Let $BM$ be a median of nonisosceles right-angled triangle $ABC$ ($\angle B = 90^o$), and $Ha, Hc$ be the orthocenters of triangles $ABM, CBM$ respectively. Prove that lines $AH_c$ and $CH_a$ meet on the medial line of triangle $ABC$. (D. Svhetsov)
Problem
Source: Sharygin Geometry Olympiad 2015 Final 9.5
Tags: geometry, concurrency, concurrent, midpoint