Prove that the line parallel to the base AC of an isosceles triangle ABC is perpendicular to its median BD.

Given: isosceles triangle ABC;
KM is parallel to the AC warp;
BD is the median.
Prove that BD is perpendicular to CM.
Proof: Consider the triangle ABC. He is isosceles. In an isosceles triangle, the median drawn to the base of the triangle is the height. Then the median BD of an isosceles triangle ABC is the height, that is, BD is perpendicular to AC. By the corollary of the theorem that two lines, perpendicular to the third, are parallel. Then BD is perpendicular to the CM in the same way that the CM is parallel to the base AC. Proven.