Given is a triangle $ABC$ with $AB>AC$. Its incircle touches $AB, AC$ at $D, E$, respectively. Let $CD$ meet the incircle at $K$ and $L$ is the foot of the perpendicular from $A$ to $CK$. If $M$ is the midpoint of $DE$ and $H$ is the orthocenter of $\triangle KML$, prove that $\angle AHK=90^{o}$. Proposed by Dominik Burek
Problem
Source: Polish MO 2nd round 2023/5
Tags: geometry
nobodyknowswhoIam
11.02.2023 19:02
Since we know $KH$ and $LM$ are perpendicular all we need to prove is $\angle MLE + \angle EKD = 90$. We all know that $ALME$ is concyclic, hence $\angle MLE = \angle MAE$. It is a known fact that $\angle EFD = 90 - \angle MAE$ so K,D and H are collinear.
We do a lil bit of angle chasing... Again. $\angle LAK = 90 - \angle LKA = 90 - \angle MKD = 180 - \angle LMK$. Therefore, $\angle LHK = 180 - \angle LMK = \angle LAK$ and $LAHK$ is concyclic so $\angle AHK = 180 - \angle ALK = 90$ and we are done.
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timon92
12.02.2023 03:20
This problem was proposed by Burii. By the way, you mistyped the original statement of the problem.