In triangle ABC, medians AM and BH. prove that АРС = BНC, if АС = ВC.

Since, by condition, AC = BC, the triangle ABC is isosceles.

The segments AP and BH are the medians of the triangle, then CH = AH, CP = BP, and therefore CH = CP.

In the triangles AРС and ВНС, the angle C is common.

The СН side of the triangle BHC is equal to the СР side of the АРС triangle.

The AC side of the APC triangle is equal to the BC side of the BHC triangle.

Then the triangles BНС and AРС are equal on two sides and the angle between them, which was required to prove.