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These four theorems concerning secants and tangents are reiterated graphically below. L # D O # Two tangents from a common exterior point are congruent – in this case, DL ≅ DY Any radius drawn to a point of tangency is perpendicular to the tangent – in this case, OL ⊥ DL and OY ⊥ DY Y H If a tangent and a secant are drawn from a common exterior point, the product of the secant’s length and the length of its external part equals the square of the length of the tangent: in this case, UE × SE = (HE) 2 [Secant-Tangent Power Theorem] # If two secants are drawn from a common point, the product of the first secant’s length and the length of its external part equals the product of the second secant’s length and the length of its external part: in this case, SC × SU = SN × SH [Secant-Secant Power Theorem] E S O # U C U S O N H 31 GEOMETRY RESOURCE DEMIDEC RESOURCES © 2001 Example: Find the measure of arc PI given that m∠P = 20°.

RF × RP = RL2 (x + 3) × 3 = 5 2 3x + 9 = 25 3x = 16 16 x= 3 5 R L 3 P y O x F D 5 4 B After we have x, we can find y using the Secant-Secant Power Theorem and exterior point B. We using the Secant-Secant Power Theorem. BP × BF = BL × BD (5 + 16 ) × 5 = (4 + y) × 4 3 ( 15 ) × 5 = 16 + 4y + 16 3 3 31 × 5 = 16 + 4y 3 155 = 48 + 4y 3 3 107 4y = 3 107 =y 12 33 GEOMETRY RESOURCE DEMIDEC RESOURCES © 2001 ABOUT THE AUTHOR, PART II …ged in a broom closet at Nimitz High School. His most noticeable accomplishments include primarily the face found in the picture here, along with passing freshman physics at Caltech.

OT2 + TM2 = OM2 (r – 4)2 + 82 = r2 r2 – 8r + 16 + 64 = r2 -8r = -80 r = 10 b) To use the Chord-Chord Power Theorem, we must draw radius OZ to create diameter ZI . ZO = r and OT = r – 4 so we know that ZT = 2r – 4. By the Chord-Chord Power Theorem, we can calculate the radius. ZT × TI = NT × TM (2r – 4) × 4 = 8 × 8 8r – 16 = 64 8r = 80 r = 10 Example: Find x and y in the given diagram. Solution: The problem requires us to employ the Secant-Secant Power Theorem and the Secant-Tangent Power Theorem in the same problem.