Question

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.

Answer 1

Given: A quadrilateral $ABCD$ in which the mid-point of the sides of it are joined in order of form parallelogram $PQRS$

To prove: $ar\left(||gmPQRS\right)=\frac{1}{2}ar\left(ABCD\right)$

Construction: Join $BD$ and draw perpendicular from $A$ and $BD$ which intersect $SR$ and $BD$ at $X$ and $Y$ respectively.

Proof: In $\Delta ABD$, $S$ and $R$ are the mid-points of sides $AB$ and $AD$ respectively.

$\therefore SR||BD$

$\Rightarrow SX||BY$

$\Rightarrow X$ is the mid-point oy $AY$ [Converse of mid-point theorem]

$\Rightarrow AX||XY$...........(1)

($\because S$ is the mid-point of $AB$ and $SX||BY$)

$SR=\frac{1}{2}BD$.......(2) ($\because$ Mid-point theorem)

Now, $ar\left(\Delta ABD\right)=\frac{1}{2}\times BD\times AY$

$ar\left(\Delta ASR\right)=\frac{1}{2}\times SR\times AX$

$ar\left(\Delta ASR\right)=\frac{1}{2}\times \left(\frac{1}{2}BD\right)\times \left(\frac{1}{2}AY\right)SR\times AX$ (Using (1) and (2))

$ar\left(\Delta ASR\right)=\frac{1}{4}\times (\frac{1}{2}BD\times AY)$ (Using (1) and (2))

$\Rightarrow ar\left(\Delta ASR\right)=\frac{1}{4}\times \left(\Delta ABD\right)$.......(3)

Similarly,

$\Rightarrow ar\left(\Delta CPQ\right)=\frac{1}{4}\times \left(\Delta CBD\right)$...........(4)

$\Rightarrow ar\left(\Delta BPS\right)=\frac{1}{4}\times \left(\Delta BAC\right)$...........(5)

$\Rightarrow ar\left(\Delta DRQ\right)=\frac{1}{4}\times \left(\Delta DAC\right)$...........(6)

Adding (3),(4),(5) and (6), we get,

$ar\left(\Delta ASR\right)+ar\left(\Delta CPQ\right)+ar\left(\Delta BPS\right)+ar\left(\Delta DRQ\right)$

$=\frac{1}{4}ar\left(\Delta ABD\right)+\frac{1}{4}ar\left(\Delta CBD\right)+\frac{1}{4}ar\left(\Delta ABC\right)+\frac{1}{4}ar\left(\Delta DAC\right)$

$=\frac{1}{4}\left[ar\right(\Delta ABD)+ar(\Delta CBD)+ar(\Delta BAC)+ar(\Delta DAC\left)\right]$

$=\frac{1}{4}\left[ar\right(ABCD)+ar(\Delta ABCD\left)\right[$

$=\frac{1}{4}\times 2ar\left(ABCD\right)$

$=\frac{1}{2}\times 2ar\left(ABCD\right)$

$ar\left(\Delta ASR\right)+ar\left(\Delta CPQ\right)+ar\left(\Delta BPS\right)+ar\left(\Delta DRQ\right)$

$=\frac{1}{2}\times 2ar\left(ABCD\right)$

$\Rightarrow ar\left(ABCD\right)-ar\left(||gmPQRS\right)=\frac{1}{2}ar\left(ABCD\right)$

$\Rightarrow ar\left(||gmPQRS\right)=ar\left(ABCD\right)-\frac{1}{2}ar\left(ABCD\right)$

$\Rightarrow ar\left(||gmPQRS\right)=dfrac12ar\left(ABCD\right)$

Hence, proved.