IRL_ABooklet_2023

First note that $\angle BAC=\angle BDC$ (inscribed angles) and $\angle BAC=\angle ACD$ (alternate angles at $AB\parallel CD$). Hence, $\angle BDC=\angle ACD$ and triangle $CDP$ is isosceles with $|CP|=|DP|$. This implies that $P$ is on the perpendicular bisector of $CD$ and that $\angle PBA=\angle BAP=45^{\circ }$ (because $AC\perp BD$). The centre $O$ of the circle is on the perpendicular bisectors of the chords $AB$ and $CD$. As $AB$ and $CD$ are parallel, their perpendicular bisectors are parallel as well. As they have $O$ in common, they coincide. This means that $OP$ is the common perpendicular bisector of $AB$ and of $CD$.

Because $\angle BEP=\angle CPD=90^{\circ }$ and $\angle PDC=\angle PBA=45^{\circ }$, triangles BEP and CPD are right angled and isosceles. In particular, $|BE|=|EP|$. Moreover, $\angle BOC=2\angle BDC=90^{\circ }$ (central angle), hence $$\angle BOE=180^{\circ }-\angle BOC-\angle COF=90^{\circ }-\angle COF=\angle OCF.$$ Since $|OB|=|OC|$, we can now use ASA to see that $△OCF$ is congruent to $△BOE$, thus $|OF|=|BE|=|EP|$.