1972 AHSME Problems/Problem 18

Problem

Let $ABCD$ be a trapezoid with the measure of base $AB$ twice that of base $DC$ , and let $E$ be the point of intersection of the diagonals. If the measure of diagonal $AC$ is $11$ , then that of segment $EC$ is equal to

$\textbf{(A) }3\textstyle\frac{2}{3}\qquad \textbf{(B) }3\frac{3}{4}\qquad \textbf{(C) }4\qquad \textbf{(D) }3\frac{1}{2}\qquad \textbf{(E) }3$

Solution

We begin with a diagram:

$[asy] pair A, B, C, D, E; A = (0, 0); B = (8, 0); C = (7, 4); D = (3, 4); E = intersectionpoint(A--C, B--D); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, 3W); label("$11$", midpoint(A--C), 2S); [/asy]$

The bases of a trapezoid are parallel by definition, so $\angle EDC$ and $\angle EBA$ are alternate interior angles, and therefore equal. We have the same setup with $\angle ECD$ and $\angle EAB$ , meaning that $\triangle ABE \sim \triangle CDE$ by AA Similarity. We could've also used the fact that $\angle BEA$ and $\angle DEC$ are vertical angles.

With this information, we can setup a ratio of corresponding sides: $\frac{AB}{CD} = \frac{AE}{CE} \implies \frac{2CD}{CD} = \frac{11 - CE}{CE}.$ And simplify from there: $\begin{align*} \frac{2CD}{CD} &= \frac{11 - CE}{CE} \\ 2 &= \frac{11 - CE}{CE} \\ 2CE &= 11 - CE \\ 3CE &= 11 \\ CE &= \frac{11}{3} = 3 \frac{2}{3}. \end{align*}$ Therefore, our answer is $\boxed{\textbf{(A) }3\textstyle\frac{2}{3}.}$

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1972 AHSME Problems/Problem 18

Problem

Solution