1962 AHSME Problems/Problem 7

Problem

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:

$\textbf{(A)}\ \frac{1}{2}(90-A)\qquad\textbf{(B)}\ 90-A\qquad\textbf{(C)}\ \frac{1}{2}(180-A)\qquad$

$\textbf{(D)}\ 180-A\qquad\textbf{(E)}\ 180-2A$

1962 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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