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C hapter 10 Circles. C hapter 10 Section 10.3 Inscribed Angles . Bell Problem. Objective. U SING I NSCRIBED A NGLES. inscribed angle. intercepted arc.

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## C hapter 10 Circles

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**USING INSCRIBED ANGLES**inscribed angle intercepted arc An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of the circle. The arc that lies in the interior of an inscribed angle and has endpoints on the angle is called the intercepted arc of the angle.**USING INSCRIBED ANGLES**THEOREM A 1 2 mADB= mAB C D B mAB= 2mADB THEOREM 10.8 Measure of an Inscribed Angle If an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc.**Finding Measures of Arcs and Inscribed Angles**W N R S C C Z C M 115° X Q T P Y mQTS = 2mQRS = 2(90°) = 180° 1 2 1 2 mZWX = 2mZYX = 2(115°) = 230° M NMP = mNP = (100°) = 50° Find the measure of the blue arc or angle. 100° SOLUTION**USING INSCRIBED ANGLES**THEOREM A B C D C D THEOREM 10.9 If two inscribed angles of a circle intercept the same arc, then the angles are congruent.**USING PROPERTIES OF INSCRIBED POLYGONS**If all of the vertices of a polygon lie on a circle, the polygon is inscribed in the circle and the circle is circumscribed about the polygon. The polygon is an inscribed polygon and the circle is a circumscribed circle.**USING PROPERTIES OF INSCRIBED POLYGONS**A B C B is a right angle if and only if AC is a diameter of the circle. THEOREMS ABOUT INSCRIBED POLYGONS THEOREM 10.10 If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of an inscribed triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle.**USING PROPERTIES OF INSCRIBED PLOYGONS**THEOREMS ABOUT INSCRIBED POLYGONS F E C D G D, E, F, and G lie on some circle, C, if and only if m D + m F = 180° and m E + m G = 180°. . THEOREM 10.11 A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.**Using an Inscribed Quadrilateral**A 2y° In the diagram, ABCD is inscribed in P. Find the measure of each angle. P 3y° D B 3x° 5x° C .**Using an Inscribed Quadrilateral**A 2y° P 3y° D B 3x° 5x° C SOLUTION ABCD is inscribed in a circle, so opposite angles are supplementary. 3x + 3y = 180 5x + 2y = 180**Using an Inscribed Quadrilateral**A 3x + 3y = 180 5x + 2y = 180 2y° P 3y° D B 3x° 5x° C To solve this system of linear equations, you can solve the first equation for y to get y = 60 – x. Substitute this expression into the second equation. 5x + 2y = 180 Write second equation. 5x + 2(60 – x) = 180 Substitute 60 – x for y. 5x + 120 – 2x = 180 Distributive property 3x = 60 Subtract 120 from each side. x = 20 Divide each side by 3. y = 60 – 20 = 40 Substitute and solve for y.**Using an Inscribed Quadrilateral**A 2y° P 3y° D B 3x° m C = 100°, andm D = 120°. 5x° som A = 80°, m B = 60°, C x = 20 andy = 40,**Homework**p617 (9-23)

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