Question

Question 2
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

Answer 1

We know that diagonals of a parallelogram bisect each other.
OA = OC
OB = OD
and $\angle AOB=\angle COD$ [vertically opposite angles]
$\therefore \Delta AOB\cong \Delta COD$ [by SAS congruence rule]
Then,
$ar\left(\Delta AOB\right)=ar\left(\Delta COD\right)$
[Since, congruent figures have equal area]

Now, in $\Delta AOP$ and $\Delta COQ$,
$\angle PAO=\angle OCQ$ [alternate interior angles]
OA = OC
OB = OD [vertically opposite angles]
$\therefore \Delta AOB\cong \Delta COD$ [by SAS congruence rule]
Then, $ar(\Delta AOB=ar(\Delta COD)$
[since, congruent figures have equal area]

Now, in $\Delta AOP$ and $\Delta COQ$,
$\therefore \angle PAO=\angle OCQ$ [alternate interior angles]
OA = OC
$\angle AOP=\angle COQ$ [vertically opposite angles]
$\therefore \Delta AOP\cong \Delta COQ$ [by ASA congruence rule]
$\therefore ar\left(\Delta AOP\right)=ar\left(\Delta COQ\right)$
[since, congruent figures have equal area]

Similarly,
$ar\left(\Delta POD\right)=ar\left(\Delta BOQ\right)$

Now,
$ar\left(ABQP\right)=ar\left(\Delta COQ\right)+ar\left(\Delta COD\right)+ar\left(\Delta POD\right)$
$=ar\left(\Delta AOP\right)+ar\left(\Delta AOB\right)+ar\left(\Delta BOQ\right)$
$\Rightarrow ar\left(ABQP\right)=ar\left(CDPQ\right)$