The diagonal 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 ABCD into two parts of equal area.
Hi!
Here is the answer to your question.
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.
Cheers!