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

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

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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