Hopefully the diagram below explains everything. Here, the red lines are perpendicular to $DP$ at $D$, $EP$ at $E$, $BP$ at $P$, and $CP$ at $C$, respectively. It suffices to prove that there exists a homothethy centered at $A$ that sends the quadrilateral $IDPE$ to $PBHC$. Indeed, if this holds, then $A$, $I$, $P$, and $H$ are collinear; but the circumcenters of $EPD$ and $BPC$ are the midpoints of $IP$ and $PH$, respectively. So this would imply the desired statement. Of course, consider the homothethy $\phi$ centered at $A$ that sends $D$ to $B$. Since $DE$ is parallel to $BC$, then $\phi(E) = C$. Next, observe that $ID$ and $BP$ are both perpendicular to $DP$, so they must be parallel. Similarly, $IE$ is parallel to $PC$. Thus, the triangles $IDE$ and $PBC$ are similar, with the same orientation; this forces $\phi(I) = P$. Finally, apply the same argument to get that the triangles $PDE$ and $HBC$ are similar with the same orientation (and thus $\phi(P) = H$). We are done.

Let $Q$ and $R$ be the circumcenter of $\triangle PDE$ and $\triangle PBC$ respectfully. Through some angle chasing, you will see that $\triangle ADQ \sim \triangle ABR$, thus $A, Q, R$ collinear. With some angle chasing, you will also see that $\angle RPC + \angle EPQ = 90^{\circ}$ thus $P, Q, R$ are collinear. So we are done.

This can also be done through homothety First, Let $O_1$ and $O_2$ be the circumcenter of triangle $EPD$ and triangle $BPC$ After some angle chasing, we'll get that $\angle O_1PO_2 = 180^{\circ}$, thus $O_1,P,O_2$ are collinear Since $DE // BC$, we can see the homothety between the circumcircle of triangle $EPD$ and $BPC$, especially since $\angle EPD+\angle BPC=180^{\circ}$, which means that there exist a point $P'$ in the circumcircle of triangle $BPC$ such that $A,P,P'$ are collinear and $O_1P // O_2P'$. Since $O_1,P,O_2$ are collinear, that means $A,O_1,P,O_2$ are all collinear, thus proving the statement

DavyDuf wrote: Let $Q$ and $R$ be the circumcenter of $\triangle PDE$ and $\triangle PBC$ respectfully. Through some angle chasing, you will see that $\triangle ADQ \sim \triangle ABR$, thus $A, Q, R$ collinear. With some angle chasing, you will also see that $\angle RPC + \angle EPQ = 90^{\circ}$ thus $P, Q, R$ are collinear. So we are done. Interesting idea. Could you please elaborate how to conclude $\triangle ADQ \sim \triangle ABR$?

removablesingularity wrote: Interesting idea. Could you please elaborate how to conclude $\triangle ADQ \sim \triangle ABR$? This one is for the case $\angle BPC > 90^{\circ}$, the other case can be solved similarly Through angle chasing $\triangle QDE \sim \triangle RBC$

Also, since $DE // BC$ then $\triangle ADE \sim \triangle ABC$ From there we can conclude $\triangle ADQ \sim \triangle ABR$

DavyDuf wrote: removablesingularity wrote: Interesting idea. Could you please elaborate how to conclude $\triangle ADQ \sim \triangle ABR$? This one is for the case $\angle BPC > 90^{\circ}$, the other case can be solved similarly Through angle chasing $\triangle QDE \sim \triangle RBC$

Also, since $DE // BC$ then $\triangle ADE \sim \triangle ABC$ From there we can conclude $\triangle ADQ \sim \triangle ABR$

And to get $\angle RPC+\angle EPQ=90^\circ$?

removablesingularity wrote: And to get $\angle RPC+\angle EPQ=90^\circ$? I'm sure there's a cleaner way to do it, but here's the elementary one Again, this one is for the case $\angle BPC > 90^{\circ}$, the other case can be solved similarly Let $\angle RBC = \alpha \ ; \ \angle BCP = \beta \ ; \ \angle ACB = \angle AED = \gamma$ \begin{align*} \angle BRP &= 2 \beta \\ \angle CRP &= 180 - 2\alpha - 2\beta \\ \angle RPC &= 180 - (180 - 2\alpha - 2\beta + \alpha + \beta) \\ &= \alpha + \beta \\ \\ \angle PCE &= \gamma - \beta \\ \angle CEP &= 90 + \beta - \gamma \\ \angle EPQ &= 180 - (90 + \beta - \gamma + \gamma + \alpha) \\ &= 90 - \alpha - \beta \\ \\ \therefore \ \angle RPC + \angle EPQ &= 90^{\circ} \end{align*}