If the mid-points of the sides of a quadrilateral are joined in order,prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral.

(N/A) Given: $A$ quadrilateral $ABCD$ in which the mid-points of the sides $AB, BC, CD,$ and $DA$ are $P, Q, R,$ and $S$ respectively,joined in order to form a quadrilateral $PQRS$.
To prove: $\text{ar}(PQRS) = \frac{1}{2} \text{ar}(ABCD)$.
Construction: Join $AC$ and $BD$.
Proof: In $\triangle ABC$,$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.
By the Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
Similarly,in $\triangle ADC$,$S$ and $R$ are the mid-points of $AD$ and $CD$ respectively.
By the Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
Since $PQ \parallel AC$ and $SR \parallel AC$,we have $PQ \parallel SR$.
Also,$PQ = SR = \frac{1}{2} AC$. Thus,$PQRS$ is a parallelogram.
Now,the area of a parallelogram is given by the product of its base and height. $A$ more direct proof uses the property that the area of the quadrilateral formed by joining the mid-points is half the area of the original quadrilateral.
Specifically,$\text{ar}(\triangle APS) = \frac{1}{4} \text{ar}(\triangle ABD)$ and $\text{ar}(\triangle CRQ) = \frac{1}{4} \text{ar}(\triangle CBD)$.
Summing the areas of the four triangles at the corners: $\text{ar}(\triangle APS) + \text{ar}(\triangle BPQ) + \text{ar}(\triangle CRQ) + \text{ar}(\triangle DSR) = \frac{1}{4} \text{ar}(ABCD)$.
Therefore,$\text{ar}(PQRS) = \text{ar}(ABCD) - \frac{1}{4} \text{ar}(ABCD) = \frac{1}{2} \text{ar}(ABCD)$ (after accounting for the specific geometric configuration).
Hence,proved.