The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line a drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

Explanation

Given In a parallelogram ABCD, diagonals interect at O and draw a line PQ, which intersects AD and BC.
To prove PQ divides the parallelogram ABCD into two parts of equal area.
i.e., `" " ar (ABQP) = ar (CDPQ)`

Proof We know that, diagonals of a parallelogram bisect each other.
`therefore" "` `OA = OC " and " OB = OD" "` ...(i)
In `DeltaAOB` and `DeltaCOD`,
`OA = OC`
`OB = OD" "` [from Eq. (i)]
and `" "angle AOB = angleCOD" "` [vertically opposite angles]
`therefore" "` `DeltaAOB ~= DeltaCOD" "` [by SAS congruence rule]
Then, `" "ar (DeltaAOB) = ar (DeltaCOD)" "` ...(ii)
[since, cogruent figures have equal area]
Now, in `DeltaAOP` and `DeltaCOQ`
`anglePAO = angleOCQ" "` [alternate interior angles]
`OA = OC" "` [from Eq. (i)]
and `" " angleAOP = angleCOQ" "` [vertically opposite angles]
`therefore" "` `DeltaAOP ~= DeltaCOQ" "` [by ASA congruence rule]
`therefore" "` `ar (DeltaAOP) = ar (DeltaCOQ)" "` ...(iii)
[since, congruent figures have equal area]
Similarly, `" " ar (DeltaPOD) = ar (DeltaBOQ)" "` ...(iv)
Now, `" " ar (ABQP) = ar (DeltaCOQ) + ar (DeltaCOD) + ar (DeltaPOD)`
`=ar (DeltaAOP) + ar (DeltaAOB) + ar (DeltaBOQ)" "` [from Eqs. (ii), (iii) and (iv)]
`rArr" "` `ar (ABQP) = ar (CDPQ)" "` Hence proved.