Given the parallelogram ABCD, $\frac{AD}{AE}$ = $\frac{BF}{FA}$ and $\frac{area of \triangle ADF}{areaof \triangle AEF}$ = $\frac{AD}{AE}$ (or in other words $\frac{area of \triangle ADF}{areaof \triangle AEF}$ = $\frac{BF}{FA}$), prove that area of $\triangle$ADF = area of $\triangle$BEF.
On
You just need to use the obvious theorem:
If two triangles have a vertex in common and the sides opposite to that vertex lie on the same line, then the ratio of their areas is equal to the ratio of the opposite sides.
Hence: $$ {area\triangle BEF\over area\triangle AEF}={BF\over FA}={AD\over AE}= {area\triangle ADF\over area\triangle AEF} $$
EDIT.
Notice that $AD/AE=/$ is the same as $BC/AE=/$, and this follows from the similarity of triangles $AEF$ and $CBF$.
Let EI the height of the triangle AEF. Also DH the height of the triangle DAF. We know that the ratio between the height of similar triangle is equal to the ratio between the sides. So $\frac{DH}{EI} = \frac{BF}{FA} $. Now we can write $ DH \cdot FA = BF \cdot EI$. $BF \cdot EI$ is the double of the area of the traingle $ EFB$; $DH \cdot FA$ is the double of the area of $ DAF$. So $ DAF = EFB$ (Area).
Do I have to post an image to clarify?