Bariz - Problem

Problem

Source: Mathcenter Contest / Oly - Thai Forum 2008 R3 p2 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

Tags: geometry, equal segments, incircle

parmenides51

10.11.2022 01:26

In triangle $ABC$ ($AB\not= AC$), the incircle is tangent to the sides of $BC$ ,$CA$ , $AB$ at $D$ ,$E$, $F$ respectively. Let $AD$ meet the incircle again at point $P$, let $EF$ and the line passing through the point $P$ and perpendicular to $AD$ intersect at $Q$. Let $AQ$ intersect $DE$ at $X$ and $DF$ at $Y$. Prove that $AX=AY$. (tatari/nightmare)

It's sufficient to prove $AQ \parallel BC$ Denote $G$ by the second intersection of $PQ$ and incircle of $\Delta ABC$, $H$ by the intersection of $ID$ and $EF$ Since $\widehat{GPD}=\widehat{QPD}=90^o$, we have $D,I,G$ are colinear. Moreover, $A,P,D$ are colinear, therefore $(DP,EF)=-1$ $\Rightarrow G(DP,EF)=-1 \Rightarrow (QH,EF)=-1$ Therefore, the polar line of $H$ with respect to $(I)$ goes through $Q$. Also $H \in EF$, by La Hire's theorem the polar line of $H$ goes through $A$. Hence, $AH$ is the polar line of $H$ with respect to $(I)$, infer $IH \perp AQ$. This leads to $AQ \parallel BC$ (Q.E.D)