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.

Name: 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.
Uploaded: Wed, 2021-01-06 01:41
Description: 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.

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.

Answers (1)

Solution.
Given in parallelogram ABCD, diagonals intersect at O and draw a line PQ, which intersects AD at P and BC at Q.
To prove: PQ divides the parallelogram ABCD into two part of equals area.
ar (ABQP) = ar(CDPQ)

We know that, diagonals of a parallelogram bisect each other.
\ OA = OC and OB = OD …(i)
In $\triangle AOB$ and $\triangle COD$
OA = OC and OB = OD [From eq. (i)]
and
ÐAOB = ÐCOD [Vertically opposite angles]
$\therefore \triangle AOB\cong \triangle COD$    [SAS congruence rule]
$ar(\triangle AOB)=ar(\triangle COD)$ …(ii) [Congruent figures have equal area]
In $\triangle AOP$ and $\triangle COQ$
ÐPAO = ÐOCQ [alternate interior angles]
OA = OC [From eq. (i)]
ÐAOP = ÐCOQ   [Vertically opposite angles]
$\triangle AOP\cong \triangle COQ$ [ASA congruence rule]
$ar(\triangle AOP)=ar(\triangle COQ)$ …(iii) [Congruent figures have equal area]
In $\triangle POD$ and $\triangle BOQ$
ÐPDO = ÐOBQ   [alternate interior angles]
OD = OB [From eq. (i)]
ÐDOP = ÐBOQ   [Vertically opposite angles]
$\triangle DOP\cong \triangle BOQ$ [ASA congruence rule]
$ar(CDPQ)=ar(\triangle COQ)+ar(\triangle COD)+ar(\triangle POD)$ …(iv) [Congruent figures have equal area]
$=ar(\triangle AOP)+ar(\triangle AOB)+ar(\triangle BOQ)$ [From ii, iii, iv]
= ar(ABQP)
$\Rightarrow$ ar(ABQP) = ar(CDPQ)
Hence Proved.