AoPS - Problem

Source: Mathematics Regional Olympiad of Mexico Center Zone 2019 P3

Tags: geometry, equal segments, equal angles

parmenides51

14.11.2021 04:37

Let $ABC$ be an acute triangle and $D$ a point on the side $BC$ such that $\angle BAD = \angle DAC$. The circumcircles of the triangles $ABD$ and $ACD$ intersect the segments $AC$ and $AB$ at $E$ and $F$, respectively. The internal bisectors of $\angle BDF$ and $\angle CDE$ intersect the sides $AB$ and $AC$ at $P$ and $Q$, respectively. Points $X$ and $Y$ are chosen on the side $BC$ such that $PX$ is parallel to $AC$ and $QY$ is parallel to $AB$. Finally, let $Z$ be the point of intersection of $BE$ and $CF$. Prove that $ZX = ZY$.

Let $\angle BAD=\theta$ $\implies$ $\angle CAD=\angle BED=\angle CFD=\angle FCD=\angle EBD=\angle FDP=\angle BDP=\angle EDQ=\angle CDQ=\theta$. Let $\angle ACB=\angle PXB=\alpha$ $\implies$ $\angle ACF=\angle ADF=\alpha-\theta$. Let $\angle ABC=\angle QYC=\beta$ $\implies$ $\angle ABE=\angle ADE=\beta-\theta$. From angles that I mentioned above we can say that $DF=DC$, $DE=DB$, $\triangle BDF\cup {P}\cong \triangle EDC\cup {Q}$, $FPXD$ and $EQYD$ are cyclic, $FP=CQ=PX$ and $EQ=BP=QY$. Also $ZB=ZC$. So that's enough to prove $BX=CY$. $\frac{BX}{\sin \alpha+\beta}=\frac{BP}{\sin \alpha}$ and $\frac{CY}{\sin \alpha+\beta}=\frac{YQ}{\sin \alpha}$ and since $BP=YQ$ we get $BX=YC$ $\blacksquare$