1986 AHSME Problems/Problem 3

Problem

$\triangle ABC$ has 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

Since $\angle C = 90^{\circ}$ and $\angle A = 20^{\circ}$ , we have $\angle ABC = 70^{\circ}$ . Thus $\angle DBC = 35^{\circ}$ . It follows that $\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}$ , which is $\boxed{D}$ .

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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