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.
 
Proof:
In ΔBOQ and ΔDOP,
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.

 

Regards

Team Topperlearning

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

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