 90
o + 12
sACdCB (2)
To verify (1) notice that ap
A
CdCbp
C
is isosceles, as two of its angles are either 120o (+)
( < 30o) or 60o + (+ ) ( > 30o). Using the fact thatapCCpCb
pC is equilateral, infer that
CdCCpC bisects a
pCb
pC and so C
pC is lying on the a
pCC
dCb
pC bisector.
To verify (2) notice that in the isosceles apCCdCb
pC
if< 30o then ACdCB= 180o 2[120o (+ )] = 60o 2= s(60o 2) ,
if> 30o then ACdCB= 180o 2[60o + (+ )] = 2 60o s(60o 2).
So for the cases < 30o and > 30o have respectively:
ACdCB= s(60o 2)and
ACpCB= 360o (+)+ (+)+ =120o = 90o + 12 (60
o 2) =90o + 12sAC
dCB.
Therefore from (1) and (2), by the Incenter Lemma ( < 30o, s = 1) or the Excenter
Lemma ( > 30o, s = 1), CpC is the incenter or the excenter ofACdCB respectively. Thus
CpCAbpC = CpCAB and CpCBapC = CpCBA.Moreover from CpCAb
pC and C
pCBa
pC infer C
pCAb
pC = + and C
pCBa
pC = + .
Deduce for =30o it holds CpCAbpC = C
pCAB= + and C
pCBa
pC = C
pCBA= + .
Assume = 30o and so += 30o. Then the choice of angles in construction ofCpCBapC
and CpCAbpC implies C
pCb
pCA= C
pCa
pCB= + = 30
o. HenceAbpC, BapC a
pCb
pC and thus
AbpC//BapC. Draw from C
pC the height of the equilateral a
pCb
pCC
pC meeting AB at M. Hence
CpCM//AbpC//Ba
pC, and also MC
pC bisects a
pCb
pC. But soM is the midpoint ofAB. Also notice
that ACpCB is right triangle as ACpCB = 360
o 60o ( +30o)+ (+30o)+ = 90o. Then
CpCM = MA = MB. Now CpCM = MA implies C
pCAM = MC
pCA. Yet C
pCM//Ab
pC implies
CpCAbpC = AC
pCM. Thus C
pCAb
pC = C
pCAMand so AC
pC is the b
pCAB bisector.
Similarly it is shown that BCpC is the apCBAbisector.
Also the choice of angles in the construction ofCpCAbpC gives
CpCAbpC = 180
o 30o (+ 30o)+ =30o + .
Deduce for = 30o it holds CpCAbpC = C
pCAB= 30
o + = + .
Similarly it is shown that CpCBapC = C
pCBA= + .
Conclude for all it holds
CpCAbpC = C
pCAB= 30
o + = + and CpCBapC = C
pCBA= + .
From Abp
CCit follows
b
p
CAC= 180o
, since from the construction choice of anglesAbpCC = and ACb
pC = , as found in the first step. But clearly b
pCAC = b
pCAC
pC +
CpCAB+ CAB= 2( + ) + A. Then A= 3and similarly B= 3. Therefore the angles
ofABCare 3, 3 and 3. Since CpCAbpC = C
pCAB = + it follows that Ab
pC and AC
pC
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are trisectors ofAin ABC.
Similarly it shown that BapC and BCpC are trisectors ofB in ABC.
Corollary 3. a) In anyABC, the angles formed by the sides of the exterior Morley
equilateralapCCpCb
pC relative to theCand the exterior trisectors ofAandBare: a
pCC
pCB=
( + )+, bpACpCA= ( + )
+, CpCapCB= C
pCb
pAA= + , while with the interior trisectors
ofC are: CpCapCC= and C
pCb
pCC= .
b) The heights of the equilateralapCCpCb
pC are: a
pCa
dC, b
pCb
dC andC
pCC
dC.
4 Implications
4.1 Companion Equilaterals of the inner Morley equilateral
Theorem 4. The two sides extensions of the inner Morley equilateral meet the correspondinginner trisectors at two points which with the two sides common vertex form an equilateral.
Fig.12
Proof. As usually ApBpCp denotes the interior Mor-
ley equilateral ofABC. Let SA be the intersection of the
extension of sideApCp with the trisector CBp and letKA be
the intersection of the extension of side ApBp with the trisec-
torBCp. Moreover letAd be the intersection of the trisectors
BCp and CBp. By Corollary 1a, ApBpC= CpApB= +
and soBpAdCp is isosceles. ThusApAd is bisector of both
BpApCp and BpAdCp.
Hence BpApAd = CpApAd and BpAdAp = CpAdAp.
Also SAAdAp = KAA
dAp as obviously SAAdCp =
KAAdBp and SAA
dAp = SAAdCp + CpAdAp while
KAAdAp = KAA
dAp + BpAdAp.
Therefore ApSAAd and ApKAA
d are equal because in
addition they have side ApAd in common. Consequently
SAApKA is equilateral.
The previous equilateral is named companion equilateral relative to vertex Ap and it will be
denoted by SAApKA. Obviously there are two more companion equilaterals relative to vertices
Bp
andCp
, denoted by SBBp
KB and SCCp
KC, respectively.
Corollary 4. For the companion equilateral relative to vertex Ap, SAApKA, it holds
BASA = CAKA = |+ |.
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In fact, the pointsSA andKA , for< 30o are outsideABC, for= 30o are onAB andAC,
respectively, and for> 30o are insideABC.
Fig.13
Proof. Corollary 1a asserts CpApB = + and
BpApC = +. Thus from KAApB and KAA
pB it
is calculated CpSABp =and CpKAB
p =, respec-
tively. Then the points Bp, Cp, SA and KA are cyclic as
BpCp is seen fromSA andKA with the same angle. Thus
ASAKA = ABpKA and AKASA = AC
pSA. From
BCpA and CBpA infer that ACpKA = + and
ABpSA = +, respectively. Hence
ASAKA = + and AKASA = +.
Next notice that SAACp = 180o ASAC
p
SACpA= 180o (60o + +) (+) = +
and similarly KAAC= +.However
SAACp = SAAB BAC
p and
KAAC= KAAC CABp,
where may be either + or depending on the location ofSA and KA with respect to AB and
ACrespectively. Therefore SAAB= KAAC= | + |.
Because + += 60o for < 30o the points SA andKA are outside ABC, for = 30o
the points SA andKA are onAB and AC, respectively and for > 30o the points SA andKA are
inside ABC.
4.2 Relation of the Morley equilaterals
Theorem 5. In any triangle the sides of Morley equilaterals are either collinear or parallel.
Fig.14
Proof. Corollary 2a claims BCpCApA = . Also
Corollary 3a confirms BCpCapC = ( +)
+. Hence
BpBCpCa
pC = B
pBC
pCA
pA+ A
pAC
pCB + BC
pCa
pC =
=60o + + ( +)+ =180o.
Thus apCCpC is extension ofB
pBC
pC.
Similarly it is shown that bpCCpC is extension ofA
pAC
pC.
As bpCapCC
pC = A
pAB
pBC
pC = 60
o, it follows
apCbpC//A
pAB
pB.
Since bpCapCC= b
pCa
pCC
pC+ C
pCa
pCC= 60
o +
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and BpApC= + then
apCbpC//A
pBp.
The previous result is mentioned as a fact in [8] and it might be in print elsewhere. At anyevent, it inspires the next theorem.
4.3 Interrelationship between central and exterior Morley equilaterals
Theorem 6. In any triangle, the exterior trisectors of its angles, proximal to the three sides re-
spectively, meet at the vertices of an equilateral, if and only if, the interior trisectors of an angle
and the exterior trisectors of the other two angles, proximal the three sides respectively, meet at
the vertices of an equilateral.
Fig.15a
Proof. LetABCbe given with A= 3, B= 3
and C= 3, where + += 600.
(=) Assume that ApABpBC
pC is the central Mor-
ley equilateral formed by the intersections of the exterior
trisectors, proximal to the sides ofABC. See Fig.15a.
Extend ApACpC and B
pBC
pC to meet the extensions of
ApABand B
pBAat A
andB respectively. ThenAB and
BA are trisectors of the exterior Aand Brespectively,
as extensions of the corresponding trisectors. Thus it suf-
fices to show that A CpCB is equilateral and also that
CA
andCB
are trisectors of the interior
C.First show that A CpCB
is equilateral.
Notice that A ApABpB = B
ApABpB because they
haveApABpB common, A
BpBApA = B
ApABpB =60
o and A ApABpB = B
BpBApA = 60
o + by
Corollary 2a. Thus A BpB =BApA. But so C
pCA
=CpCB
, as CpCApA = C
pCB
pB since A
pAB
pBC
pC
is equilateral. Also A CpCB = ApAC
pCB
pB = 60
o and hence A BpBCpC is equilateral.
Next show that CA and CB are trisectors of the interiorC.
From ApAABpB and A
pAB
BpB it is easily calculated that ApAA
BpB = + and ABBpB =
+ . Hence ApA, BpB, A
, B are cyclic. But BpBApAC = and A
pAB
pBC = , by Corollary
2a. So from ApACBpB, A
pACB
pB = 180
o(+ ). But so C is also on this circle. Thus
CA
Ap
A = Ap
ABp
BC = and CB
Bp
B = Bp
BAp
AC = . Finally note that inBA
C andCB Ait holds respectively A CB= (+) = and B CA= ( +) = .
ConcludeCA and CB are trisectors of the interiorC, as C= 3.
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Fig.15b
(=) Assume that the exterior triangle apCCpCb
pCis
equilateral formed by the intersections of the trisectors of
the interior Cand the exterior Aand B. See Fig.15b.
Extend ap
C
Cp
C
and bp
C
Cp
C
to meet the extensions of
apCBandbpC A atA
andB respectively. ThenAB and
BA are trisectors of the exterior Aand Brespectively,
as extensions of the corresponding trisectors.
Thus it suffices to show that A CpCB is equilateral
and that CA and CB are trisectors of the interior C.
Note A apCbpC = B
apCbpC, because they have
apCbpC common, A
BpBApA = B
ApABpB = 60
o and
A ApABpB = B
BpBApA = 60
0 + + by Corollary 3a. Thus A bpC =BapC. Since a
pCC
pCb
pC
is equilateral then CpCA =CpCB
and so A CpCB is equilateral.
Also from apC
BCand bpC
CAit is calculated that apC
CB= bp
CCA= and so Cap
Cand
CbpC are trisectors ofC.
5 The non Equilateral Morley Triangles
Fig.16
For a given ABCthere are in general 64 Morley triangles, as the
trisectors of its three angles meet at many points. Among them are
the inner, the central and the exterior Morley equilaterals.
A number of authors (see for example [4] or [5]) have wondered:
Are there more Morley equilaterals forABC?
This part examines all the remaining Morley triangles ofABCsystematically and shows that none of them is equilateral.
In the sequel the following easily proved lemma is
used.
The Equilateral Center Lemma. The incenter of an equi-
lateral is the unique interior point from which its sides are seen with
120o. Similarly the excenter relative to an angle is the unique exterior point from which the side
opposite to the angle is seen with120o while the other two sides are seen with60o.
5.1 Morley triangles by trisectors of interior angles
This section treats the non equilateral Morley triangles formed by the trisectors of the interior
angles ofABC. The proximal to the sides trisectors meet at Ap, Bp and Cp and ApBpCp
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90 Spiridon A. Kuruklis CUBO16, 2 (2014)
denotes the inner Morley equilateral.
5.1.1 The Interior Morley triangle of distal vertices
Fig.17
The interior Morley triangle of distal vertices is
denoted byAdBdCd whereAd,Bd andCd are
the meeting points of the distal trisectors with re-
spect to the sides BC, CA and AB, respectively,
as shown in Fig.17. IfABCis equilateral then
AdBdCd is equilateral as well. Thus in the fol-
lowing we assume that ABCis not equilateral.
From Corollary 1b, we have that ApAd,
BpBd andCpCd are the heights of the inner Mor-
ley equilateral ApBpCp.
Let M be the center ofApBpCp. Thus
ApMBd=BdMCp= CpMAd=AdMBp
= BpMCd=CdMAp= 60o
So AdMBd = BdMCd = CdMAd = 120o.
Hence the sides ofAdBdCd are seen from M
with 120o.
Assume towards a contradiction that AdBdCd is equilateral. Then, by the Equilateral
Center Lemma, Ap
Ad
is a height ofAd
Bd
Cd
. Thus Ap
Ad
bisectsBd
Cd
. Hence Ap
Ad
bisectsBdApCd and so BApC. Since Ap is the incenter ofBAdC, ApAd bisects also BAdC. But
so the exterior angles ofAdApB and AdApCat vertexAp are 12BApC= 1
2BAdC + and
12BApC = 1
2BAdC+ . Hence = . Similarly it is shown that = . ThusABC is
equilateral contrary to the assumption.
Conclude that AdBdCd cannot be equilateral (ifABC is not equilateral).
5.1.2 Interior Morley triangles with one proximal and two mix vertices
There are three interior Morley triangles with one proximal and two mix vertices denoted by
ApBC, BpCA andCpAB. We will study onlyApBC as the other two are similar.
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SinceAp is the intersection of the proximal trisectors, B
must be the intersection of the remaining trisector CBp
(proximal to CA) with ACp as distal. Then C is the
intersection of the left trisectorsBCp
(distal toAB) andABp (proximal). So B is on ACp and C is on ABp.See Fig.18. Corollary 1a asserts ACpBp = + andABpCp = +. Hence ACpAp = 60o ++ < 180o
and ABpAp =60o ++ < 180o. Therefore the quad-rangle ABpApCp is convex and so BApC is insideBpApCp.
Fig.18
Conclude that BApC < 60o and thus ApBC cannot be equilateral.
5.1.3 Interior Morley triangles with one distal and two mix vertices
Fig.19
There are three interior Morley triangles with
one distal and two mix vertices which are de-
noted byAdBC,BdCA andCdAB.
We will study only AdBC as the other two
are similar. SinceAd is the intersection of the
distal trisectors, B must be the intersection of
the remaining trisectorsCAp (distal toCA) and
ABp, as proximal. So C is the intersection of
the left trisectors ACp and BAp (mix to AB).
First notice that if = then AdBC
is isosceles, because Corollary 1a with = yields BpApB = CpApC and so ApB =
ApC which implies AdApB =AdApC.
Thus if= = then AdBC is equi-
lateral.
Assume ABC is not equilateral. Then it
has two sides not equal and thus in the following we may assume < . Fig.19.
Suppose, towards a contradiction, that AdB = AdC. Let Z be the symmetric point ofC
with respect to ApAd. We will fist show that Z is inside AdApB.
From Corollary 1b,ApAd is height of the equilateralApBpCp and soBp andCP are symmetric
with respect to ApAd. Consequently BpApZ = CpApC and since CpApC = + inferBpApZ= +. Since BpApB = + and by assumption< , deduce BpApZ < BpApB.
Moreover ApBpZ= ApCpC =+ (+ ) =+while ApBpB =+ (+) = + .
So ApBpZ < ApBpB.
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Since Z is inside AdApB then AdZB > AdApB = AdApBp + BpApB = 30o +
+ =90o + . However, the assumptionAdB =AdC implies AdZ= AdB and so AdZB =
AdBZ. Thus AdZB + AdBZ > 2(90o +) > 180o. Hence two angles ofAdZB have
sum greater than 180o, which is a contradiction.
Conclude that AdBC cannot be equilateral (ifABC is not equilateral).
5.2 Morley triangles by trisectors of exterior angles
This section treats the non equilateral Morley triangles formed by the trisectors of the exterior
angles. So throughout this section trisectors mean trisectors of the exterior angles ofABC.
The proximal trisectors meet at the points ApA, BpB and C
pC and so A
pAB
pBC
pC denotes the
central Morley equilateral. Notice that the trisectorsBCpC and CBpB are parallel iff
BpBCB + CpCBC= 180
o 2(+ ) + 2(+) =180o.
So for = 30o BCpC//CBpB. In this case the distal trisectors with respect to BCdo not intersect
and hence the distal to BC vertexAdA is not determined. Also if30o >then AdA andApA are onthe same side ofBC while BAdAC= 60
o 2. If30o 30o)
The Morley triangle of distal vertices is denoted by AdABdBC
dC, where A
dA, B
dB and C
dC are
the meeting points of the distal trisectors of the exterior angles with respect to the sides BC, CA
andAB, respectively. In Fig.20a and Fig.20b the different locations ofAdAB
dBC
dC with respect to
ApAB
pBC
pC are illustrated. Note thatA
pAA
dA, B
pBB
dB and C
pCC
dC, by Corollary 1b, are the heights
of the central Morley equilateral ApABpBC
pC and let N be their intersection.
Notice that ifABCis equilateral then AdABdBC
dCis equilateral as well. Thus in the following
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we assume that ABCis not equilateral. Also suppose towards a contradiction that AdABdBC
dC
is equilateral.
IfABCis an acute triangle the equilateral ApABpBC
pC is inside A
dAB
dBC
dC. Hence
AdANBdB = BdNCdC = CdCNAdA = 120o.
IfABCis an obtuse triangle (assume > 30o) AdA and ApA are on different sides ofBC.
Hence
AdANBdB = A
dANC
dC = 60
o and BdBNCdC = 120
o.
Thus, by the Equilateral Center Lemma, N is the incenter (acute) or the excenter (obtuse) of
the assumed equilateralAdABdBC
dC. HenceA
pAA
dA is a height ofA
dAB
dBC
dCand soA
pAA
dAbisects
BdBCdC and B
dBA
pAC
dC.
In the case of the acute triangle, notice in BAdACthatApAis the incenter while the bisector
Ap
AAdA bisects BA
p
AC. As a result AdABC= A
dACB 2(+) =2(+ ) = .
In the case of the obtuse triangle, note that since ApAAdA bisects B
dBC
dC it bisects BA
pAC.
Also it bisects BAdAC as a height ofApAB
pBC
pC. ThenA
pABA
dA = A
pACA
dA and so A
pAB =
ApAC.
Therefore += + and so = . Similarly we show that = .
Deduce that ABCis equilateral contrary to the assumption that it is not.
Conclude that AdABdBC
dC cannot be equilateral (ifABCis not equilateral).
5.2.2 The Central Morley triangles with one proximal and two mix vertices
Fig.21
There are three Morley triangles of ABC formed
by exterior trisectors with one proximal and two
mix vertices denoted by ApAB
BC
C, BpBC
CA
A and
CpCA
AB
B. We will study only ApAB
BC
Cas the other
two are similar.
As vertexApA is the intersection of the proximal to
BC trisectors, vertex BB is the intersection of the re-
maining trisector CBPB (proximal to CA) with ACPC, as
distal. Then vertexCC is the intersection of the left tri-
sectors ABPB (distal to AB) and BCPC (proximal). Thus
BB is on CBPB while CC is on BCPC. Using the anglevalues between the sides ofApAB
pBC
pC and the trisec-
tors ofABCgiven by Corollary 2a it is easily deduced
that all the angles of the quadrangles BBBpBA
pAC
pC and
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CCCpCA
pAB
pB are less than180
o and so they are convex.
For instance
CCCpCA
pA = 180
o BCpCA
pA =180
o ,
whileABpBA
pA = 60
o + .
Since BBBpBA
pAC
pC is convex infer A
pAC
C is inside BpBA
pAC
pC. Also since C
CCpCA
pAB
pB is
convex infer ApAB
B is inside BpBA
pAC
pC.
Conclude that BCApAC
C < LBpBA
pAC
pC = 60
o and so ApAB
BC
C cannot be equilateral.
5.2.3 The Central Morley triangles with one distal and two mix vertices
There are three Morley triangles formed by exterior trisectors with one distal and two mix vertices
which are denoted by AdAB
BC
C, BdBC
CA
A andCdCA
AB
B. We will study onlyAdAB
BC
C as
the other two are similar.
SinceAdA is the intersection of the distal to BCtrisectorsBCpCand CB
pB,B
B is the intersection
of the remaining trisectorCApA (proximal to CA) withABPB as distal. Then C
C is the intersection
of the left trisectors ACpC (proximal to AB) and BApA (distal).
Fig.22
If= 30o then AdA is not determined.
If 30o > then AdA is on the same side of
BC with ApA and BAdAC = 60
o 2. If
30o < then AdA and ApA are on different
sides ofBC while BAdAC= 2 60o.
Consider the case = . Then
BpBApAB
B = CpCA
pAC
C by Corol-
lary 2a. So ApAB
B = ApAC
C and in
turn AdAApAB
B = AdAA
pAC
C. Thus
AdAB
BC
Cis isosceles. So ifABCis equi-
lateral then AdAB
BC
C is equilateral.
Next consider that ABC is not equi-
lateral and let > .
Since the location ofAdAdepends on the
value of, assume that 30o > . Suppose,
towards a contradiction that AdABBCC isequilateral. Let Wbe the symmetric point
ofCC with respect to ApAA
dA and let V be
the intersection ofAdAWand B
BBpB. In the
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sequel we find the location ofW.
From Corollary 2b, ApAAdA is height of the equilateral A
pAB
pBC
pC and so A
pA and B
pB are
symmetric with respect to ApAAdA.
Also by Corollary 2a the angles between the trisectors ofABC and the central Morleyequilateral are as depicted in Fig.22. Consequently,
AdACpCW= A
dAC
pCC
C = 180o 60o = 60o +and AdAB
pBB
B = 180o 60o =
60o + . Therefore BBBpBW < A
dAB
pBB
B and the line BpBWis inside CB
pBB
B.
Furthermore BpBApAW= C
pCA
pAC
C = and since > , ApAWis inside B
pBA
pAB
B. Thus
Wis inside CBpBB
B.
ClearlyCCis outside ABCpC. Thus we may set C
pCA
dAC
C = . Then by symmetry BpBA
dAW=
. Also set AdAWB
B = . Since AdAB
BC
C is assumed equilateral then C
CAdAB
B =60o and
so WAdAB
B = 2( ).
HoweverAdAB
B = AdAWand soA
dAC
C = AdAB
BimpliesAdAW= A
dAB
B. Thus WB
BAdA =
and fromAdA
WBB
, 2+ 2() =180o = 90o +. Given thatWis insideCBpB
BB
inferCBpBB
B, > AdAVB
B.
But from AdAVB
B we deduce AdAVB
B = AdAB
pBB
B+ BpBA
dAV=60
o + + .
Thus > 60o + += > 60o + + ( + 90o)= + < 30o which contradicts
the assumption 30o >.
The case > 30o is similar and it is omitted.
Conclude that AdAB
BC
C cannot be equilateral (ifABCis not equilateral).
5.3 Morley triangles by trisectors of one interior and two exterior angles
This section deals with the non equilateral Morley triangles formed by the trisectors of one interior
and two exterior angles. Even crude figures of these triangles indicate clearly that they are too
asymmetric to be equilaterals. Nevertheless it must be shown rigorously that they are not. We will
consider only those formed by the interior trisectors ofC and the exterior trisectors ofA and
B, as the other two cases are similar. apcCpCb
pc denotes the exterior Morley equilateral relative
to C.
Notice that the trisectors AbpC and BapC are parallel iff
apCBA + bpCAB= 180
o 2(+) + 2( +) = 180o = 30o.
In this case the distal trisectors with respect to AB do not intersect and hence the distal vertex
CdC is not determined. Also if30
o
> thenCdC and C
p
Care on the same side ofAB with ACdCB=
60o 2. If30o
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intersects the trisector BCpC betweenB and CpC. Moreover Corollary 3a implies that the extension
ofACpC is inside apCC
pCBand soCa
pC intersectsAC
pC insideC
pCBa
pC. In additionAC
pC intesects
BapC between B and apC. Similarly the trisectror Cb
pC intersects the trisector AC
pC between A and
CpC and the trisector BCpC inside C
pCAb
pC. AlsoBC
pC intersects Ab
pC between A and b
pC.
5.3.1 The Morley triangle of distal vertices
This is denoted by adCCdCb
dC. VertexC
dC is the intersection of the distal to AB trisectors Ab
pC
andBapC and is determined iff =30o. VertexadC is the intersection of the distal to BC trisectors
BCpC and CbpC and hence it is inside C
pCAb
pC. Vertexb
dC is the intersection of the distal to AC
trisectorsCapC andACpC and hence it is inside C
pCBa
pC. See Fig.23.
Thus, for C =90o CdCis determined whileadCand b
dCare inside AC
dCB. But so a
dCC
dCb
dC 30o)
5.3.2 The Morley triangles with one proximal and two mix vertices
There are three such triangles denoted byapCb
CC
C, bpCa
CC
C and CpCa
Cb
C.
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Fig.24
a. apCb
CC
C : Vertex apC is the intersection of
the proximal to BCtrisectors BapC andCapC. Vertex
bC must be the intersection of the remaining interior
trisectorCbpC (proximal to ) with the exterior trisec-
tor AbpC, as distal. Hence CC is the intersection ofthe left trisectors, BCpC (proximal to AB) and Ab
pC
(distal). See Fig.24.
So bC is on ACpC and it is between A and C
pC.
Also CC is onAbpC and it is between A and b
pC. No-
tice apCbpCC
pC = a
pCC
pCb
pC = 60
o while, by Corol-
lary 3a, CpCbpCA= + and b
pCC
pCA= (+)
+. Thus apCCpCA < 180
o and apCbpCA < 180
o.
Hence the quadrangle AbpCapCC
pC is convex. Therefore b
CapCC
C is inside bpCa
pCC
pC and so
bCapCC
C < 60o.
Conclude that apCb
CC
C is not equilateral.
b. bpCa
CC
C : It is shown as above that it is not equilateral.
c. CpCa
Cb
C : VertexCpC is the intersection of the proximal to AB exterior trisectors. Thus
aC is the intersection of the remaining exterior trisector BapC (proximal to BC) with the interior
trisectorCbpC, as distal. Then b
C is the intersection of the left trisectors CapC (distal to AC) and
AbpC (proximal).
So aC and CpC are on the same side ofBC iff
BCbpC+ CBapC < 180
o 2 +2(+) + 3< 180o < .
If = then aC is not determined as BapC//Cb
pC.
Also b
C and CpC are on different sides ofAC iffACapC+ CAb
pC < 180
o 2 + 2( +) + 3< 180o < .
If = then bC is not determined as AbpC//Ca
pC.
Since aC, b
C and CpC are outside ABC while a
C and b
C are on AbpC and Ba
pC respectively we
deduce
aC and CpC are on the same side ofAB iff<
while
bC and CpC are on the same side ofAB iff< .
The above conditions correlate the ranges of,CpC and with the different locations ofa
C
and bC and vise versa.
Recall that AbpC and BapC intersect at CdC iff =30o, with ACdCB = |60o
2|, while CpC andCdC are on the same side ofABiff< 30
o.
Next all the different locations ofaC and b
C are considered.
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98 Spiridon A. Kuruklis CUBO16, 2 (2014)
Case 1: aC andb
C are on the other side ofAB from CpC.
This happens iff 30o (and so < and < ) or < 30o with< and < . Fig.25a,b.
If= 30o
then Abp
C//Bap
C, while for =30o
Abp
C andBap
C meet at CdC.
But so, aC and b
C are on the extensions (to the other side ofAB fromCpC) ofCb
pC,Ba
pC and
CapC, bpCA respectively.
SinceCpC is inside apCC
pCb
pC, then a
CCpCb
C encompasses a
CCb
C . Therefore
aCCpCb
C < a
CCb
C. However a
CCb
C = . Deduce a
CCpCb
C< 60o.
Fig.25a (> 30o) Fig.25b (< 30o,< ,< )
Case 2 : aC and b
C are on the same side ofAB with CpC.
This happens iff< 30o, > and > . Fig.25c.
Then CdC and CpC are on the same side ofAB. Note that Ca
pC intersects sides AB and BC
dC of
ACdCBinternally and so, by Paschs axiom, it intersects the third side ACdC externally. Thus b
C
is on the extension ofACdC. Similarlya
C is on the extension ofBCdC. Since C
pC is inside AC
dCB
then aCCpCb
C encompasses a
CCdCb
C. Therefore a
CCpCb
C < a
CCdCb
C. But a
CCdCb
C =
ACdCB= 60o 2. Deduce aCC
pCb
C< 60o.
Case 3: aC andb
C are on different sides ofAB. Fig.25d.
This happens iff < 30o with > and < or with < and > .
Next consider the case < 30o with > and < .
ThenCpC is on the same side with CdC. HenceCpC is inside ACdCBand alsoCpCis insideapCCbpC.By Paschs axiom on ACdCB , since Ca
pC intersects sides AB and BC
dC at interior points, infer
CapC intersects the third sideACdC at an exterior point. Thusb
C is on extensions ofACdC and Ba
pC
on the same side ofAB withCpC. SoCpC is inside b
pCCb
C on the other side ofapCb
pCfromC
dC and
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Fig.25c (< 30o,> ,> ) Fig.25d (< 30o, < , < )
bC. ConsequentlyapCC
dC intersects b
CCpC between b
C and CpC. Hence a
CapC is inside a
CCpCb
C.
ConsequentlyapC and B are inside a
CCpCb
C. Therefore a
CCpCb
C encompasses apCC
pCBand so
aCCpCb
C> apCC
pCB.
However Corollary 3a asserts apCC
pCB= ( +)
+
. Deduce b
CCpCa
C > 60o
.
The case < 30o with > and < is similar and it is omitted.
Conclude that CpCa
Cb
C is not equilateral.
5.3.3 The Morley triangles with one distal and two mix vertices
These Morley triangles are denoted byCdCa
Cb
C, bdCC
Ca
C and adCc
CB
C.
Fig.26
a. CdCa
Cb
C: Vertex CdC is the in-
tersection of the distal to AB trisectors
AbpC and BapC. Thus vertexa
C is deter-
mined by the intersection of the remain-
ing trisectorCPC , distal toBC, withCapC,
as proximal. VertexbC, is determined by
the left trisectors ACPCand CbpCwhich are
distal and proximal to CA, respectively.
VertexCdC is determined iff =30o
with CPC and CdC to be on the same side
ofAB iff> 30o.
Moreover aC is always located be-
tween B and CPC while bC is always lo-cated between A and CPC. Fig.39 depicts the case for C
PC and C
dC to be on the same side ofAB.
Regardless the location ofCdC, verticesa
Cand b
C are inside ACdCB. Thus a
CCdCb
C< ACdCB.
Since ACdCB= |60o 2|< 60o then aCC
dCb
C< 60o.
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Conclude that CdCa
Cb
C is not equilateral.
b. bdCC
Ca
C: Vertex bdC is the intersection of the distal to CA trisectors Ca
pC and AC
pC.
Hence vertex CC is determined by the intersection of the remaining trisector bp
C
, distal to AB,
with BCpC, as proximal.
Vertex aC is determined by the intersection of the left trisectors CbpC and Ba
pC which are
distal and proximal toBC, respectively.
Trisectors CapC andACpC always intersect each other and sob
dC is located on the same side ofAB
withCpC. AlsoAbpC andBC
pC always intersect each other and so C
C is located on the same side of
AB with CpC. However a
C is not always determined as CbpC//Ba
pC iffb
pCCB+ b
pCCa
pC =180
o
= . In fact aC is on the same side with bdC and C
C iff= . It should also be noted that
30o
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Fig.27c (< 30o, < )
Thus adCbpCC
C is exterior angle in bpCa
CC
C. Hence bpCa
CC
C < adCb
pCC
C. But
adCbpCC
C = CbpCA. InCb
pCAit is calculated Cb
pCA= and so b
pCa
CC
C < .
Infer BaCC
C < () + < 60o.
Conclude that bdCa
CC
C is not equilateral.
c. adCB
Cc
C: It is shown as above that it is not equilateral.
5.4 Morley triangles by trisectors of one exterior and two interior angles
Eventually the non equilateral Morley triangles formed by trisectors of one exterior and two interior
angles are treated. Obviously these Morley triangles have one vertex in the interior and two in
the exterior ofABC. As previously we will consider only those formed by the trisectors of the
exterior Acombined with the interior trisectors ofBand Cas the other two cases are similar.
5.4.1 The Morley triangle of proximal vertices
This is denoted by ApbpAcpA . Vertex A
p isthe intersection of the proximal to BC interiortrisectors, vertex bpA is the intersection of CB
p
with the exterior trisector ofAproximal toAC,while vertex cpA is the intersection of BC
p withthe exterior trisector ofAproximal to AB.
Fig.28
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Consider the companion equilateral SAApKArelative to vertex A. In Fig.28 the case< 30
o
is depicted for which Corollary 4 asserts that BASA = CAKA = + while SA and KA
are outside ofABC. Consider the intersections of lineApCp with the sides ofbpAAdcpA. Then
ApCp intersects sideAdcpA at Cp and so externally, while it intersects side AdbpA atSA internally
since BAbpA = 2(+ ) and BASA = + . Thus, by Paschs axiom, ApCp intersects
the third side bpAcpA internally. Similarly line A
pBp intersectsbpAcpA internally. Thus b
pAA
pcpAencompasses SAA
pKA and so SAApKA < Lb
pAA
pcpA.
But SAApKA = 60
o and so bpAApcpA > 60
o. Therefore for< 30o the ApbpAcpA is not
equilateral. The cases > 30o and = 30o are similar.
Conclude that ApbpAcpA cannot be equilateral.
Note that the non equilateralApbpAcpA fails the original statement of Morleys theorem.
5.4.2 The Morley triangle of distal vertices
This is denoted by AdbdAcdA. Vertex b
dA is the
intersection ofCAp, the distal to CA trisector of the
interior
C, and the distal to CA trisector of theexterior A. Vertex cdA is the intersection of BAp,
the remaining trisector ofB(distal toAB) with thedistal to AB trisector of the exterior A. Also it iseasily seen that bdA and c
dA are determined iff =.
Hence, for =, Ad is inside AcdABand CbdAA.
From CbdAAit is calculated thatCbdAA= 180
o3(+)2 =2 and similarly
fromAcdAB, AcdAB= 2. Since ++= 60
o,at least one of and is less than30o. Thus eitherbdAc
dAA
d or cdAbdAA
d is less than 60o.Conclude that AdbdAc
dA is not equilateral.
Fig.29
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5.4.3 The Morley triangles with a proximal and two mix vertices
These triangles are denoted by ApbAc
A, bpAA
cA and cpAA
bA.
Fig.30
a. ApbAc
A : VertexAp is the inter-
section of the proximal to BC interior trisec-
tors. Hence bA and c
A are the intersections
of the two remaining interior trisectors,CBp
and BCp, with the trisectors of the exterior
A. Since each of these interior trisectors is
proximal to the side it belongs, it must be
paired with the distal to the corresponding
side exterior trisector.
Consider the companion equilateral
SAApKA relative to vertex A
p. In Fig.30
the case > 30o is depicted for which Corol-
lary 4 asserts that vertices SA and KA are
inside ABC.
Consider the intersections of line ApCp with the sides ofbpAAdcpA. A
pCp intersects side
AdcpA at Cp and so externally, while ApCp intersects side AdbpA at SA and so internally. Thus,
by Paschs axiom, ApCp intersects the third side AbA internally. Similarly ApBp intersects
AcA internally. Hence bpAA
pcpA encompasses SAApKA and so SAA
pKA < bpAA
pcpA. But
SAApKA = 60
o and so bAApcA > 60
o. Therefore for > 30o ApbAc
A is not equilateral.
The cases < 30o and = 30o are similar and they are omitted.
Conclude that ApbAc
A is not equilateral.
b. bpAAcA: Vertex b
pA is the intersection of the proximal to ACtrisectors, which are CB
p
and the corresponding trisector of the exterior A. Thus A is the intersection of the remaining
interior trisector ofC, CAp, which is proximal to BC, with BCp as distal to BC. Then cA is the
intersection of the left trisectors BAp, which is distal to AB, with the proximal to AB trisector of
the exterior A.
Notice that the last two trisectors are parallel iff2 = + = . ThuscA exists iff
=.
Also if > thenbp
A andc
A are on the same side ofAC, while for < , bp
A and c
A are ondifferent sides ofAC.
Case > : We will show bpAc
AA < 60o.
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104 Spiridon A. Kuruklis CUBO16, 2 (2014)
Notice that A is inside ABAp and so cAAp is a right bound for the right side cAA
of
bpAc
AA.
In following we will find a left bound for the left side cAb
A ofbpAc
AA.
Let bA be the intersection ofCAp withAbp
A
and note that the pointsA, Ap, cA, andb
A are
cyclic, because fromBApCit follows thatbAApcA =+and sob
Ac
A is seen from A and Ap
with angle +.
The extension ofApCp meets the exterior trisector AbA between Aand b
A. Then it crosses
the circle, say at T. We will show that a left bound for side cAb
A ofbpAc
AA is the bisector of
bAc
AA.
Note that in a triangle the bisector of an angle crosses its opposite side at a point which is
between the sides middle point and the sides common vertex with the shortest of the other two
sides.
Let G be the intersection of the bAc
A A bisector with Ab
A. Also letM be the middle of
AbA
.
CbpA is angle bisector in ACb
A. It is easily calculated from ACb
Athat Ab
AC= +and
so AbAC < b
AAC. Thus CA < Cb
A. Hence bpA is between A and M.
cAGis angle bisector in Ab
Ac
A. Obviously b
AAc
A = +. Also c
Ab
AA= c
Ab
AC+
CbAAwhile c
Ab
AA= c
Ab
AC + Cb
AA. But c
Ab
AC= c
AAAp = cAAB + BAA
p =
(+)+BAAp. So bAAc
A < c
Ab
AAand thusb
Ac
A < b
AA. HenceGis betweenb
Aand M.
ThereforebpA is onAb
A and it is betweenA and G. So Gc
AAp encompasses bpAc
AA and thus
bpAc
AA < GcAA
p .
In the sequel we calculate GcAAp. Notice that GcAA
p = GcAA+ Ac
AAp whereas
AcAAp = and GcAA=
12bAc
AA= 12bAA
pA= 12
[bAApT+ TApA].
ButbAA
pT = BApCp BApA =+ ( +) = + .
Also
TAbA = TApbA = +and A
pTA= AcAAp = .
Then from TAAp it is calculated
TApA= + CpAAp .
Thus
GcAA= + 12TApA
and so GcAAp =+ 1
2CpAAp where0 < CpAAp bpAAcA is not equilateral.
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Fig.31a (> )
Fig.31b
Case < : We will show that bPAAcA >
60o.
Consider the intersections ofAcA withBCp and
CBp denoted bycpAand b
A respectively. Notice that
bPAAcA encompasses b
AAcpA. So it suffices to
show that bAAcpA > 60
o.
Observe that b
A, A
, C and c
p
A are cyclic, be-cause sidebAA is seen from cpA andC with angle
as from AcpAB it is calculated AcpAB= . Thus
bAAcpA = b
ACcpA.
Moreover bACcpA = A
dCcpA, since Cb
A
passes through Ad. But
AdCcpA = AdCA+ACc
pA = +ACc
pA and so
bACcpA = + ACc
pA.
In aditionA, Ad, C and cpA are also cyclic since
AAd is seen from C and cpA with angle .
ConsequentlyACcpA = AAdcpA and so
bACcpA = + AA
dcpA.
Yet from AAdB infer
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106 Spiridon A. Kuruklis CUBO16, 2 (2014)
AAdcpA = ABAd + AdAB.
However AdAB > CpABand thus
AAdcpA > ABAd + CpAB= + .
Therefore bAAcpA > + + = 60
o.
Conclude that for < , bpAAcA is not equilateral.
c. cpAA
bA : It is showed that it is not equilateral similarly as bpAA
cA.
5.4.4 The Morley triangles with one distal and two mix vertices
These triangles are denoted byAdbAc
A, cdAA
bA andbdAA
cA.
Fig.32
a. AdbAc
A: ObviouslyAd is the intersection
ofCBp and BCp. Then b
A is the intersection of theremaining interior trisectorsCAp (distal toAC) with
the proximal to AC exterior trisector ofA. Notice
these two lines are parallel iff += 2 = .
Moreover cA is the intersection of the left trisectors,
the interior BCp (proximal) with the distal to AB
trisector of the exterior A. Notice these lines are
parallel iff = . Therefore AdbAc
A is deter-
mined iff = . FromAcAB and Ab
AC it fol-
lows ApbAA= ApbAA= || and so b
A and
cA are on the same side ofAC.
We will consider only the case > as the other
one is similar.
Notice that Ap is inside BAdC. Thus
AdcAb
A encompasses ApcAb
A and so
AdcAb
A > ApcAb
A.
Also notice thatA,Ap, bAand c
A are cyclic asAAp is seen frombAand c
A with angle.
Thus ApcAb
A = ApAbA. But A
pAbA = ApAC + CAbA = A
pAC + (+). Moreover
ApAC > BpAC= and so ApAbA > + ( +) =60o. Therefore AdcAb
A > 60o.
Conclude that Ad
b
Ac
A is not equilateral.
b. bdAAcA : Obviously b
dA is the intersection of CA
p with the distal to AC exterior
trisector of A. Then A is the intersection of the remaining trisector CBp (proximal to AB)
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Trisectors like Bisectors with equilaterals instead of Points . . . 107
Fig.33
with the BAp (distal). Thus cA is the intersection of the left trisectors,BCp and the distal to AB
exterior trisector ofA.
Notice that bdA and c
A exist iff =. We will show that bdAA
cA > 60o.
Note that Bp is inside bdAAcA and so b
dAA
cA encompasses bpAA
cA. Thus it suffices
to prove bpAAcA > 60
o.
For this we use a symmetric argument to the proof ofbAAcpA > 60
o (5.4.4.a case < ).
Let bpA be the intersection of Ac
A with CA. Notice that cA, A
, B, bpA are cyclic as
cAA is seen from B and bpA with angle . Thus c
AAbpA = c
ABbpA. Moreover c
AAbpA =
AdBbpA, since Bc
A passes through Ad. But AdBbpA = A
dBA+ ABbpA = + ABb
pA and
so cAAbpA= + ABb
pA.
HoweverA, Ad, B, bpA are cyclic as AAd is seen from B andbpA with angle . Consequently
ABbpA = AAdbpA and so c
AAbpA = + AA
dbpA.
Yet from AAdC infer AAdbpA = ACAd + AdAC. However AdAC > BpAB and
thus AAdcp
A
> ACAd + BpAB= + . Therefore cAAbp
A
> + + = 60o.
Conclude that bdAAcA is not equilateral.
c. cdAAbA: This case is similar to the above and it is omitted.
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108 Spiridon A. Kuruklis CUBO16, 2 (2014)
6 Analogy between Bisectors and Trisectors in a triangle
The essence of the previous work is portrayed in the following two figures illustrating the analogy
between the (well understood) structure of angle bisectors and the (under study) structure of angle
trisectors in a triangle.
The structure of angle bisectors The structrure of angle trisectors
Fig.34a Fig.34b
The interior angle bisectors pass through a
unique point (incenter).
The interior angle trisectors proximal to the tri-
angle sides pass through the vertices of a unique
equilateral (inner Morley equilateral).
The bisector of an interior angle and the bi-
sectors of the other two exterior angles pass
through a unique point (excenter).
The trisectors of an interior angle and the trisec-
tors of the other two exterior angles proximal to
the triangle sides pass through the vertices of a
unique equilateral (exterior Morley equilateral).
The exterior bisectors pass through the vertices
of a unique triangle with orthocenter the interior
angle bisectors common point (incenter). +
The exterior trisectors proximal to the triangle
sides pass through the vertices of a unique equi-
lateral (central Morley equilateral). +
+ This fact follows from the previous one
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Trisectors like Bisectors with equilaterals instead of Points . . . 109
This structural similarity suggests that the triangle trisectors with the proper pairing meet
at equilaterals which correspond to the triangle bisectors common points. The perception that
trisectors behave like bisectors with equilaterals instead of points invites further exploration. New
results could be inspired from the vast variety of the angle bisectors point-line-circle theorems
revealing more exciting analogies between the two structures.
This