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 equal parts of equal area.
Asked by Sthitaprajna Mishra
| 4th Feb, 2014,
08:17: PM
Expert Answer:
Consider the following diagram.
Now consider the triangles DOP and BOQ.
OD = OB [diagonals of a parallelogram bisect at O]
Thus, by Angle-Side-Angle criterion of congruence,
.
Since congruent triangles have equal areas, we have
...(1)
The diagonal BD divides the parallelogram into two triangles,
of equal areas.
Therefore, we have
...(2)
Now consider the quadrilateral QCDP.
Thus PQ divides the parallelogram ABCD into two equal quadrialterals QCDP and AQCB.


OD = OB [diagonals of a parallelogram bisect at O]






Answered by
| 5th Feb, 2014,
11:34: AM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Sign Up
Verify mobile number
Enter the OTP sent to your number
Change