# 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

Since we know $BF$ is equal to the short side of both triangles and that $DB$ is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line $DF$ (as seen below), all four triangles in $\triangle ACE$ are congruent.

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

Then, since we know the sum of all four triangles is $128$, we can solve the easy division problem $128/4$ and see that the area of $\triangle ABF = \boxed{32}$