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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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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 = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**Standard 21 A = BAC m (3x+5)° m BC 1 2 B 1 2 (40°) 40° (3X+5) = C**

3X = 15 X=5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**Standard 21 L J K (2x+7)° 54° m JL 1 2 = JKL m 1 2 (54°) (2X+7) =**

2X = 20 X=10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**Standard 21 4X° (6X-10)° L N M K 4X° + (6X-10)° = 90° 10X-10 = 90**

10X = 100 X=10 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**Standard 21 A B P D C K E F N Q L H G M**

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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**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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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