In triangle ABC, points D and E lie on the sides AB and BC, respectively, and AD = CE b AE = CD. Prove that ABC is isosceles.

Docs 1) First, let’s consider triangles ADC and AEC They have: 1) AD = CE (by condition) 2) AE = CD (by condition) 3) AC – common side Therefore, the triangles ADC and AEC are equal (on three sides – 3 sign of rab-va treug-kov). Equality implies that the angles BAC and BCA are equal. 2) Now we will prove that the triangles ABE and CBD are equal They have: 1) AE = CD (according to conv) 2) AC- common side 3) Angle BAC = angle of BCA (from the above proved) These triangles are equal in 1 sign of the rab-va triangles From the equality it follows that AB = BC => Triangle ABC is isosceles, and it was required to prove.
(Although already from the equalization of the angles BAC and ICA it was possible to conclude that the triangle is isosceles)