# 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 $\boxed{\textbf{(E) }15^{\circ}}$.