Problem: A B A B A D A D B C B C A C A C B D B D A D = a A D = a B C = b , a ≠ b B C = b , a = b
Answer Choices:
A. ∣ a − b ∣ ∣ a − b ∣
B. 1 2 ( a + b ) 2 1 ( a + b )
C. a b a b
D. a b a + b a + b a b
E. 1 2 a b a + b 2 1 a + b a b Solution:
From similar triangles x y = a d y x = d a x d − y = b d d − y x = d b
∴ x 2 y ( d − y ) = a b d 2 . ∴ y ( d − y ) x 2 = d 2 a b . x 2 = y ( d − y ) . ∴ 1 = a b d 2 ∴ d = a b x 2 = y ( d − y ) . ∴ 1 = d 2 a b ∴ d = a b
OR
Let the degree measure of arc AP be m m
Then ∠ D = 90 − m 2 ∠ D = 9 0 − 2 m ∠ C = m 2 ∠ C = 2 m
∴ d b = tan m 2 ∴ b d = tan 2 m d a = tan ( 90 − m 2 ) = 1 tan m 2 a d = tan ( 9 0 − 2 m ) = tan 2 m 1
∴ 2 m 2 = a b ∴ d 2 b 2 = a b , d 2 = a b , d = a b ∴ tan 2 2 m = b a ∴ b 2 d 2 = b a , d 2 = a b , d = a b
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