P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD show that ar(∆APB) = ar ∆(BQC).

Extend CB to G and drop perpendiculars from point P and Q on AB and BG respectively as shown

If we consider AB as the base of parallelogram ABCD then PF is the height and if we consider BC as the base of parallelogram ABCD then BG is the height

So we can write area of parallelogram ABCD in two ways

Area of parallelogram = base × height

Considering AB as base

⇒ Area of parallelogram ABCD = AB × PF …(i)

Considering BC as base

⇒ Area of parallelogram ABCD = BC × QH …(ii)

Now consider ΔABP

PF is the height

Base = AB

Area of triangle = × base × height

⇒ Area of ΔABP = × AB × PF

Using (i)

⇒ Area of ΔABP = × area of parallelogram ABCD …(iii)

Now consider ΔCQB

QH is the height

Base = BC

Area of triangle = × base × height

⇒ Area of ΔCQB = × BC × QH

Using (ii)

⇒ Area of ΔCQB = × area of parallelogram ABCD …(iv)

From (iii) and (iv)

Area of ΔABP = Area of ΔCQB