, 2025

As shown in the figure, draw a line through $D$ parallel to $AB$, intersecting $BF$ at $K$. Then:
$$\angle DKB = 180^\circ - \angle ABF = 180^\circ - \angle DEC = \angle DEB,$$
so points $B, K, E, D$ are concyclic.
Since $AB \perp AD$ and $DK \parallel AB$, we have $AD = BK \cdot \sin \angle ABK$. Let $\omega$ be the circumcircle of triangle $CDE$ with radius $r$, then $DC = 2r \cdot \sin \angle DEC$. From $AD = DC$ and $\angle ABK = \angle DEC$, we get $BK = 2r$.
Let $FO$ intersect circle $\omega$ at two points $U$ and $V$, with $F, U, O, V$ in order. By the power of a point theorem:
$$FU \cdot FV = FE \cdot FD = FK \cdot FB.$$
Since $UV = 2r = BK$ and $FB > FK$, we conclude $FV = FB$. Given $FP = FB$ and both $P, V$ lie on the extension of $FO$, it follows that $P$ coincides with $V$.
From $AD = DC$, the sum of directed angles from $\overrightarrow{AC}$ to $\overrightarrow{AD}$ and $\overrightarrow{DC}$ is $0^\circ$. From $FP = FB$, the sum of angles from $\overrightarrow{PB}$ to $\overrightarrow{PF}$ and $\overrightarrow{FB}$ is $0^\circ$ (mod $360^\circ$). This shows

that as directed angles (mod $180^\circ$):
$$\angle AQB = \angle (AC, PB) = \frac{1}{2}\angle (AD, FB) + \frac{1}{2}\angle (DC, PF).$$
Since $\angle (AD, FB) = 90^\circ - \angle FBA = 90^\circ - \angle CED$, and connecting $PE$ gives:
$$\angle (DC, PF) = \angle CDP + \angle DPF = \angle CEP + \angle DPF,$$
we have:
$$\begin{aligned} \angle AQB &= \frac{1}{2}(90^\circ - \angle CED + \angle CEP + \angle DPF) \\ &= \frac{1}{2}(90^\circ - \angle PED + \angle DPF) \\ &= \frac{1}{2}(\angle DPO + \angle DPF) = \angle DPF. \end{aligned}$$
This completes the proof. $\Box$