Let be $ABC$ an acute-angled triangle. Consider point $P$ lying on the opposite ray to the ray $BC$ such that $|AB|=|BP|$. Similarly, consider point $Q$ on the opposite ray to the ray $CB$ such that $|AC|=|CQ|$. Denote $J$ the excenter of $ABC$ with respect to $A$ and $D,E$ tangent points of this excircle with the lines $AB$ and $AC$, respectively. Suppose that the opposite rays to $DP$ and $EQ$ intersect in $F\neq J$. Prove that $AF\perp FJ$.
Problem
Source: Czech and Slovak Olympiad 2019, National Round, Problem 4
Tags: geometry, excircle, Triangle, perpendicular lines, national olympiad
MessingWithMath
07.06.2019 03:55
Notice that $\angle AEJ=\angle ADJ=90^o$, so it suffices to prove that $ADFE$ cyclic. This is equivalent to $\angle DFE=180-\angle A$. Let $X$ be the tangency point of the $A$-excircle with $BC$. Then, by SAS, triangle $PBD$ is congruent to $ABX$. Thus, $\angle DPB=\angle BAX$. Similarily, $\angle EQC=\angle CAX$. Thus, we have $\angle PFQ=180-\angle FPQ-\angle FQP=180-\angle BAX-\angle CAX = 180-\angle A$. This proves the problem.
Flash_Sloth
07.06.2019 04:14
By SAS, $\triangle JBA \cong \triangle JBP$, $\triangle JCA \cong \triangle JCQ$. Thus $JQ= JP$. Moreover, by easy angle chasing $\angle PJQ = \angle PJA + \angle AJQ = 2 \angle BJA + 2\angle AJC = 180^\circ - \angle A = \angle DJE$, thus $\triangle JPQ \sim \triangle JDE$, implying that $\triangle JDP \cong \triangle JEQ$. Hence $\angle JEF = 180^\circ - \angle JEQ = \angle JDF$ implying that $D,J,F,E$ are concyclic.
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zuss77
07.06.2019 20:31
With $T$ - touchpoint on $BC$, $PT=TQ=AD=AE$. Consider rotation about $J$ sending $D$ to $E$. You can see on diagram that it send $PD$ to $QE$.
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