Let $BM$ be a median of right-angled nonisosceles triangle $ABC$ ($\angle B = 90$), and $H_a$, $H_c$ be the orthocenters of triangles $ABM$, $CBM$ respectively. Lines $AH_c$ and $CH_a$ meet at point $K$. Prove that $\angle MBK = 90$.
Source: Sharygin geometry olympiad 2015, grade 10, Final Round, Problem 5
Tags: geometry
Let $BM$ be a median of right-angled nonisosceles triangle $ABC$ ($\angle B = 90$), and $H_a$, $H_c$ be the orthocenters of triangles $ABM$, $CBM$ respectively. Lines $AH_c$ and $CH_a$ meet at point $K$. Prove that $\angle MBK = 90$.