Orthocenter property

[ins-]

It is given a triangle ABC. A1 and B1 are the feets of the perpendiculars from the vertices A and B to the sides BC and AC respectively. There are drawn circles with diameters AA1 and BB1 intersecting each other at the points C1 and C2. Prove that orthocenter of the triangle ABC lies on the line C1C2.

[vvsss]

It is given a triangle ABC. A1 and B1 are the feets of the perpendiculars from the vertices A and B to the lines BC and AC respectively. There are drawn circles with diameters AA1 and BB1. Prove that H orthocenter of the triangle ABC lies on the radical axes of the circles.

Более удобна обща и проста для решения следующая формулировка. Доказать что ортоцентр треугольника лежит на радикальной оси окружностей, диаметрами которых являются высоты треугольника.

Все свелось к пропорции для нахождения степени точки

[ins-]

It is very interesting solution. The same is true if we replace heights with angle bisectors or medians. Can the problem be generalized further?