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Incenter is the foot of altitude
Sadigly   0
19 minutes ago
Source: Azerbaijan JBMO TST 2023
Let $ABC$ be a triangle and let $\Omega$ denote the circumcircle of $ABC$. The foot of altitude from $A$ to $BC$ is $D$. The foot of altitudes from $D$ to $AB$ and $AC$ are $K;L$ , respectively. Let $KL$ intersect $\Omega$ at $X;Y$, and let $AD$ intersect $\Omega$ at $Z$. Prove that $D$ is the incenter of triangle $XYZ$
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Sadigly
19 minutes ago
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