PQ//BC wanted, FB = BD, DC = CE

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PQ//BC wanted, FB = BD, DC = CE

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Source: 2022 Czech-Polish-Slovak Match Junior, team p3 CPSJ

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#1 h Sep 3, 2022, 11:23 AM

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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.

This post has been edited 1 time. Last edited by parmenides51, May 10, 2024, 5:29 PM

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#2 h Dec 14, 2023, 5:20 PM

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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 $PQ\parallel BC$ . $\blacksquare$

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#3 h Dec 14, 2023, 7:12 PM

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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.

This post has been edited 1 time. Last edited by bluelinfish, Dec 14, 2023, 7:13 PM

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#4 h Dec 15, 2023, 7:08 PM

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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,
$\begin{align*} \angle FQI = \angle EQI = 90 - \angle QEF = 90 - \angle AIF = \angle FPI \end{align*}$

This post has been edited 4 times. Last edited by Shreyasharma, Dec 15, 2023, 7:12 PM

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