MathDB

2

Part of 2020 Mexico National Olympiad

Problems(1)

<PDQ=90° in an incenter configuration

Source: Mexico National Olympiad 2020 P2

11/11/2020
Let ABCABC be a triangle with incenter II. The line BIBI meets ACAC at DD. Let PP be a point on CICI such that DI=DPDI=DP (P≠IP\ne I), EE the second intersection point of segment BCBC with the circumcircle of ABDABD and QQ the second intersection point of line EPEP with the circumcircle of AECAEC. Prove that ∠PDQ=90∘\angle PDQ=90^\circ.
Proposed by Ariel García
geometryincentercircumcircle