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Iberoamerican Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Ibero-American Mathematical Olympiad

Problem:

$ABC$ is a triangle. $BD$ is an angle bisector. $E$ , $F$ are the feet of the perpendiculars from $A$ , $C$ respectively to the line $BD$ . $M$ is the foot of the perpendicular from $D$ to the line $BC$ . Show that $\angle DME = \angle DMF$ .

Solution

Solution:

Let $H$ be the foot of the perpendicular from $D$ to $AB$ . $\angle AHD = \angle AED = 90^\circ$ , so $AHED$ is cyclic. Hence $\angle DAE = \angle DHE$ . But $M$ is the reflection of $H$ in the line $BD$ , so $\angle DME = \angle DAE$ .

$AE$ is parallel to $CD$ , so $\angle DAE = \angle DCF$ . $\angle DFC = \angle DMC$ , so $DMCF$ is cyclic. Hence $\angle DCF = \angle DMF$ . Hence $\angle DME = \angle DMF$ .

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