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Problem 1095

AIME late

Geometry Difficulty 5.0 Prove it Saudi Arabian Mathematical Competitions · Saudi Arabia · 2015

Let $ABC$ be a triangle and $D$ a point on the side $BC$ . Point $E$ is the symmetric of $D$ with respect to $AB$ . Point $F$ is the symmetric of $E$ with respect to $AC$ . Point $P$ is the intersection of line $DF$ with line $AC$ . Prove that the quadrilateral $AEDP$ is cyclic.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $\alpha = \angle BAC$ and $\theta = \angle BAD$ . Because $E$ is the symmetric of $D$ with respect to $AB$ , we have $AD = AE$ and $DE$ is perpendicular to $AB$ . We deduce that $\angle EAD = 2\theta$ and $\angle DEA = 90^\circ - \theta$ .

Because $F$ is the symmetric of $E$ with respect to $AC$ , we have $AE = AF$ and $EF$ is perpendicular to $AC$ . We deduce that
$\angle CAF = \angle EAC = \angle EAB + BAC = \theta + \alpha$
and
$\begin{aligned} \angle PDA & = 90^\circ - \frac{1}{2} \angle DAF = 90^\circ - \frac{1}{2}(\angle DAC + CAF) \\ & = 90^\circ - \frac{1}{2}((\alpha - \theta) + (\alpha + \theta)) = 90^\circ - \alpha. \end{aligned}$
We deduce that
$\angle DPC = \angle PDA + \angle DAC = (90^\circ - \alpha) + (\alpha - \theta) = 90^\circ - \theta = \angle DEA.$
This proves that quadrilateral $AEDP$ is cyclic.

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