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
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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.