In triangle ABC, the height of AH is equal to the median BM. Find the angle of the MBC.

MG – triangle height AMH

Triangle AHC is rectangular and point M is the midpoint of hypotenuse AC.

Therefore, M is the center of the circle circumscribed around AHC and | MA | = | MH | = | MC |

Triangle AMH is isosceles because | MA | = | MH |, which means | AG | = | GH |.

Lines MG and BC are perpendicular to AH and therefore parallel, which means

angles MBC and BMG are equal.

Let the angle MBC = a. Then | HF | = | BF | * sin (a), because BFH is rectangular.

And | FG | = | FM | * sin (a), because MGF is rectangular.

Then | HG | = | HF | + | FG | = | BF | * sin (a) + | FM | * sin (a) = (| BF | + | FM |) * sin (a) = | BM | * sin (a) and

sin (a) = | HG | / | BM |

But 2 * | HG | = | AH | since triangle AMH is isosceles.

By the problem statement | AH | = | BM |. So 2 * | HG | = | BM | and | HG | / | BM | = 0.5

Hence sin (a) = 0.5 and a = 30 degrees.