Chords $AB$ and $CD$ of a circle $\omega$ meet at point $E$ in such a way that $AD = AE = EB$. Let $F$ be a point of segment $CE$ such that $ED = CF$. The bisector of angle $AFC$ meets an arc $DAC$ at point $P$. Prove that $A$, $E$, $F$, and $P$ are concyclic.
Problem
Source: Sharygin Finals 2022 9.5
Tags: geometry
02.08.2022 07:54
First, since $\angle CBE = \angle CDA = \angle DEA = \angle CEB$, we have that $CE = CB$. Redefine $P$ as the point so $ABCP$ is an isosceles trapezoid. Also, $\angle PAD = \angle DAE + \angle PAE = 180 - 2 \angle ABC + \angle ABC = 180 - \angle ADC$ so $APCD$ is also an isosceles trapezoid. Redefine $F$ as $(APE) \cap CE$, so it suffices to show that $CF = DE$. But we have $\angle PFC = \angle PAB = \angle ADE$ so $AD || PF$ and we have $AE || CP$ and $AE = AD = PC$ due to isosceles trapezoids so triangles $ADE$ and $PFC$ are congruent, giving $CF = DE$. $\blacksquare$
02.08.2022 08:26
Cute, indeed. Here is my solution, which has a bit different finish. Redefine $P$ such that $ADCP$ is an isosceles trapezoid, and due to the equal segments, $AEFP$ also will be isosceles trapezoid. Hence we have to prove that $PF$ is bisector of $\angle AFC$. We have that $\angle PFC = \angle ADE= \angle AED = \varphi$ and also $ADFP$ is parallelogram (congruent triangles $ADE$ and $PFC$). Hence we want $\angle AFP = \angle FAD= \varphi$, so it is sufficient to show that $\triangle AED$ and $\triangle AFD$ are similar, or equivalently, $DA^2=DE.DF$. The last one rewrites as $EA.EB=ED.EC$, which is true due to PoP.