The acute angles of a right-angled triangle are 69 degrees and 21 degrees. Find the angle between the bisector CH

The acute angles of a right-angled triangle are 69 degrees and 21 degrees. Find the angle between the bisector CH and the median CD, drawn from the top of the right angle.

Consider a triangle CDB: angle DBC (angle B) = 21 degrees (by condition), CD is the median of triangle ABC drawn from a right angle, therefore it is equal to half of the hypotenuse (property of the median of a rectangle), then: CD = AB / 2 = AD = BD. Since CD = BD, the triangle CDB is isosceles, CD and BD are the sides, CB is the base, the angles DCB and DBC are the angles at the base, so the angle DBC = angle DCB = 21 degrees.
Since CH is the bisector of a right angle, it divides the angle C into angles that are equal to 90/2 = 45 degrees. Then:
angle ACН = angle BCN = 45 degrees.
The BCH angle consists of two angles:
angle BCH = angle DCB + angle DCH;
45 = 21 + angle DCH;
angle DCН = 45 – 21;
angle DCH = 24 degrees.
Answer: angle DCH = 24 degrees.