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Romanian Mathematical Olympiad

Geometry Difficulty 4.8 AIME Prove it Romania

Let $ABC$ be a right triangle with legs $AB$ and $AC$ . The bisector of the angle $ACB$ intersects $AB$ in $D$ and the perpendicular in $B$ on $BC$ in $E$ . Denote $F$ the reflection of $E$ across $B$ and $P$ the intersection of the lines $DF$ and $BC$ . Prove that $EP \perp CF$ .
Cătălin Cristea

Solution

In $\triangle BEC$ , $m(\angle CEB) = 180^\circ - m(\angle EBC) - m(\angle ECB) = 90^\circ - m(\angle ECB)$ . In $\triangle ADC$ , $m(\angle ADC) = 180^\circ - m(\angle DAC) - m(\angle ACD) = 90^\circ - m(\angle ACD)$ . Since $m(\angle ACD) = m(\angle ECB)$ , it follows $\angle ADC \equiv \angle CEB$ .
Now $\angle ADC \equiv \angle EDB$ . This yields $[BD] = [BE]$ . Since $[BE] = [BF]$ , we infer that $[BD] = [BE] = [BF]$ , so $\triangle DEF$ has the right angle $D$ .
In $\triangle CEF$ , $CB$ and $FD$ are altitudes, hence $P$ is the orthocenter. In conclusion, $EP$ is an altitude, that is $EP \perp CF$ .

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