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1 INSCRIBED ANGLES PROBLEM 1a CONGRUENT AND INSCRIBED INSCRIBED TO A SEMICIRCLE PROBLEM 2 INSCRIBED AND CIRCUMSCRIBED PROBLEM 3 Standard 21 PROBLEM 4 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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3 C B A ABC is an inscribed angle L M K KML m = m KL 1 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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4 C B A (3x+5)° 40° (3X+5) = 1 2 (40°) = BAC m m BC 1 2 3X + 5 = 20 -5 3X = 15 3 X=5 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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5 L J K (2x+7)° 54° (2X+7) = 1 2 (54°) = JKL m m JL 1 2 2X + 7 = 27 -7 2X = 20 2 X=10 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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6 D B C A P If ADB and ACB intercept same arc AB then ADBACB 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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7 A B C P IfABC 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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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8 4X° (6X-10)° L N M K 4X° + (6X-10)° = 90° 10X-10 = 90 +10 10X = 100 10 X=10 Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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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 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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10 B A D C P L K N M Q H G F E Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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11 B A 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 F E g Line g is TANGENT to circle X at point R. Línea g es tangente al círculo X en punto R. R X Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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12 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 m = 4X+10 and Find the following:

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13 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 m = 4X+10 and Find the following:

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14 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 m = 4X+10 and Find the following:

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15 and FGDE is a rhombus so all sides C B A D E F H I G EB 1. ED m = ? Since is inscribed to SEMICIRCLE EFB then EFB m = 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: EBF m EBD m -3X+45 = 4X+10 -45 -3X = 4X – 35 -4X - 7X = - 35 -7 X=5 EBF m =-3X+45 EBF m =-3( )+45 5 = -15+45 So: =30° 30° EBD m = 30° 30° and D E B EDB And 60° Take notes 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 m = 4X+10 and Find the following: EF ED 60° 30°

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16 C B A D E F H I G 1. ED m = ? 30° 60° EBD m = m ED 1 2 If then: 30° = (2) m ED 1 2 30° = m ED = 60° 2. FE m = ? Since thenand m ED 1 2 EF ED m FE = 60° FE ED 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 m = 4X+10 and Find the following:

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17 C B A D E F H I G 1. ED m = ? 30° 60° EBD m = m ED 1 2 If then: 30° = (2) m ED 1 2 30° = m ED = 60° 2. FE m = ? Since thenand m ED 1 2 EF ED m FE = 60° 3. BED m = ? m = From figure 4. BGD m = ? FE ED 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 m = 4X+10 and Find the following:

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18 C B A D E F H I G 1. ED m = ? 30° 60° EBD m = m ED 1 2 If then: 30° = (2) m ED 1 2 30° = m ED = 60° 2. FE m = ? Since thenand m ED 1 2 EF ED m FE = 60° 3. BED m = ? m = From figure 4. BGD m = ? FE ED 60° Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved DGE m DEG m = because EFGD is a rhombus and then 60° + BGD m = 180° -60° BGD m = 120° Take notes EGD m + BGD m = 180° Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi. EBF m = -3X+45 EBD m = 4X+10 and Find the following:

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19 C B A D E F H I G 60° 5. DB m = 120° Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved 120° Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi. EBF m = -3X+45 EBD m = 4X+10 and Find the following:

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20 C B A D E F H I G 60° 5. DB m = 6. DEB = m 60°+60°+120° = 240° 7. AIB m = 120° Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved 120° 90° Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi. EBF m = -3X+45 EBD m = 4X+10 and Find the following:

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21 Statements Reasons a. b. c. d. e. f. g. C B D A Given: ABDC Prove: AD BC Given CDB m ABD m = Alternate interior are S S have the same measure h. m AD 1 2 m BC 1 2 = Transitive Property. AD m = BC m Division Property of Equality AD BC Arcs with the same measure are ABDC ABD CDB m AD 1 2 ABD m = m BC 1 2 CDB m = An inscribed is half its intercepted arc Standard 21 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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