In isosceles ΔABC, the point K and M are the midpoints of the lateral sides AB

In isosceles ΔABC, the point K and M are the midpoints of the lateral sides AB and BC, respectively. BD is the median of the triangle. Prove that ΔAKD = ΔCMD.

Since the triangle ABC is isosceles, the angles at the base of the AC are equal, the angle BAC = BCA, and the sides are equal, AB = BC.

Points K and M are the middle of the lateral sides, then AK = ВK, CM = ВM, then AK = CM.

Median BD divides the AC side in half, then AD = CD.

Then in triangles AKD and CMD:

AK = CM, AD = CD, angle KAD = MCD, then the triangles are equal on two sides and the angle between them, which was required to prove.