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 $\boxed{\textbf{(B)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS