1968 AHSME Problems/Problem 18

Problem

Side $AB$ of triangle $ABC$ has length 8 inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC$ , and $E$ is on segment $BC$ . Line $AE$ extended bisects angle $FEC$ . If $DE$ has length $5$ inches, then the length of $CE$ , in inches, is:

$\text{(A) } \frac{51}{4}\quad \text{(B) } 13\quad \text{(C) } \frac{53}{4}\quad \text{(D) } \frac{40}{3}\quad \text{(E) } \frac{27}{2}$

Solution

Draw a line passing through $A$ and parallel to $BC$ . Let $\angle FEC = 2n$ . By alternate-interior-angles or whatever, $\angle BAE = n$ , so $BAE$ is an isosceles triangle, and it follows that $BE = 8$ . $\triangle ABC \sim \triangle DEC$ . Let $CE = x$ . We have $\frac{8}{8+x} = \frac{5}{x} \Rightarrow 40+5x = 8x \Rightarrow x = CE =\boxed{\frac{40}{3}}.$

Thus, we choose answer $\fbox{D}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1968 AHSME Problems/Problem 18

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