Question

In the adjoining figure, ABCD is a || gm and O is a point on the diagonal AC. Prove that ar(∆AOB) = ar(∆AOD).

Answer 1

Given: ABCD is a parallelogram and O is the point on diagonal AC.

To prove: ar(∆AOB) = ar(∆AOD)
Construction: Join DB to intersect AC at P.
Proof: Join DB to intersect AC at P.
Since the diagonals of a parallelogram bisect each other, P is the mid point of AC as well as that of BD.
We know that the median of a triangle divides it into two triangles of equal areas.
In ∆ABD, AP is the median.
i.e., ar(∆PAB) = ar(∆PAD) ...(i)
In ∆OBD, OP is the median.
i.e., ar(∆OPB) = ar(∆OPD) ...(ii)
Adding (i) and (ii), we get:
​ar(∆PAB) + ​ar(∆OPB) = ​ar(∆PAD) + ​ar(∆OPD)
⇒ ar(∆AOB) = ar(∆AOD)
​Hence, proved.