Bariz - Problem

We'll show that $PQCA$ is an isosceles trapezoid which implies the cyclicity. First let $H = (AOE) \cup (COD)$ we prove $H\in (ABC)$. $$\angle AHC=\angle AHO+ \angle OHC=\angle AEF+\angle ODC $$$$=\angle FEB+\angle BDG=180^{\circ}-\angle ABC.$$ Now we're ready to prove $PQ \parallel AB$; $OH$ is the radical axis of $(AOE)$ and $(COD)$ Therefore $OH\perp PQ$ so it suffices to prove $AC\perp OH$. Let $I$ be the midpoint of $AC$ we prove $AI=IC$ or since $OA=OC$ it's enough to prove $\angle IOC=\angle IOA$. $$\angle AOI=\angle AOH=\angle AEH=180^{\circ}-\angle AEB$$$$=180^{\circ}-\angle BDC=\angle HDC=\angle IOC$$ What's left now is $AP=CQ$ or equivalently $OP=OQ$ or (since $OI\perp PQ$) $\angle IOP=\angle IOQ$ but $$\angle IOP=\frac{1}{2}(180^{\circ}-\angle OPH)=90^{\circ}-\angle OAH=\angle FEH-90^{\circ}=90^{\circ}- \angle FEB=\angle ABE$$Similarly $\angle IOQ =\angle EBC$ And the problem says they're equal. Thank you for posting this good problem.