1986 AHSME Problems/Problem 3

Problem

$\triangle ABC$ is a right angle at $C$ and $\angle A = 20^\circ$ . If $BD$ ( $D$ in $\overline{AC}$ ) is the bisector of $\angle ABC$ , then $\angle BDC =$

$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$

Solution

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1986 AHSME Problems/Problem 3

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Solution

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