MathDB

Let the bisectors of the exterior angles at

B

and

C

of triangle

ABC

meet at

D.

Then, if all measurements are in degrees, angle

BDC

equals: (A)\ \frac {1}{2} (90 \minus{} A) \qquad (B)\ 90 \minus{} A \qquad (C)\ \frac {1}{2} (180 \minus{} A) \qquad (D)\ 180 \minus{} A \qquad (E)\ 180 \minus{} 2A

Problem Statement