Question

The diagonals $AC$ and $BD$ of a parallelogram $ABCD$ meet at $O$. From mid-point of $AD$, a line $MH$ is drawn parallel to $DB$ meeting $AO$ at $H$ and a line $MK\parallel AO$ meeting $DO$ at $K$. Prove that $ar$ (parallelogram$MHOK$)=$\frac{1}{8}ar$ (parallelogram $ABCD$)

Answer 1

$MK\parallel AO$ [ Given ]

So, $MK\parallel HO$ ---- ( 1 )

$\square ABCD$ is a parallelogram and $BD$ is its diagonal.

$\therefore$ $ar\left(ABD\right)=ar\left(BCD\right)=\frac{1}{2}ar\left(ABCD\right)$

Now, $M$ is the mid-point of $AD$ and $MH\parallel DO$. --- ( 2 )

From ( 1 ) and ( 2 )

$MKOH$ is a parallelogram.

$\Rightarrow$ $H$ is the mid-point of $AO$

Similarly, $k$ is the mid-point of $DO$

Now, $AO$ is the median of $△ABD$

$\Rightarrow$ $ar\left(AOD\right)=\frac{1}{2}ar\left(ABD\right)$

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

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

Then, $ar\left(AOM\right)=\frac{1}{2}ar\left(AOD\right)$

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

$=\frac{1}{8}ar\left(ABCD\right)$

$\Rightarrow$ $ar\left(MHO\right)=\frac{1}{2}\times \frac{1}{8}ar\left(ABCD\right)$

$=\frac{1}{16}ar\left(ABCD\right)$ ----- ( 3 )

Similarly $ar\left(MKO\right)=\frac{1}{16}ar\left(ABCD\right)$ ---- ( 4 )

Adding ( 3 ) and ( 4 ),

$\Rightarrow$ $ar\left(MHO\right)+ar\left(MKO\right)=\frac{1}{16}ar\left(ABCD\right)+\frac{1}{16}ar\left(ABCD\right)$

$\Rightarrow$ $ar\left(MHOK\right)=\frac{1}{8}ar\left(ABCD\right)$

$\Rightarrow$ $ar\left({\parallel }^{gm}MHOK\right)=\frac{1}{8}ar\left({\parallel }^{gm}ABCD\right)$