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DE = AC + BC , related to collinear points, and 2 intersecting circumcircles

Source: 49th Austrian Mathematical Olympiad National Competition (Final Round, part 2) 1st June 2018 p4

May 25, 2019

geometrycircumcirclecollinear

Problem Statement

Let

ABC

be a triangle and

P

a point inside the triangle such that the centers

M_B

and

M_A

of the circumcircles

k_B

and

k_A

of triangles

ACP

and

BCP

, respectively, lie outside the triangle

ABC

. In addition, we assume that the three points

A, P

and

M_A

are collinear as well as the three points

B, P

and

M_B

. The line through

P

parallel to side

AB

intersects circles

k_A

and

k_B

in points

D

and

E

, respectively, where

D, E \ne P

. Show that

DE = AC + BC

.
(Proposed by Walther Janous)

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