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Geometry Difficulty 6.3 National Olympiad Prove it Taiwan

Let $\Omega$ be the circumcircle of an isocles trapezoid $ABCD$ , in which $AD$ is parallel to $BC$ . Let $X$ be the reflection point of $D$ with respect to $BC$ . Point $Q$ is on the arc $BC$ of $\Omega$ that does not contain $A$ . Let $P$ be the intersection of $DQ$ and $BC$ . A point $E$ satisfies that $EQ$ is parallel to $PX$ , and $EQ$ bisects $\angle BEC$ . Prove that $EQ$ also bisects $\angle AEP$ .

Solution

Let $R$ be the intersection of $AE$ and $BC$ . To prove the original statement, it suffices to prove that $\odot(EBC)$ and $\odot(ERP)$ are tangent at $E$ (because this tells us that $EP$ , $EA$ are antiparallel with respect to $EB$ , $EC$ , and the latter is preserved with respect to $EQ$ , $EQ$ antiparallel).

Let $AE$ intersect $\Omega$ , $\odot(EBC)$ at $S$ , $T$ respectively, then by Reim's theorem, $P$ , $Q$ , $R$ , $S$ are concyclic. Since $EQ$ bisects $\angle BEC$ , the intersection point $M$ of $EQ$ and $\odot(EBC)$ is the midpoint of arc $BC$ on $\odot(EBC)$ . From
$\angle ETM = \angle (EM, BC) = \angle (PX, BC) = \angle (CB, DP) = \angle ADP = \angle ASQ,$
we obtain $TM \parallel SQ$ , hence $\odot(ETM)$ and $\odot(ESQ)$ are tangent at $E$ .

Note that the radical center of the three circles $\Omega$ , $\odot(EBC)$ , $\odot(ESQ)$ is the intersection point $U$ of $BC$ and $SQ$ , hence from the tangency of $UE$ to $\odot(ESQ)$ we know $\overline{UE}^2 = US \cdot UQ = UR \cdot UP$ , that is, $UE$ is tangent to $\odot(ERP)$ . Combining with $UE$ tangent to $\odot(EBC)$ , we obtain that $\odot(EBC)$ and $\odot(ERP)$ are tangent. $\square$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from en; metadata (topic, difficulty) added by this project.