Question

Prove that the straight line joining the mid-points of the opposite sides of a parallelogram divides it in to two parallelograms of equal area.

Answer 1

Given: parallelogram $ABCD$ in which $E$ is the midpoint of $AD$ and $F$ is the midpoint of $BC$; segment $EF$ is drawn.

To prove: area $\left(ABFE\right)=$ area $\left(EFCD\right)$.

Construction: Draw $AP\perp CB$ produced. Join $AF$.

Proof: First we show that $ABFE$ and $EFCD$ are parallelograms. Observe

$AE=\frac{AD}{2}=\frac{BC}{2}=BF$.

If you join $AF$, you get two triangles $AEF$ and $FBA$. In these triangles

$AE=BF$ (proved)

$\angle EAF=\angle AFB$ (alternate angles)

$AF=AF$ (common)

Hence $△AEF\cong △FBA$ (SAS postulate). Hence $AB=EF$. Thus in the quadrilateral $ABFE$, the opposite sides are equal. It follows that $ABEF$ is a parallelogram. Similarly $EFCD$ is also a parallelogram.

The two parallelograms $ABFE$ and $EFCD$ have equal bases $BF$ and $FC$, and have the same height $AP$. Hence they have equal areas. Thus area$\left(ABFE\right)=$ area$\left(EFCD\right)$.