1
INSCRIBED ANGLES
PROBLEM 1a PROBLEM 1a
CONGRUENT AND INSCRIBED
INSCRIBED TO A SEMICIRCLE
PROBLEM 2
INSCRIBED AND CIRCUMSCRIBED
PROBLEM 3
Standard 21
PROBLEM 4
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2
Standard 21:
Students prove and solve problems regarding relationships among chords, secants, tangents, inscribed angles, and inscribed and circumscribed polygons of circles.
Los estudiantes prueban y resuelven problemas relacionados con cuerdas, secantes, tangentes, ángulos inscritos y polígonos inscritos y circunscritos a círculos.
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3
CB
A
ABC is an inscribed angle
LM
K
KMLm = m KL1
2
Inscribed angles are angles formed by two chords whose vertex is on the circle.
Ángulos inscritos son ángulos formados por dos cuerdas cuyo vértice esta en el circulo
If an angle is inscribed in a circle then the measure of the angle equals one-half the measure of its intercepted arc.
Si un ángulo en un círculo es inscrito entonces la medida de el ángulo es igual a la mitad de su arco intersecado.
Standard 21
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4
C
B
A
(3x+5)°
40° (3X+5) =1
2(40°)
=BACm m BC1
2
3X + 5 = 20-5 -5
3X = 153 3
X=5
Standard 21
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5
L
JK
(2x+7)°
54°(2X+7) =
1
2(54°)
=JKLm m JL1
2
2X + 7 = 27-7 -7
2X = 202 2
X=10
Standard 21
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6
D
B
C
A
P
If ADB and ACB intercept same arc AB
then ADB ACB
If two inscribed angles of a circle or congruent circles intercept congruent arcs, or the same arc, then the angles are congruent.
Si dos ángulos inscritos de un círculo o de círculos congruentes intersecan el mismo arco o arcos congruentes entonces los ángulos son congruentes.
Standard 21
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7
A
B
CP
If ABC intercepts semicircle AC
then ABC=m 90°
If an inscribed angle intercepts a semicircle, then the angle is a right angle.
Si un ángulo inscrito interseca a un semicírculo entonces el ángulo es recto.
Standard 21
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8
4X°
(6X-10)°
L
N
M
K4X° + (6X-10)° = 90°
10X-10 = 90
+10 +10
10X = 10010 10
X=10
Standard 21
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9
These are concentric circles and all circles are similar.
Estos son círculos concéntricos y todos los círculos son semejantes.
Standard 21
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10
BA
D C
P
L
K
N
M
Q
H G
FE
Standard 21
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11
BA
D C
P
Quadrilateral ABCD is inscribed to circle P.
Cuadrilatero ABCD esta inscrito al círculo P.
Quadrilateral EFGH is circumscribed to circle Q, having sides to be TANGENT at points K, L, M and N.
Cuadrilátero EFGH esta circunscrito al círculo Q, teniendo los lados TANGENTES en los puntos K, L, M y N.
L
K
N
M
Q
H G
FE
g
Line g is TANGENT to circle X at point R.
Línea g es tangente al círculo X en punto R.
R
X
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
12
C
B
A
D
E
F
H IG1. EDm = ?
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
13
C
B
A
D
E
F
H IG1. EDm = ?
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
14
C
B
A
D
E
F
H IG1. EDm = ?
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
15
and FGDE is a rhombus so all sides C
B
A
D
E
F
H IG
EB EB
1. EDm = ?
Since is inscribed to SEMICIRCLEEFB
then EFBm =
EAB
90° and then are right,
E
F
B
EFB
are congruent
therefore ; and because the
Reflexive Property then: EFB EDB
by HL. EBF EBD by CPCTC.
Then = So:EBFm EBDm
-3X+45 = 4X+10-45 -45
-3X = 4X – 35 -4X -4X
- 7X = - 35-7 -7
X=5
EBFm =-3X+45
EBFm =-3( )+455= -15+45
So:
=30°
30°
30°
EBDm = 30°
30°
and
D
E B
EDB
And
60°
60°
Take notesStandard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
EF ED
60°
60° 30°
16
C
B
A
D
E
F
H IG1. EDm = ? 30°
30°60°60°
EBDm =
m ED1
2
If
then: 30° =
(2)(2) m ED1
230° =
m ED = 60°
2. FEm =? Since then and
m ED1
2
EF ED m FE = 60°FE ED
60°
60°
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
17
C
B
A
D
E
F
H IG1. EDm = ? 30°
30°60°60°
EBDm =
m ED1
2
If
then: 30° =
(2)(2) m ED1
230° =
m ED = 60°
2. FEm =? Since then and
m ED1
2
EF ED m FE = 60°
3. BEDm = ? BEDm =From figure
4. BGDm = ?
FE ED
60°
60°
60°
60°
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
18
C
B
A
D
E
F
H IG1. EDm = ? 30°
30°60°60°
EBDm =
m ED1
2
If
then: 30° =
(2)(2) m ED1
230° =
m ED = 60°
2. FEm =? Since then and
m ED1
2
EF ED m FE = 60°
3. BEDm = ? BEDm =From figure
4. BGDm = ?
FE ED
60°
60°
60°
60°
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
DGEmDEGm = because EFGD is a rhombus
and then
60° + BGDm = 180°-60° -60°
BGDm = 120°
Take notes
EGDm + BGDm = 180°
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
19
C
B
A
D
E
F
H IG
60°
60°
5. DBm = 120°
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
120°
120°
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
20
C
B
A
D
E
F
H IG
60°
60°
5. DBm =
6. DEB =m 60°+60°+120°
= 240°
7. AIBm =
120°
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
120°
120°
90°
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBFm = -3X+45 EBDm = 4X+10and
Find the following:
21
Statements Reasons
a. a.
b. b.
c. c.
d. d.
e. e.
f. f.
g. g.
C
B
D
A
Given:
AB DC
Prove:
AD BC
Given
CDBmABDm =
Alternate interior are
S
S have the same measure
h. h.
m AD1
2m BC
1
2= Transitive Property.
ADm = BCm Division Property of Equality
AD BC Arcs with the same measure are
AB DC
ABD CDB
m AD1
2ABDm =
m BC1
2CDBm =
An inscribed is half its intercepted arcAn inscribed is half its intercepted arc
Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved