Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 806.

## Wednesday, September 26, 2012

### Problem 806: Triangle, Points on sides or Extension, Circle, Circumcenters, Similar triangles

Labels:
circumcenter,
similarity,
triangle

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Let the intersection of the three circles be P.

ReplyDeleteSince

A'B' bisects arc DP, A'C' bisects arc PF,

thus

∠B'A'P = 1/2 ∠DA'P

∠C'A'P = 1/2 ∠FA'P

So

∠B'A'C' = 1/2 ∠DA'F = ∠BAC

Similarly,

∠B'C'A' = ∠BCA

∠A'B'C' = ∠ABC

Hence,

∆ABC ~ ∆A'B'C'

To Jacob

Deleteyour statement "Let the intersection of the three circles be P" may need some explanation.

Let P be the intersection of circle A' and circle C'.

DeleteA,F,P,D concyclic => ∠DPF = 180 - ∠A

C,F,P,E concyclic => ∠EPF = 180 - ∠C

∠DPE

= 360 - (180 - ∠A) - (180 - ∠C)

= ∠A + ∠C

= 180 - ∠B

=> B,D,P,E concyclic

=> Circle C' passes through P

=> Three circles concurrent at P

Special thanks to W Fung for the proof.