In triangle ABC, angle C = 90, angle A = 70, CD-bisector. find the corners of the triangle BCD.

1. Find the degree measure of the angle B in the triangle ABC:
angle A + angle B + angle C = 180 degrees (according to the theorem on the sum of the angles of a triangle);
70 + angle B + 90 = 180;
angle B = 180 – 160;
angle B = 20 degrees.
2. Since CD is a bisector, it divides the angle C into two equal triangles:
angle ACD = angle DCB = angle C / 2 = 90/2 = 45 (degrees).
3. In triangle BCD: angle DCB = 45 degrees, angle DCB (angle B) = 20 degrees.
By the theorem on the sum of the angles of a triangle:
DCB angle + DVS angle + CDB angle = 180 degrees;
45 + 20 + CDB angle = 180;
angle CDB = 180 – 65;
angle CDB = 115 degrees.
Answer: DCB angle = 45 degrees, CDB angle = 115 degrees, DWS angle = 20 degrees.