the diagonals AC and BD of a ||gm ABCD meet at O. from mid-point of AD, a line MH is drawn || to DB meetiong AO at H and a line MK||AO meeting DO at K. prove that ar(||gm MHOK) = 1/16 ar(||gm ABCD)

Your question appears to be incorrect.

If the question is prove that:

ar (parallelogram MHOK) = ar (parallelogram ABCD) 

Then the solution is:

ABCD is a parallelogram and BD is its diagonal.

Since, a diagonal in a parallelogram divides it in two triangles of equal area,

So, ar(ΔABD) = ar(ΔBCD) 

Since ar(ABCD) = ar(ΔABD) + ar(ΔBCD) 

⇒ar(ΔABD) = ar(ΔBCD) =  ar(ABCD) 

Since, diagonals of a parallelogram bisect each other,

So, AO is the median of ΔABD.

So, ar(ΔAOD) = ar(ΔABD) 

Now, M is the midpoint of AD and MH is parallel to OD, 

So, by using the converse of mid point theorem, H is the mid point of AO.

In ΔAOD, OM is the median.

Since a median divides the triangle in two triangles of equal area,

ar(ΔAMO) =  ar(ΔOMD) = ar(ΔAOD) = ar(ΔABD) =  ar(ABCD) = ar(ABCD)

Now, MH is the median for ΔAOM 

So, ar(ΔMHO) = ar(ΔMKO) = ar(ΔAOM) = ar(ABCD) = ar(ABCD)

ar (MHOK)  =ar(ΔMHO) + ar(ΔMKO) = ar(ABCD) = ar(ABCD)