In triangle ABC, angle A is 64 °. Find the angle BDC between the bisectors of angles B and C.

ABC is a triangle, ВK is the bisector of angle B, CM is the bisector of angle C, D is the point of intersection of ВK and CM.
1. In triangle ABC we denote angle B as x, angle C as y. By the theorem on the sum of the angles of a triangle:
angle A + angle B + angle C = 180 degrees;
64 + x + y = 180;
x + y = 180 – 64;
x + y = 116.
2. Since BK is the bisector of angle B, then:
angle ABK = angle KBC (aka angle DBC) = angle B / 2 = x / 2.
Since CM is the bisector of angle C, then:
angle ВСМ (aka angle ВСD) = angle МСА = angle С / 2 = у / 2.
3. Consider a triangle ВDС: angle DBС= x / 2, angle ВСD = y / 2.
By the theorem on the sum of the angles of a triangle:
DVS angle + ВСD angle + ВDC angle = 180 degrees;
x / 2 + y / 2 + angle ВDC = 180;
angle ВDC = 180 – (x + y) / 2.
Since x + y = 116, then:
angle ВDC = 180 – 116/2;
angle ВDC = (360 – 116) / 2;
angle ВDC = 244/2;
angle ВDC = 122 degrees.
Answer: angle ВDC = 122 degrees.