1986 AHSME Problems/Problem 27


In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$, with $\angle AED = \alpha$. The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ is

[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(-1,0), B=(1,0), E=(0,-.4), C=(.6,-.8), D=(-.6,-.8), E=(0,-.8/(1.6)); draw(unitcircle); draw(A--B--D--C--A); draw(Arc(E,.2,155,205)); label("$A$",A,W); label("$B$",B,C); label("$C$",C,C); label("$D$",D,W); label("$\alpha$",E-(.2,0),W); label("$E$",E,N); [/asy]

$\textbf{(A)}\ \cos\ \alpha\qquad \textbf{(B)}\ \sin\ \alpha\qquad \textbf{(C)}\ \cos^2\alpha\qquad \textbf{(D)}\ \sin^2\alpha\qquad \textbf{(E)}\ 1-\sin\ \alpha$


$ABE$ and $DCE$ are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in $AD$. Because $AB$ is a diameter, $\angle ADB=\angle ADE=90^{\circ}$. Thus, \[\frac{DE}{AE}=\cos\alpha\] So \[\frac{DE^2}{AE^2}=\cos^2\alpha\] The answer is thus $\fbox{(C)}$.

See also

1986 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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