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Iranian Mathematical Olympiad

Geometry Difficulty 5.7 AIME, harder Prove it Iran

In isosceles trapezoid $ABCD$ where $BC = AD$ and $AB \parallel CD$ , point $P$ is the intersection of diagonals $AC$ and $BD$ . Let $X$ be the second intersection point of $BC$ and circumcircle of triangle $APB$ and let $Y$ be a point on $AX$ such that $DY \parallel BC$ ( $C$ and $Y$ are on different sides of line $AD$ ). Prove that $\angle YDA = 2 \times \angle YCA$ .

Solution

Since $YD \parallel BC$ , it's concluded that $\angle YDP = \angle CBP = \angle XAP$ . So $YAPD$ and $ABCD$ are cyclic quadrilaterals. Therefore
$\angle DYP = \angle DAP = \angle DAC = \angle DBC = \angle PDY.$
Thus $PY = PD$ . On the other hand it's clear that $PD = PC$ . From these two it's obtained that $PY = PC$ and finally,
$2\angle YCA = 2\angle YCP = \angle APY = \angle YDA,$
hence the claim of the problem.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.