Question

Through the vertex $A$ of a parallelogram $ABCD$, line $AEF$ is drawn to meet $BC$ at $E$ and $DC$ produced at $F$. Show that the area $\left(\Delta BEF\right)$=area $\left(\Delta DCE\right)$.

Answer 1

Construction: Join $AC$

Proof:

$△ABC$ and $△ABF$ lie outer square Base $AB$ and between the square parallels $AB$ and $DF$

$\therefore ar\left(ABC\right)=ar\left(ABF\right)$

$\Rightarrow ar\left(ABE\right)+ar\left(AEC\right)=ar\left(ABE\right)+ar\left(BEF\right)$

$\therefore ar\left(ABC\right)=ar\left(BEF\right)-----\left(I\right)$

$ar\left(AEC\right)=ar\left(DEF\right)----\left(II\right)$

Triangles on the square base and between the square parallel

From $\left(I\right)$ & $\left(II\right)$ we get

$ar\left(BEF\right)=ar\left(DEC\right)$

Hence , proved