In an isosceles triangle ABC BD, the height drawn to the base, points M and H belong to sides AB and BC

In an isosceles triangle ABC BD, the height drawn to the base, points M and H belong to sides AB and BC, respectively, ray BD bisector of angle MDH prove AM = HC

Consider triangles AMD and СНD.

The height of the ID of an equilateral triangle ABC is also its median, then BP = СD = AС / 2.

Angle АDВ = СDВ = 900, since ВD is the height of the triangle.

By condition, the angle MDВ = НDВ, since ВM is the bisector of the MDН angle. Then the angle ADM = СDН = (90 – MDВ).

Angle ВAD = BCD as angles at the base of an isosceles triangle.

Then the triangles AMD and СНD are equal in side and two adjacent angles, and therefore AM = CH, which was required to be proved.