In $\triangle ABC$, points $P,Q$ satisfy $\angle PBC = \angle QBA$ and $\angle PCB = \angle QCA$, $D$ is a point on $BC$ such that $\angle PDB=\angle QDC$. Let $X,Y$ be the reflections of $A$ with respect to lines $BP$ and $CP$, respectively. Prove that $DX=DY$.
Problem
Source: : 2023 China Western Mathematical Olympiad Grade p7 CWMI
Tags: geometry, equal segments, equal angles
D_S
06.04.2024 09:08
your drawing does not match the question I think
eulerleonhardfan
15.06.2024 17:43
Clearly, $P, Q$ are isogonal conjugates in $\triangle ABC$. Let $Q'$ be the reflection of $Q$ over $BC$, and $D'$ be the reflection of $D$ over $PC$. Note that $P, D, Q'$ are collinear. Since $\angle Q'BC = \angle ABP$ and $\angle Q'CB = \angle ACP$, $A, Q'$ are isogonal conjugates in $\triangle BPC$. This means that $180-\angle APB = \angle Q'PC$. So $\angle APX = 2(180-\angle APB) = 2 \times \angle Q'PC$. Because $\angle YPD = \angle APD' = \angle APX + \angle XPD' = \angle Q'PC + \angle XPD' = \angle XPD$, and with the fact that $YP = AP = XP$, it follows that $\triangle YPD \equiv \triangle XPD$, therefore $XD = YD$.