# Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 5"

## Problem

In the figure $BD$ is parallel to $AE$ and also $BF$ is parallel to $DE$. The area of the larger triangle $ACE$ is $128$. The area of the trapezoid $BDEA$ is $78$. Determine the area of triangle $ABF$.

$[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); MP("A",(4,0),SE);MP("C",(1,2),N);MP("E",(0,0),SW); MP("D",(.5,1),W);MP("B",(2.5,1),NE);MP("F",(2,0),S); [/asy]$

## Solution

In general, if $G$ and $H$ are similar triangles with sides $s_1$ and $s_2$ , then $\frac{Area G}{Area H}=\frac{(s_1)^2}{(s_2)^2}$. Here, $\triangle{BCD}$ is $\frac{50}{128}=\frac{25}{64}$ of $\triangle{ACE}$, so BC is $\frac{5}{8}$ of AC or AB is $\frac{3}{8}$ of AC then $\triangle{ABF}$ is $\frac{9}{64}$ of $\triangle{ACE}$ so $\triangle ABF = \boxed{18}$