An isosceles triangle ABC is inscribed in a circle, with $AB = AC$. Show that the altitude from A to BC passes through the centre of the circle.                                                                                                                                              [Page No. 101]

Let O be the centre of the circle. Since ABC is isosceles with AB = AC, A is equidistant from B and C.
So, A lies on the perpendicular bisector of BC.

Since O is the centre, OB = OC (radii), so O also lies on the perpendicular bisector of BC.

Therefore, A and O both lie on the perpendicular bisector of BC, which means the line AO is the perpendicular bisector of BC.

The altitude from A to BC (in an isosceles triangle with AB = AC) coincides with the perpendicular bisector of BC.

Hence, the altitude from A passes through O, the centre of the circle.