Let $BH$ be an altitude of right angled triangle $ABC$($\angle B = 90^o$). An excircle of triangle $ABH$ opposite to $B$ touches $AB$ at point $A_1$; a point $C_1$ is defined similarly. Prove that $AC // A_1C_1$.
Source: Sharygin Finals 2022 9.1
Tags: geometry
Let $BH$ be an altitude of right angled triangle $ABC$($\angle B = 90^o$). An excircle of triangle $ABH$ opposite to $B$ touches $AB$ at point $A_1$; a point $C_1$ is defined similarly. Prove that $AC // A_1C_1$.