1963 AHSME Problems/Problem 6

Problem

$\triangle BAD$ is right-angled at $B$ . On $AD$ there is a point $C$ for which $AC=CD$ and $AB=BC$ . The magnitude of $\angle DAB$ is:

$\textbf{(A)}\ 67\tfrac{1}{2}^{\circ}\qquad \textbf{(B)}\ 60^{\circ}\qquad \textbf{(C)}\ 45^{\circ}\qquad \textbf{(D)}\ 30^{\circ}\qquad \textbf{(E)}\ 22\tfrac{1}{2}^{\circ}$

Solution

$[asy] draw((0,0)--(0,10)--(17.321,0)--(0,0)); draw((0,0)--(8.661,5)); dot((0,0)); label("$B$",(0,0),SW); dot((0,10)); label("$A$",(0,10),NW); dot((17.321,0)); label("$D$",(17.321,0),SE); dot((8.661,5)); label("$C$",(8.661,5),NE); [/asy]$ Note that because $C$ is the midpoint of $AD$ , $C$ is the circumcenter of the triangle, so $BC = AC = AB$ . Thus, $\triangle ABC$ is equilateral, so $\angle DAB = 60^{\circ}$ , which is answer choice $\boxed{\textbf{(B)}}$ .

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1963 AHSME Problems/Problem 6

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