# CONGRUENT AND INSCRIBED

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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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