Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 661.
Friday, August 26, 2011
Problem 661: Intersecting Circles, Centers, Secant, Parallelogram
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∠ADB = (1/2) ∠AO’B (in circle O')
ReplyDelete= (1/2)[(1/2)∠AOB] (in circle O)
= (1/2) ∠AOC (in circle O)
= ∠AEC ( ,, )
= ∠CFB ( ,, )
(AO'O and BO'O are congruent isosceles ∆les,
∠AO'O = ∠BO'O, arc AC = arc BC)
So DE∥FC, DF∥AC and DECF is a parallelogram