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 \[\frac{8}{8+x} = \frac{5}{x} \Rightarrow 40+5x = 8x \Rightarrow x = CE =\boxed{\frac{40}{3}}.\]


See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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