Otrhocenter property 2

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In the triangle ABC with A1 and B1 are denoted intersection points of the angle bisectors of the angles A and B with the sides BC and AC respectively. Circles with diameters AA1 and BB1 intersects each other in the points C1 and C2. Prove that the orthocenter of the triangle ABC lies on C1C2.

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Даны треугольник АВС и точка D. A' и B' - точки пересечения AD c BC и BD c AC. Докажите, что радикальная ось окружностей с диаметрами AA' и BB' содержит ортоцентр АВС.

It is given a triangle ABC and a point D. A' and B' are the points of intersection AD and BC, BD and AC. Prove that the radical axis of circles with diameters AA 'and BB' contain orthocenter ABC.


Окружность AA' содержит основание высоты AF, BB' - BE. Так как AH ⋅ HF = BH ⋅ HE, степень точки Н относительно окружностей одинакова.

A circle AA'(BB') includes a base of height AF(BE). It is known, that AH ⋅ HF = BH ⋅ HE. Hence the degree of H relative to the circles is the same.