Problem

Source: ELMO Shortlist 2012, G5; also ELMO #5

Tags: geometry, parallelogram, trigonometry, conics, ellipse, geometric transformation, reflection



Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. Calvin Deng.