The points $D, E, F$ lie respectively on the sides $BC$, $CA$, $AB$ of the triangle ABC such that $F B = BD$, $DC = CE$, and the lines $EF$ and $BC$ are parallel. Tangent to the circumscribed circle of triangle $DEF$ at point $F$ intersects line $AD$ at point $P$. Perpendicular bisector of segment $EF$ intersects the segment $AC$ at $Q$. Prove that the lines $P Q$ and $BC$ are parallel.
Problem
Source: 2022 Czech-Polish-Slovak Match Junior, team p3 CPSJ
Tags: geometry, parallel, equal segments
14.12.2023 20:20
Lemma:$I$ is the circumcenter of $(DEF)$. Proof:Let $O$ be the center of $(DEF)$. Then, $BO\perp DF,CO\perp DE$ so $\angle FBO=\angle OBF$ and $\angle CDO=\angle OCE$. Lemma:$A,I,D$ are collinear. Proof:Let $DI$ and $AB$ intersect at $A'$. $\angle IAF+\frac{\angle C}{2}=\frac{\angle A+\angle C}{2}=\angle DA'F+\angle FDA'=\angle DA'F+\angle FDI=\angle IA'F+\frac{\angle C}{2}$ so $A'=A$. $PF$ is tangent to $(DEF)$ so $\angle PFI=90$. Lemma:$P,Q,I,F$ are cyclic. Proof:$\angle FIA=\angle C\implies \angle IPF=90-\angle C$ $QF=QE\implies \angle QFE=\angle C \implies \angle IQF=90-\angle C$ $\angle IQF=90-\angle C=\angle IPF\implies F,I,P,Q$ are cyclic. We get $\angle IQP=90\implies PQ\perp IQ\perp EF$ so $PQ\parallel BC$.$\blacksquare$
14.12.2023 22:12
Looks like I can only do easy geo nowadays. Let $I$ be the incenter of $\triangle ABC$. Since the perpendicular bisectors of $DF$ and $DE$ intersect at $I$, $I$ is the circumcenter of $\triangle DEF$. In addition, $D$ lies on $AI$ because \[ \frac{BD}{DC} = \frac{BF}{CE} = \frac{AB}{AC}. \] Then \[ \angle IPF = 90^{\circ} - \angle AIF = 90^{\circ} - (180^{\circ} - \angle FID) = 90^{\circ} - (180^{\circ} - 2\angle FED) = 90^{\circ} - \angle C\]and $\angle IQF = 90^{\circ} - \angle QFE = 90^{\circ} - \angle C$, so $IPQF$ is cyclic. Since $\angle IQP = 180^{\circ} - \angle IFP = 90^{\circ}$ and $QI \perp BC$ we are done.
15.12.2023 22:08
Note that by ratios $AD$ is the angle bisector of $\angle A$. Also noting that the perpendicular bisectors of $\overline{DE}$ and $\overline{DF}$ are the angle bisectors of $\angle C$ and $\angle B$, we conclude $I$ is the circumcenter of $\triangle DEF$. Now let $K$ be the incenter of $\triangle AEF$. Clearly $K \in (DEF)$ by the incenter-excenter lemma. Now we can rephrase the problem. Rephrased problem wrote: Let $\triangle AEF$ have incenter $K$ and $A$-excenter $I$. Let the tangents from $F$ to $(KEF)$ meet $\overline{AK}$ at $P$. Let the perpendicular bisector of $\overline{EF}$ meet $\overline{AE}$ at $Q$. Show that $\overline{PQ} \parallel \overline{EF}$. Note in this equivalent phrasing it suffices to show $Q \in (PFI)$. However this is obvious as, \begin{align*} \angle FQI = \angle EQI = 90 - \angle QEF = 90 - \angle AIF = \angle FPI \end{align*}