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.

Solution:

Given, ABCD is a parallelogram

The diagonals AC and BD intersect at a point O.

A line through O is drawn to intersect AD at P and BC at Q.

We have to show that PQ divides the parallelogram into two parts of equal area.

We know that the diagonals of a parallelogram bisect each other.

So, OA = OC

OB = OD

Considering triangles AOB and COD,

OA = OC

OB = OD

We know that the vertically opposite angles are equal.

∠AOB = ∠COD

By SAS criteria, the triangles AOB and COD are similar.

We know that congruent triangles have equal area

So, ar(AOB) = ar(COD) ----------------------- (1)

Considering triangles AOP and COQ,

We know that the alternate interior angles are equal

∠PAO = ∠OCQ

OA = OC

We know that the vertically opposite angles are equal

∠AOP = ∠COQ

ASA criterion states that two triangles are congruent, if any two angles and the side included between them of one triangle are equal to the corresponding angles and the included side of the other triangle.

By ASA criteria, the triangles AOP and COQ are congruent.

We know that congruent triangles have equal area

So, ar(AOP) = ar(COQ) ----------------------- (2)

Similarly, ar(POD) = ar(BOQ) ---------------- (3)

Now, ar(ABQP) = ar(COQ) + ar(COD) + ar(POD)

From (1), (2) and (3),

ar(ABQP) = ar(AOP) + ar(AOB) + ar(BOQ)

Therefore, ar(ABQP) = ar(CDPQ)

✦ Try This: In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar (△ AOD) = ar (△ BOC).

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9

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NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 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.

Summary:

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. It is shown that PQ divides the parallelogram into two parts of equal area

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