Other Angle Relationships in Circles Geometry Section 8 Day 3

Theorem 10.11 If a tangent and a chord intersect at a point on the circle, then the measure of angle formed is one half the measure of its intercepted arc. 𝑚∠1= 1 2 𝑚 𝐴𝐵 𝑚∠2= 1 2 𝑚 𝐵𝐶𝐴 Geometry S8 Day 3

Theorem 10.12- Angles Inside the Circle Theorem Geometry S8 Day 3 If two chords intersect inside a circle, the measure of each angle is one half the sum of the measures of the arcs intercepted by the angle and its vertical angle 𝑚∠1= 1 2 𝑚 𝐷𝐶 +𝑚 𝐴𝐵 𝑚∠2= 1 2 (𝑚 𝐵𝐶 +𝑚 𝐴𝐷 )

Theorem 10.13- Angles Outside the Circle Theorem If a tangent and a secant, two tangents, or two secants intersect outside a circle, then the measure of the angle formed is one half the difference of the measures of the intercepted arcs 𝑚∠1= 1 2 𝑚 𝐵𝐶 −𝑚 𝐴𝐶 𝑚∠2= 1 2 𝑚 𝑃𝑄𝑅 −𝑚 𝑃𝑅 𝑚∠3= 1 2 (𝑚 𝑋𝑌 −𝑚 𝑊𝑍 ) Geometry S8 Day 3

Example 1: Solve for the ? 68°= 1 2 80°+𝑚 𝑊𝑀 68°=40°+ 1 2 𝑚 𝑊𝑀 68°= 1 2 80°+𝑚 𝑊𝑀 68°=40°+ 1 2 𝑚 𝑊𝑀 68°−40°= 1 2 𝑚 𝑊𝑀 28°= 1 2 𝑚 𝑊𝑀 2∙28°=𝑚 𝑊𝑀 56°=𝑚 𝑊𝑀 ?=56° Geometry S8 Day 3

Example 2: Solve for x 40°= 1 2 124°− 5𝑥−6 2∙40°=124°− 5𝑥−6 40°= 1 2 124°− 5𝑥−6 2∙40°=124°− 5𝑥−6 80°=124°−5𝑥+6 80°−124°−6=−5𝑥 −50=−5𝑥 −50 −5 = −5𝑥 −5 𝑥=10 Geometry S8 Day 3

Homework Geometry S8 Day 3 Assignment 8-3

Segment Lengths in Circles Geometry Section 8 Day 3

Theorem 10.14- Segments of Chords Theorem Geometry S8 Day 3 If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord 𝐸𝐴∙𝐸𝐵=𝐸𝐶∙𝐸𝐷

Theorem 10.15 If two secant segments share the same endpoint outside a circle, then the product of the lengths of one secant segment and its external segment equals the product of the lengths of the other secant segment and its external segment. EA∙𝐸𝐵=𝐸𝐶∙𝐸𝐷 Geometry S8 Day 3

Theorem 10.16- Segments of Secants and Tangents Theorem If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the lengths of the secant segment and its external segment equals the square of the length of the tangent segment. 𝐸 𝐴 2 =𝐸𝐶∙𝐸𝐷 Geometry S8 Day 3

Example 3: Solve for x 24∙14=16∙𝑥 336=16𝑥 336 16 = 16𝑥 16 𝑥=21 Geometry S8 Day 3

Example 4: Find the indicated segment 20 2 =16∙ 16+ 2𝑥−3 400=16∙ 16+2𝑥−3 400=16∙ 13+2𝑥 400=208+32𝑥 400−208=32𝑥 192=32𝑥 192 32 = 32𝑥 32 𝑥=6 𝐺𝐸=16+2 6 −3 =16+12−3=𝟐𝟓 Geometry S8 Day 3

Homework Geometry S8 Day 3 Assignment 8-4