1. CONGRUENT AND INSCRIBED
Standard 21
INSCRIBED ANGLES
PROBLEM 1a
PROBLEM 1a
CONGRUENT AND INSCRIBED
INSCRIBED TO A SEMICIRCLE
PROBLEM 2
INSCRIBED AND CIRCUMSCRIBED
PROBLEM 3
PROBLEM 4
END SHOW
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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. Standard 21
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 C B A
ABC
is an inscribed angle
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. L M K m KL 1 2
KML m =
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4. Standard 21 A = BAC m (3x+5)° m BC 1 2 B 1 2 (40°) 40° (3X+5) = C
3X = 15
X=5
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5. Standard 21 L J K (2x+7)° 54° m JL 1 2 = JKL m 1 2 (54°) (2X+7) =
2X = 20
X=10
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6. Standard 21 D B C A P ADB ACB AB If and intercept same arc then ADB
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.
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7. intercepts semicircle
Standard 21 A B C P AC
ABC If
intercepts semicircle
ABC= m
90°
then
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.
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8. Standard 21 4X° (6X-10)° L N M K 4X° + (6X-10)° = 90° 10X-10 = 90
10X = 100
X=10
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9. These are concentric circles and all circles are similar.
Standard 21
These are concentric circles and all circles are similar.
Estos son círculos concéntricos y todos los círculos son semejantes.
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10. Standard 21 A B P D C K E F N Q L H G M
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11. Quadrilateral ABCD is inscribed to circle P.
Cuadrilatero ABCD esta inscrito al círculo P. R X g
Line g is TANGENT to circle X at point R.
Línea g es tangente al círculo X en punto R. L K N M Q H G F E
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.
Standard 21
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12. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G 1. ED m = ?
Standard 21
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13. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G 1. ED m = ?
Standard 21
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14. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G 1. ED m = ?
Standard 21
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15. and FGDE is a rhombus so all sides EFB
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G
EAB
and FGDE is a rhombus so all sides
EFB
90°
EBF m
EBD m
are congruent 1. ED m = ?
60°
30°
EB EB
60°
30°
Since
is inscribed to SEMICIRCLE
then
EFB m =
and then E F B
EFB
and D E B
EDB
are right,
because the
Reflexive Property
therefore
EF ED
; and
then:
EFB
EDB
by HL.
And
EBF
EBD
by CPCTC.
60°
30°
Then =
So:
-3X+45 = 4X+10
EBF m
=-3X+45
60°
30°
-3X = 4X – 35
EBF m
=-3( )+45 5
-4X -4X =
- 7X = - 35
=30°
So:
EBD m
= 30°
Take notes
X=5
Standard 21
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16. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G
FE ED
60°
EF ED
EBD m m ED 1 2 1. m ED = ? If
60°
30° =
60°
30° m ED 1 2
then:
60°
30° = m ED 1 2
30° =
(2) m ED
= 60° 2. FE m = ?
Since
then
and m FE
= 60°
Standard 21
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17. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G
60° 1. m ED = ? m ED 1 2 If
60°
30°
EBD m = 3.
BED m = ?
60°
60°
30° m ED 1 2
then:
60°
30° = m ED 1 2
30° =
(2) m ED
= 60° 2. FE m = ?
Since
EF ED
then m FE
FE ED
and
= 60°
From figure
BED m = 4.
BGD m = ?
Standard 21
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18. because EFGD is a rhombus 60°
Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G
because EFGD is a rhombus
60° 1. m ED = ? m ED 1 2 If
60°
30°
EBD m =
DEG m =
EGD m +
BGD =
180°
60°
60°
DGE m
30° m ED 1 2
then:
60°
30° = m ED 1 2
30° =
(2) m ED
= 60° 2. FE m = ?
Since
EF ED
then
FE ED
and m FE
= 60° 3.
BED m = ?
From figure
BED m = 4.
BGD m = ?
and then
60° +
BGD m =
180°
Take notes
-60° °
BGD m =
120°
Standard 21
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19. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G
60°
120° 5. DB m =
120°
60°
120°
Standard 21
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20. Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.
EBF m =
-3X+45
EBD
4X+10
and
Find the following: C B A D E F H I G
120°
60°
90° 5. DB m =
120°
60° 6.
DEB = m
60°+60°+120°
120°
= 240° 7.
AIB m =
Standard 21
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21. Alternate interior are S CDB b.
Statements
Reasons m BC 1 2
ABD
ABD m = AB DC C a.
Given A m AD 1 2
Alternate interior are S
CDB b.
CDB m = c.
CDB m
ABD = S
have the same measure D d.
An inscribed is half its intercepted arc
Given: AB DC
An inscribed is half its intercepted arc e.
Prove: m AD 1 2 BC = f.
AD BC
Transitive Property. g. AD m = BC
Division Property of Equality
Arcs with the same measure are h.
AD BC
Standard 21
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