Problem

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Tags: geometry, circumcircle, LaTeX, geometric transformation, reflection, incenter, parallelogram



in a triangle $ABC$, $I$ is the incenter. $BI$ and $CI$ cut the circumcircle of $ABC$ at $E$ and $F$ respectively. $M$ is the midpoint of $EF$. $C$ is a circle with diameter $EF$. $IM$ cuts $C$ at two points $L$ and $K$ and the arc $BC$ of circumcircle of $ABC$ (not containing $A$) at $D$. prove that $\frac{DL}{IL}=\frac{DK}{IK}$.(25 points)