In an isosceles triangle abc, the medians are AE and CD. Prove that the angle AE = CD.

In an isosceles triangle, the lengths of the sides are equal, then AB = BC.

Since CD and AE are medians, they divide the sides AB and BC in half, then AD = BD = BE = CE.

The angles at the base of the AC are equal, the angle BAC = BCA.

Then, in triangles ACD and ACE, the side AC is common, the segments AD and CE are equal and the angles DAC and ECA are equal, therefore the triangles AC and ACE are equal on both sides and the angle between them.

Then AE = CD, as required.