A medial line parallel to the side $AC$ of triangle $ABC$ meets its circumcircle at points at $X$ and $Y$. Let $I$ be the incenter of triangle $ABC$ and $D$ be the midpoint of arc $AC$ not containing $B$.A point $L$ lie on segment $DI$ in such a way that $DL= BI/2$. Prove that $\angle IXL = \angle IYL$.
Problem
Source: Sharygin Finals 2022 9.3
Tags: geometry
WizardMath
31.07.2022 17:46
Let $E$ be the midpoint of $BI$. Note that if $I_B$ is the $B-$excenter of $ABC$, then $L$ is the midpoint of $BI_B$. Let $X'$ be the reflection of $X$ in $BI$. Since $XY \parallel BC$, $BD$ bisects $\angle XBY$, so $B, X', Y$ are collinear. The problem reduces to showing that $BX \cdot BY = BI \cdot BL$. Since $BX, BY$ are isogonal, if $BX$ meets $AC$ at $W$, then $2 \cdot BX \cdot BY = BW \cdot BY = BA \cdot BC = BI \cdot BI_B$, and we are done.
Infinityfun
12.03.2023 17:08
Motivation can be found easily by taking inversion that swaps $X$ with $Y$ It swaps $A$ with midpoint of $BC$, $I$ with $L$ etc.
Mahdi_Mashayekhi
02.09.2024 16:24
Let $X'$ be reflection of $X$ w.r.t $BI$. Now we need to prove $X'YLI$ is cyclic. Note that $XY \parallel AC$ so $B,X',Y$ are collinear. Let $BY$ meet $AC$ at $T$. Since $I_B$ is the reflection of $B$ across $l$ and $T$ is the reflection of $B$ across $Y$ we can instead prove $IXTI_B$ is cyclic. Since $BI.BI_B = BA.BC$ we need to prove $BX'.BT = BA.BC$ which is true since $XAB$ and $CTB$ are similar by angle chasing.