A bisector of an acute angle is drawn in a right-angled triangle; the segment connecting its base with

A bisector of an acute angle is drawn in a right-angled triangle; the segment connecting its base with the point of intersection of the medians is perpendicular to the katitu. Find the sharp corners of the triangle.

Segments CM and AK of the median of triangle ABC.

By the property of the medians, the point of their intersection divides them in a ratio of 2/1 starting from the top.

Let OM = X cm, BO = 2 * X cm, BK = Y cm, BC = 2 * Y cm.

The BCM triangle is rectangular, the OH segment is perpendicular to the BC, which means it is parallel to the AC.

Then, according to Thales’ theorem, the segment OH and the leg AC cuts off proportional segments on the segments BM and CК.

ОМ / ОВ = СН / ВН = X / 2 * X = 1/2.

AH is the bisector of angle A, then, by its property: AC / CH = AB / BН.

AC * ВН = CH * AB.

CH / BH = AC / AB = 1/2.

AB = 2 * AC. The hypotenuse AB is twice the leg AC, then the angle ABC = 30.

Angle BAC = (90 – 30) = 60.

Answer: The acute angles of the triangle are 30 and 60.