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Difference between revisions of "1972 AHSME Problems/Problem 18"
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<math>\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</math>
 
<math>\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</math>
 
==Solution==
 
==Solution==
−
<math>\fbox{E}</math>
+
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 <math>\angle EDC</math> and <math>\angle EBA</math> are alternate interior angles, and therefore equal. We have the same setup with <math>\angle ECD</math> and <math>\angle EAB</math>, meaning that <math>\triangle ABE \sim \triangle CDE</math> by AA Similarity. We could've also used the fact that <math>\angle BEA</math> and <math>\angle DEC</math> are vertical angles.
 +
 
 +
With this information, we can setup a ratio of corresponding sides:
 +
<cmath>\frac{AB}{CD} = \frac{AE}{CE} \implies \frac{2CD}{CD} = \frac{11 - CE}{CE}.</cmath>
 +
And simplify from there:
 +
<cmath> \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*} </cmath>
 +
Therefore, our answer is <math>\boxed{\textbf{(A) }3\textstyle\frac{2}{3}.}</math>

Latest revision as of 03:54, 23 June 2022

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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