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:

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

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

Solution

Calculating for angles $\angle DBC$ and $\angle DCB$, we get

$\angle DBC$ = $90 - \frac{B}{2}$ and $\angle DCB$ = $90 - \frac{C}{2}$.

In triangle $BCD$, we have

$\angle BDC$ = $180 - (90 - \frac{B}{2}) - (90 - \frac{C}{2})$ = $\frac{B+C}{2}$ = $\frac{1}{2}\cdot(180 - A)$, so the answer is $\fbox{C}$.

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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