MathDB
Orthocenter and Incenter

Source: Bulgarian IMO TST 2005, Day 2, Problem 2

July 7, 2013
geometryincentercircumcircletrigonometrytrapezoidgeometric transformationreflection

Problem Statement

Let ABCABC, ACBCAC \not= BC, be an acute triangle with orthocenter HH and incenter II. The lines CHCH and CICI meet the circumcircle of ABC\bigtriangleup ABC at points DD and LL, respectively. Prove that CIH=90\angle CIH = 90^{\circ} if and only if IDL=90\angle IDL = 90^{\circ}