Question 140
Let $ABC$ be a triangle whose three angles are acute. Let $D \in [BC]$ be such that $\widehat{BAC} = \widehat{ADB}$. Let $H$ be the foot of the altitude from $B$ in $ABC$. The perpendicular to $(BC)$ through $H$ meets $(AD)$ at $K$. Suppose that $K$ lies inside triangle $ABC$. Let $M$ be the midpoint of $[AC]$. Show that $MH = MK$.