WLOG $AB<AC$. Let $X$ be the center of the Euler circle of $ABC$ and $I$ the center of the incircle of $ABC$. We want to show that $XP=XD$. First, we have $\widehat{XPD}=90-(AI,XP)=90-\widehat{EAI}=90-(\widehat{BAI}-\widehat{BAE})=180-\frac{\widehat{BAC}}2-\widehat{ABC}=90-\frac12(\widehat{ACB}-\widehat{ABC})$ since $(XP)||(AE)$. Then, since the homothety of center $H$ the orthocenter of $ABC$ and of ratio $2$ sends $X$ onto $O$, the circumcircle onto the Euler circle of $ABC$, $E$ onto $F=AE\cap (ABC)$ and $D$ onto $A'$ the symmetrical of $A$ with respect to $O$, (this all arises from the proof that the Euler circle does pass through the nine points), we get that $\widehat{EXD}=\widehat{FOA'}=2\widehat{HAO}=2(\widehat{HAC}-\widehat{OAC})=2(\widehat{ABC}-\widehat{ACB})$. Now since $XP$ is the perpendicular bissector of $DE$, $\widehat{PXD}=\frac12\widehat{EXD}=\widehat{ABC}-\widehat{ACB}$. Combining the two results, we get $\widehat{XDP}=90-\frac12(\widehat{ACB}-\widehat{ABC})$, so $XPD$ is indeed isosceles, and we are done.

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