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Centroamerican Math Olympiad 2017 - P4

Source: OMCC 2017

June 23, 2017
OMCCOMCC 2017CENTROgeometrygeometric transformationreflectionangle bisector

Problem Statement

ABCABC is a right-angled triangle, with ∠ABC=90∘\angle ABC = 90^{\circ}. B′B' is the reflection of BB over ACAC. MM is the midpoint of ACAC. We choose DD on BM→\overrightarrow{BM}, such that BD=ACBD = AC. Prove that B′CB'C is the angle bisector of ∠MB′D\angle MB'D.
NOTE: An important condition not mentioned in the original problem is AB<BCAB<BC. Otherwise, ∠MB′D\angle MB'D is not defined or B′CB'C is the external bisector.
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