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)