Difference between revisions of "1968 AHSME Problems/Problem 18"

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== Solution ==
 
== Solution ==
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== See also ==
 
== See also ==

Revision as of 03:31, 29 September 2014

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

$\fbox{D}$

See also

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