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Point P on incircle with <APE = <DPB

Source: IOM 2018 #6, Du&scaron;an Djukić

September 6, 2018

geometryincircle

Problem Statement

The incircle of a triangle

ABC

touches the sides

BC

and

AC

at points

D

and

E

, respectively. Suppose

P

is the point on the shorter arc

DE

of the incircle such that

\angle APE = \angle DPB

. The segments

AP

and

BP

meet the segment

DE

at points

K

and

L

, respectively. Prove that

2KL = DE

.
Dušan Djukić

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