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  • 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.

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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.

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