In triangle

*ABC*,

*DB* and

*DC* are angle bisectors and the angle

*BAC* = 60°. If angle

*DCB* is 40°, what is the degree measure of angle

*BDC*? (

**Note**: Figure not drawn to scale.)

A. 130°

B. 120°

C. 100°

D. 80°

E. 75°

(B) First of all remember that the sum of all angles in a triangle is equal to 180°. We are told that angle

*BAC* = 60°. That leaves 120° to be split between angles

*ABC* and

*ACB*. We know that

*DC* and

*DB* are angle bisectors (an angle bisector can be described as a line segment that divides an angle into two equal angles), so the sum of angles DBC and DCB is 120°/2 = 60°. BDC is the third angle in triangle BDC. Therefore it equals 180° – 60° = 120°. The correct answer is B.

----------

I do not understand why the answer is 120° and not 100°. If as you say, DB and DC are angle bisectors and DCB is 40° degrees (as stated in the question) CBD should also be 40° (which lets CBA and ACB as 60 degrees angles). If all the angles must be 180 degrees 180° – 80° = 100°.