Question

Prove that the radical centre of the circles described on the sides of a triangle as diameters is the orthocentre of the triangle.

Answer 1

The circles described on AB and AC as diameters (see Fig.) will intersect at a point D on the side BC and because  ΔADB  = ΔADC =- 90°.

We have that AD is an altitude of ΔABC which is also the radical axis of the two circles with AB and AC as diameters. Similarly, the other altitudes BE and CF are the radical axes of the pairs described on AB, BC and AC, BC. Thus, the altitudes of  ΔABC are radical axes. Hence, the orthocentre is their radical centre.