1957 AHSME Problems/Problem 44

Problem

In $\triangle ABC, AC = CD$ and $\angle CAB - \angle ABC = 30^\circ$ . Then $\angle BAD$ is:

$[asy] defaultpen(linewidth(.8pt)); unitsize(2.5cm); pair A = origin; pair B = (2,0); pair C = (0.5,0.75); pair D = midpoint(C--B); draw(A--B--C--cycle); draw(A--D); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE);[/asy]$

$\textbf{(A)}\ 30^\circ\qquad\textbf{(B)}\ 20^\circ\qquad\textbf{(C)}\ 22\frac{1}{2}^\circ\qquad\textbf{(D)}\ 10^\circ\qquad\textbf{(E)}\ 15^\circ$

Solution

Because $\triangle ACD$ is isosceles with $AC=CD$ , $\measuredangle CAD=\measuredangle CDA=\theta$ , where $\theta$ is some angle measure. Because $\angle CDA$ and $\angle ADB$ form a straight angle, $\measuredangle ADB = 180^{\circ}-\theta$ . Thus, because the interior angles of a triangle add to $180^{\circ}$ , $\measuredangle BAD + \measuredangle ABC = \theta$ , so $\measuredangle ABC = \theta - \measuredangle BAD$ . Notice that $\measuredangle CAB = \theta + \measuredangle BAD$ . With all of this information and recalling that, from the problem, $\measuredangle CAB - \measuredangle ABC = 30^{\circ}$ , we see that: \begin{align*} \measuredangle CAB - \measuredangle ABC &= 30^{\circ} \\ \theta + \measuredangle BAD - (\theta - \measuredangle BAD) &= 30^{\circ} \\ 2\measuredangle BAD &= 30^{\circ} \\ \measuredangle BAD &= 15^{\circ} \end{align*} Thus, our answer is $\boxed{\textbf{(E) }15^{\circ}}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 43	Followed by Problem 45
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1957 AHSME Problems/Problem 44

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