In an isosceles triangle abc with base ac on the sides ab and bc, points M and N are marked, respectively, so that AM = CN.

In an isosceles triangle abc with base ac on the sides ab and bc, points M and N are marked, respectively, so that AM = CN. The segments CM and AN intersect at point O. Prove that triangle AOC is isosceles.

Let us prove the equality of the triangles ANС and CMA.

Since the triangle ABC is isosceles, the angle MAC = NСA. In the triangles ANС and СMA, the AC side is common. By condition, the segment AM = MN.

Then the triangles ANС and СMA are equal on two sides and the angle between them, and then the angle NAC = MCA.

In the AOC triangle, the angle OAC = OCA, and therefore the AOC triangle is isosceles with the base of the AC, which was required to be proved.