Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 1.6 Junior Find the answer

$\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$ . If $BD$ ( $D$ in $\overline{AC}$ ) is the bisector of $\angle ABC$ , then $\angle BDC =$
$\textbf{(A)}\ 40^\circ$ $\textbf{(B)}\ 45^\circ$ $\textbf{(C)}\ 50^\circ$ $\textbf{(D)}\ 55^\circ$ $\textbf{(E)}\ 60^\circ$

Your answer

Multiple choice: answer with the letter of the option you want.

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 $\boxed{D}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.