Problem

1957 AHSME Problem 44

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

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.


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Category: AHSME
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