Let $AH_1, BH_2$ and $CH_3$ be the altitudes of a triangle $ABC$. Point $M$ is the midpoint of $H_2H_3$. Line $AM$ meets $H_2H_1$ at point $K$. Prove that $K$ lies on the medial line of $ABC$ parallel to $AC$.
We need to prove KB = KH2.
Let P be projection of H3 on AH2. We have H3P || BH2. M is midpoint of H2H3 so ∠H2MP = 2∠H3PM = 2∠PH3M = 2∠H3H2B = H3H2K so PM || H2K. AM/AK = AP/AH2 = AH3/AB ---> MH3 || BK. So triangles BH2K and H3PM are similar and we know H3PM is isosceles so BH2K is isosceles as well and as we wanted KB = KHN and we're Done.