diagonals of a parallelogram ABCD intersect at point O. through O a line is drawn to intersect AD at P and BC at Q. show that PQ divides the parallelogram into 2 parts of equal areas

Asked by Ranjith Kumar Ranjith Kumar | 7th Mar, 2011, 09:55: PM

Expert Answer:

Dear Student,
Given: Parallelogram ABCD where the diagonals intersect at O. PQ is a line segment passing through O.
To prove: PQ divides the parallelogram ABCD into two parts of equal area.
OB = OD  (Diagonals of parallelogram bisect each other)
∠BOQ = ∠DOP  (Vertically opposite angles)
∠OBQ = ∠ODP  (Alternate interior angles)
∴ ΔBOQ ≅ ΔDOP  (ASA Congruence criterion)
⇒ Area (ΔBOQ) = Area (ΔDOP)  … (1)
The diagonal BD of parallelogram ABCD divides it into two triangles ΔBCD and ΔABD of equal areas
Area (ΔBCD) = Area (ΔABD)  … (2)
Now, area (QCDP) = area (ΔBCD) + area (ΔDOP) - area (ΔBOQ)
⇒ area (QCDP) = area (ΔABD) + area (ΔBOQ) - area (ΔDOP)  [Using (1) and (2)]
   = area (APQB)


Hope! You got the answer.



Team Topperlearning

Answered by  | 7th Mar, 2011, 04:25: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.