2025 International Mathematical Olympiad China National Team Selection Test

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: $$\begin{array}{rl}\angle AQB& =\frac{1}{2}(90^{\circ }-\angle CED+\angle CEP+\angle DPF)\\ & =\frac{1}{2}(90^{\circ }-\angle PED+\angle DPF)\\ & =\frac{1}{2}(\angle DPO+\angle DPF)=\angle DPF.\end{array}$$ This completes the proof. $\square$