Prove that triangle ABC is isosceles if its bisector BD is the median.

1. Let’s conduct the ВM in such a way that BD = MD.

2. Triangle ABD = triangle CMD on two sides and the angle between them.

Then <ABD = <CMD.

3. Consider the BCM triangle. It is isosceles because the angles at the BM are equal.

DC is the median of this isosceles triangle by construction, which means that it is also the height by the property of an isosceles triangle. Then <CDB = 90 °.

4. Consider the triangle ABC. By the condition of the problem, BD is a bisector, but it is also the height of the triangle ABC, since <CDB = 90 ° from (4).

Consequently, the triangle ABC is isosceles by the property of the bisector of an isosceles triangle, as required.