In the figure (3) given below, ABCD is a parallelogram. O is any point on the diagonal AC of the parallelogram. Show that the area of ∆AOB is equal to the area of ∆AOD.

Join BD, which meets AC at P.

In ∆ABD, AP is the median (As P is mid-point of BD because diagonals of || gm bisect each other).

Since, median of triangle divides it into two triangles of equal area.

∴ Area of ∆ABP = Area of ∆ADP …….(i)

Similarly,

PO is median of ∆BOD,

∴ Area of ∆BOP = Area of ∆POD …….(ii)

Now, adding (i) and (ii), and we get

⇒ Area of ∆ABP + Area of ∆BOP = Area of ∆ADP + Area of ∆POD

⇒ Area of ∆AOB = Area of ∆AOD.

Hence, proved that area of ∆AOB = area of ∆AOD.