In the isosceles triangle ABC with the base of the AC, the median BE is drawn. Prove that ∆ABE = ∆CBE.

Since triangle ABC is isosceles, AB = BC, the median BE, drawn to the base, is also the height and bisector of the triangle.

Then triangles ABE and ВСЕ are rectangular.

In triangles ABE and ВСЕ, the side BE is common. Since BE is the height and median, AE = CE.

Then the triangles ABE and ВСЕ are equal in two legs – the first sign of equality of right-angled triangles. Q.E.D.