+1533 I get 10, but that's not the right answer.
Quadrilateral $ABCD$ is a parallelogram. Let $E$ be a point on $\overline{AB},$ and let $F$ be the intersection of lines $DE$ and $BC.$ The area of triangle $EBF$ is $5,$ and the area of triangle $EAD$ is $5.$ Find the area of parallelogram $ABCD$.
+129849 See this
AE = EB = 2.5
Height of EAD = Height of EBF = 4
Area of each triangle = (1/2) ( 4) (2.5) = 5
Triangle FDC has a base and height twice that of EAD.....so....its area = 2^2 * 5 = 20
[ FDC ] - [ EBF] = 20 - 5 = 15 = area of EBCD
[ EAD ] + [ EBCD ] = 5 + 15 = 20 = [ ABCD ]
