In the right-angled triangle KMN (KN – hypotenuse), the MC height and the MA median are drawn.

In the right-angled triangle KMN (KN – hypotenuse), the MC height and the MA median are drawn. The MP segment divides the CMA angle in half. Prove that МР is the bisector of the KMN angle.

Since AM is the median of the triangle KMN, then the triangle AKM is isosceles, AK = AM as the radii of the circumscribed circle, then the angle AKM = AMK.

Triangles MCH and MCH are rectangular and similar in acute angle, then the angle MCH = CHM.

Then the angle АМК = СМН.

Angle KMR = AMK – AMR.

Angle НМР = СМН – СМР.

And since АМК = СМН, and АМР = СМР, since МР is the bisector of the angle СМА, then the angle КМР = НМР, and therefore МР is the bisector of the right angle, which was required to be proved.