If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a '4-gon') are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)                                                                      [Page No. 142]

Let ABCD be the given 4-gon.

Let P, Q, R and S be the midpoints of sides AB, BC, CD and DA respectively.

Join P, Q, R and S in order.

So, PQRS is the parallelogram formed.

Now, draw diagonal AC of the 4-gon.

In triangle ABC, P and Q are the midpoints of AB and BC.

Therefore, by the midpoint theorem: $PQ \parallel AC$ and $\frac{PQ}{AC} = \frac{1}{2}$.

In $\Delta$ADC, S and R are the midpoints of AD and DC.

$$\text{Therefore, } SR \parallel AC \text{ and } SR = \frac{1}{2} AC$$

So, PQ || SR and PQ = SR

Hence, PQRS is a parallelogram.

Now, the four corner triangles are $\Delta$APS, $\Delta$BPQ, $\Delta$CQR and $\Delta$DRS.

Each of these triangles has half the base and half the height of the corresponding triangle into which the 4-gon is divided.

So, the total area of the four corner triangles is half the area of the original 4-gon.

Therefore, the remaining middle parallelogram PQRS has the other half of the area.

Hence, $\text{area}(\text{parallelogram } PQRS) = \frac{1}{2} \times \text{area}(4\text{-gon } ABCD)$

Hence proved.