3.2 More Neutral Theorems

Pasch’s Axiom-If a line l intersects Δ PQR at point S such that P-S-Q, then l intersects

Crossbar Theorem-If X is a point in the interior of triangle Δ UVW, then intersects at a point Y such that W-Y-V

RQPR or

UX

WV

3.3 Congruence Conditions

We already had SAS congruence for triangles (postulate 15)

Can also have ASAAASSSS

Isosceles Triangle Theorem

If 2 sides of a triangle are congruent then the angles opposite them are congruent.

Converse Isosceles Theorem: If 2 angles of a triangle are congruent then the

sides opposite them are congruent.

Converse Isosceles Proof

Suppose we have ΔABC with <A ≡ <C. Now draw as the bisector of <ABC. Let D be the intersection of with

Want to Show that ΔABD ≡ ΔCBD. Since <BAC ≡ <BCA, <ABD ≡ <CBD, and

≡ , we know that ΔABD ≡ ΔCBD by AAS, therefore ≡

Hence, the sides opposite the angles are congruent.

BDBD AC

BDBD BCBA

Inverse Isosceles Theorem

If 2 sides of a triangle are not congruent, then the angles opposite those sides are not congruent, and the larger angle is opposite the larger side.

Similarly, If two angles of a triangle are not congruent, then the sides opposite them are not congruent and the larger side is opposite the larger angle. Proof

Triangle Inequality Theorem

Triangle Inequality Theorem: The sum of the measures of any 2 sides of a

triangle is greater than the measure of the third.

The Hinge Theorem: Draw picture

Do #6 and #10

Assign #4, #7

3.4 The Place of Parallels

We have not talked about parallel lines because we have only assumed SMSG postulates 1-15

However, in Neutral Geometry it can be shown that at least one line can be drawn parallel to a given line through a point not on that line.

Alternate Interior Angle Thrm

If two lines are intersected by a transversal such that a pair of alternate interior angles formed is congruent, then the lines are parallel. Read Proof

Note: the converse (what you were probably given in high school geometry), which states that if two parallel lines are intersected by a transversal then the pairs of alternate interior angles are congruent is NOT a theorem in neutral geometry.

Corollary 3.4.2

Two lines perpendicular to the same line are parallel. Proof: Given lines l, m, n such that l ┴ n, m ┴ n. Then wts: l

is parallel to m. By definition, the intersection of perpendicular lines

produces four right angles. Since l and m are both intersected by n at right angles, the

alternate interior angles are congruent. By Alt. Interior Angle Thrm, l is parallel to m. Therefore two lines perpendicular to the same line are

parallel.

Corollary 3.4.3

If two lines are intersected by a transversal such that a pair of corresponding angles formed is congruent, then the lines are parallel . Proof: Given n is a transversal of l and m, and

corresponding angles <2 and <6 are congruent. wts: l is parallel to m Since <2 and <4 are vertical angles, <2 ≡ <4. Since <2 ≡ <4 and <2 ≡ <6, by transitivity of congruence,

<4 ≡ <6. Since <4 and <6 are alternate interior angles, by alt. int.

angle thrm. we have that l is parallel to m.

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Corollary 3.4.4

If two lines are intersected by a transversal such that a pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.

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Proof Given n is a transversal of l and m and angles 1 and 6

are interior angles on the same side that are supplementary.

wts: l is parallel to m Since <1 and <6 are supplementary, m<1+m<6 =180.

We know that <6 and <7 are supplementary by definition, so m<6+m<7=180.

This implies that m<1=m<7 and so <1 ≡ <7.

Since <1 and <7 are congruent alt. interior angles, by alt. int. angle theorem, I is parallel to m

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#3 Prove: If two lines are intersected by a transversal

such that a pair of alternate interior angles is congruent, then the lines have a common perpendicular. Proof: Given n is a transversal of l and m and <1 ≡<7 , and

<4 ≡ <6. wts: l and m have common perp, i.e. let n be perpendicular to

l then show n is perpendicular to m Since n is perp to l, then m<1=m<2=m<3=m<4=90 Since <1 ≡<7 , and <4 ≡ <6 , m<6=m<7=90. So m<8=m<5=90 since they are vertical angles. Therefore n is perpendicular to m. So they share a common perpendicular.

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Equivalences to Euclidean Parallel Postulate

Theorem (3.4.5) Euclid’s 5th postulate is equivalent to the Euclidean parallel postulate.

Euclid’s 5th: That, if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.

Proof (→) Assume Euclid’s 5th and prove EPP. Given line m and point A not on m, draw line t perp to m through

A which intersects m and B. Let l be a line through A perp to t. By Cor 3.4.2, l || m wts: there is not another line through A || to m Let n be another line through A || to m Since n ≠ l then m∩n = Ø and E is interior to <DAB So m<EAB < m<DAB and so <EAB is acute. Therefore m<EAB + m<ABC < 180 and m∩n ≠ Ø which is a

contradiction. So there is only one parallel line through A.

(←) Assume EPP and prove Euclid’s 5th. Assume m<1 + m<2 < 180. Prove l intersects m on the same side of t. Draw ray AD such that <BAD ≡ <3. So ray AD || m (Alt. Int. Angle Thrm) So since l ≠ AD, the EPP guarantees l∩m ≠ Ø In order to show I and m meet on the same side of m (on the

RHS), we will assume l∩m on the LHS of t. Then <1 is exterior angle of ΔABC. This implies <1 < <DAB which implies <1 < <3 This is a contradiction, so l∩m on the RHS.

Other equivalences to EPP

The EPP is equivalent to1. The converse of the Alt. Int. Angle Theorem2. If a line intersects one of two parallel lines,

then it intersects the other.3. If a line is perpendicular to one of two

parallel lines, then it is perp to the other.4. Given ΔPQR and any line segment AB,

there exists ΔABC having a side ≡ to AB and ΔABC is similar to ΔPQR.

Assign #5, 6, 10, 11 #5 and #10 for turn in (due Monday Sep 24)

3.5 The Saccheri-Legendre Theorem

Saccheri studied a specific class of quadrilaterals now called Saccheri quadrilaterals.

ABCD where AB ≡ CD, <A ≡ <D =90ם So either <B and <C are right angles, acute

angles or obtuse angles He tried to show that if they were acute/obtuse

you get contradictions

Saccheri-Legendre Theorem: The angles sum of any triangle is less than or

equal to 180 Lemma 3.5.2

The sum of any 2 angles of a triangle is less than 180

Lemma 3.5.3 For any ΔABC, there exists ΔA’B’C’ having the

same angle sum but m<A’ ≤ ½ m<A To prove theorem, reused this lemma repeatedly

Cor 3.5.4

The angle sum of any convex quadrilateral is less than or equal to 360

Convex: םABCD is convex if a pair of opposite sides exist AB and CD such that CD is contained in one of the half-planes bounded by line AB and AB is in one of the half planes bounded by line CD

Proof

wts: m<A + m<B +m<C + m<D ≤ 360 Assume we have convex םABCD. Draw line segment BD. It separates םABCD into ΔABD and ΔBDC Thrm 3.5.1 says sΔABD ≤ 180 and sΔBDC ≤

180 So sΔABD + sΔBCD ≤ 360 From angle addition postulate <ADB + <BDC=<D

and <ABD + <CBD =<B So m<A + m<B +m<C + m<D ≤ 360

#3

State and prove the converse of Euclid’s 5th. If 2 straight lines meet on a particular side of

a transversal, then the sum of the interior angle on that side is less than 2 right angles (180)

Proof

Let l and m be lines such that l∩m = A and let t be a transversal of I and m.

wts: <1 + <2 < 180 Let B = l∩t and C = t∩m and consider ΔABD By lemma 3.5.2 m<C +m<B < 180 So the sum of the interior angles on the same

side of a transversal is < 180.

From this Saccheri easily showed that in the quadrilateral <B and <C were not obtuse

A contradiction to acute angles still hasn’t been found.

Homework: #2

3.6 The Search for a Rectangle

For SQ םABCD: AD, BC called sides or legs AB called base DC called summit <C, <D called summit angles

His goal: show <C=<D=90 So he looked at ways to maybe construct a

rectangle in neutral geometry rectangle: quadrilateral with 4 right angles

Theorem 3.6.1

The diagonals of a Saccheri quadrilateral are congruent

Proof: Let םABCD be a SQ. So then AD ≡ BC, <A ≡ <B

=90 Draw diagonals AC and DB and consider ΔABC

and ΔBAD Since AB ≡ AB, and <A ≡ <B, AD ≡ BC Then ΔABC ≡ ΔBAD by SAS So DB ≡ AC.

Theorem 3.6.2 The summit angles of a SQ are congruent

Theorem 3.6.3 The summit angles of a SQ are not obtuse, thus

are both acute or both right Theorem 3.3.4

Trying to prove two quadrilaterals are congruent, need to show SASAS

Theorem 3.6.4

The line joining the midpoints of both the summit and the base of a SQ is perpendicular to both.

Proof: Given SQ םABCD, let MN join the midpoints of

AB and DC wts: MN ┴ AB and MN ┴ CD By def, AD ≡ BC and <A ≡ <B =90 Also, <C ≡ <D by theorem 3.6.2

By def of MN, AM=MB and DN=NC So םAMND ≡ םBMNC by SASAS theorem This implies <NMB ≡ <NMA or m<NMB=m<NMA We know m<NMA + m<NMB =180 So then m<NMB =½(180) and m<NMA=½(180) Similarly, m<MND=½(180) and m<MNC=½(180) Therefore, MN ┴ AB and MN ┴ CD

Cor. 3.6.5 The summit and base of a SQ are parallel

Theorem 3.6.6 In any SQ, the length of the summit is greater

than or equal to the length of the base.

Lambert

Lambert Quadrilateral: quadrilateral with at least 3 right angles

Lambert’s Theorem 3.6.7 The fourth angle of a LQ is not obtuse, thus is

either acute or right.

Proof

Let םPQRS be a LQ Then <P = <Q = <R =90 wts: <S ≤ 90 Since <P + <Q + <R + <S ≤ 360 by Cor 3.5.4

and <P + <Q + <R =270 Then <S ≤ 90 So it is either acute or right.

Theorem 3.6.8 The measure of the side included between 2 right

angles of a LQ is less than or equal to the measure of the side opposite it.

Proof: Let םPQRS be a LQ. Then <P=<Q=<R=90 and <S ≤ 90 wts: PQ ≤ SR and QR ≤ PS Assume PQ > SR and choose P’ such that P’Q = SR. Then םP’QRS is a SQ So then <1 ≡ <2 and m<1 = m<2 ≤ 90 This is a contradiction by exterior angle theorem to ΔPP’S So PQ ≤ SR Similarly QR ≤ PS (you try to finish this one)

Theorem 3.6.9 The measure of the line joining the midpoints of

the base and summit of at SQ is less than or equal to the measure of it sides.

Proof: in book

Lots of theorems and corollaries about when triangles and rectangles exist that you should look at

Theorem: The EPP is equivalent to saying the angle sum of

every triangle is 180 Theorem:

The Pythagorean Theorem implies the existence of a rectangle.

Theorem: The Pythagorean Theorem is equivalent to EPP

#25 If a quadrilateral is both SQ and LQ then it is a rectangle.

Proof: Given םABCD such that AB ≡ BC and

<A=<B=<C=90 wts: םABCD is a rectangle. By SQ theorem, <C ≡<D So then <D=90 So םABCD is a rectangle.

Homework: #1, 3, 4, 6, 7, 12, 14, 26 Turn in #3, 6, 14