Point M is chosen in an isosceles triangle ABC with base AC on the median BD. Prove that triangle ABM is equal to CBM.

In an isosceles triangle, the median is also the bisector of the angle. Those. ВD is the bisector ∠ ABС and, accordingly, ∠ ABD = ∠ CBD
Let us compare ∆ ABM and ∆ CBM:
AB = BC, because ∆ ABC is isosceles by the condition of the problem
BM – common side for two triangles
∠ ABM = ∠ CBM (since BD is the bisector)
Hence, ∆ ABM = ∆ CBM (according to the first sign of equality of triangles – 2 sides and the angle between them)
Q.E.D