Point A lies on the median drawn to the base of the isosceles triangle. Prove that it is equally distant

Point A lies on the median drawn to the base of the isosceles triangle. Prove that it is equally distant from the tops of the base.

Let ВСD be an isosceles triangle with base ВС, and DH its median.

The median of an isosceles triangle is also its height and bisector.

Then triangles ABH and ACH are rectangular.

Since DH is the median, then BH = CH = BC / 2.

In right-angled triangles ABH and ACH, the leg AH is common, BH = CH, then the triangles ABH and ACH are equal in two legs, then AB = AC, which was required to be proved.