show that PA+PB=PC

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Source: Singapore TST round 1

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#1 h Apr 12, 2014, 12:52 AM • 1 Y

Y by Adventure10

Given triangle ABC, with E,F on AC,AB respectively such that BE,CF bisect angle B and C respectively. P,Q are on the minor arc AC of the circumcircle of triangle ABC such that AC//PQ and BQ//EF. Show that PA+PB=PC

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#2 h Apr 12, 2014, 3:37 AM • 2 Y

Y by Adventure10, Mango247

A fun proof:

Let $I$ be the incenter of $ABC$ and $CF\cap(ABC)=G$ .

Lemma: E lies on the radical axis of $(AGF),(FBI)$

Proof: Let $(AGF)\cap AC=K$ . Since $\angle KFI=\angle CAG=\angle CFB$ and $CG$ bisects $\angle KCB$ . Therefore by $ASA$ congruence $B,K$ are reflexive over $CF$ $\Rightarrow \angle CKI=\angle IBC=\angle ABI\Rightarrow A,K,I,B$ are concyclic and hence $I$ lies on the radical axis of $(AGF),(FBI)$ .

Main problem: Reflect $B$ over the perpendicular bisector of $AC$ which lands on $(ABC)$ , denote it $B'$ . Hence $PB+PA=PC\iff QB'+QC=QA$ . Now let $L$ be a point of segment $AQ$ such that $AL=QB'$ Through $L$ construct a line parallel to $BQ$ which intersects $AC$ at $B''$ . Since $\angle ALB''=\angle AQB=\angle QCB'$ and $\angle QAC=\angle QB'C$ , therefore by $ASA$ $\triangle ALB'',\triangle B'QC$ are congruent $\Rightarrow AB''=CB'=AB,CQ=LB''$ , hence we now just need to prove $QL=AQ-AL=QA-BQ'=QC\iff QB''$ bisects $\angle AQB\iff Q,B'',F$ are collinear.

Let $QG\cap FE=J$ , since $\angle GQB=\angle GJF=\angle GAF$ , therefore $A,G,F,J$ are concyclic. By our lemma we know that $B,I,J,F$ are concyclic. Hence $\angle FIB=\angle FIB=90-\angle A/2=\angle AB''B\Rightarrow J,E,B'',B$ are concyclic. Since $\angle EJB''=\angle EBB''=90-\angle A/2-\angle B/2=\angle C/2=\angle GJF$ , therefore $G,J,B'',Q$ are collinear and we are done.

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#3 h Apr 12, 2014, 5:44 AM • 1 Y

Y by Adventure10

There is more to say about this configuration; P is actually one of the Feuerbach points of the antimedial triangle of ABC. Thus, it follows that either PC=PA+PB or PB=PA+PC, according to whether P is on arc AC or arc AB. This is an extraversion of the incircle case point [Feuerbach point of a triangle; FY + FZ = FX].

Let $\triangle XYZ$ be the antimedial triangle of $\triangle ABC$ (X,Y,Z againts A,B,C) and its X-excircle $(J)$ touches its 9-point circle $(O) \equiv \odot(ABC)$ at its Feuerbach point $F_X.$ Parallel from $F_X$ to $AC$ cuts $(O)$ again at $D$ $\Longrightarrow$ $BD, BF_X$ are isogonals WRT $\triangle ABC$ $\Longrightarrow$ $BD$ has the direction of the isogonal conjugate of $F_X$ WRT $\triangle ABC.$

It's known that $EF$ is perpendicular to the line connecting $O$ and the A-excenter of $\triangle ABC,$ hence $EF$ is also parallel to the line connecting the circumcenter $K$ of $\triangle XYZ$ with $J.$ It's also known that $JK$ is the Steiner line of $F_X$ WRT $\triangle ABC$ (this is again an extraversion of the incircle case), thus $\perp JK$ is direction of the isogonal conjugate of $F_X$ WRT $\triangle ABC$ $\Longrightarrow$ $BD \perp JK$ $\Longrightarrow$ $BD \parallel EF$ $\Longrightarrow$ $D \equiv Q,$ $F_X \equiv P,$ as desired.

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