Point E is marked in an isosceles triangle ABC with base BC on the median AD. Prove that: 1) Δ AEB = Δ AEC; 2) ΔBED = Δ CED.

According to the condition, the triangle ABC is isosceles, AB = AC, then its median AD is also its height drawn to the base of the BC, then the triangles AВD and AСD are rectangular, and since point E lies at the height of AD, then the triangles ВED and СED are also rectangular …

In right-angled triangles ВED and СED, the leg ED is common, and the legs of the ВD and СD are equal, since the median BP divides the ВС into equal segments.

Then the ВED triangle is equal to the СED triangle along two legs – the first sign of equality of right-angled triangles, which was required to be proved.

In triangle ABE and ACE AB = AC, side AE ​​is common, and BE = CE, since triangles ВED and СED are equal, then triangles ABE and ACE are equal on three sides, which was required to be proved.