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2020 EGMO P5: P is the incentre of CDE

Source: 2020 EGMO P5

April 18, 2020

geometryincenterEGMO 2020EGMO

Problem Statement

Consider the triangle

ABC

with

\angle BCA > 90^{\circ}

. The circumcircle

\Gamma

of

ABC

has radius

R

. There is a point

P

in the interior of the line segment

AB

such that

PB = PC

and the length of

PA

is

R

. The perpendicular bisector of

PB

intersects

\Gamma

at the points

D

and

E

.
Prove

P

is the incentre of triangle

CDE

.

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