Question

In the given figure, the diagonals AC and BD intersect at point O. If OB = OD and AB//DC,
show that:
(i) Area (Δ DOC) = Area (Δ AOB).
(ii) Area (Δ DCB) = Area (Δ ACB).
(iii) ABCD is a parallelogram.

Answer 1

(i)  The ratio of the area of triangles with the same vertex and bases along the same line is equal to the ratio of their respective bases. So, we have:

`["Area  of "Δ"DOC"]/["Area of"  Δ"BOC" ] = [DO]/[BO]` = 1 .....(i)

Similarly

`["Area of "Δ"DOA"]/["Area of" Δ"BOA" ] = [DO]/[BO]` = 1 ......(ii)

We know that the area of triangles on the same base and between the same parallel lines are equal.
Area of Δ ACD = Area of Δ BCD
Area of Δ AOD + Area of Δ DOC = Area of Δ DOC + Area of Δ BOC
⇒ Area of Δ AOD = Area of Δ BOC                      .....(iii)

From 1, 2 and 3 we have
Area (Δ DOC) = Area (Δ AOB)
Hence Proved.

(ii) Similarly, from 1, 2 and 3, we also have
Area of Δ DCB = Area of Δ DOC + Area of Δ BOC = Area of Δ AOB + Area of Δ BOC = Area of Δ ABC
So Area of Δ DCB = Area of Δ ABC
Hence Proved.

(iii) We know that the area of triangles on the same base and between the same parallel lines are equal.
Given: triangles are equal in the area on the common base, so it indicates AD || BC.
So, ABCD is a parallelogram.
Hence Proved.