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

Labels:
center,
intersecting circles,
parallelogram,
secant

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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