Question

In a trapezium ABCD, AB || DC and the diagonals AC and BD intersect at point ‘O’. Prove that area (ΔАOD) = area (ΔBОС).

Answer 1

Given: ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect at O.

To Prove: Area (ΔAOD) = Area (ΔBOC).

Proof (Step-wise):

1. Since AB || DC, alternate interior angles give ∠OAB = ∠OCD and ∠OBA = ∠ODC.

2. Therefore, ΔOAB ∼ ΔOCD.   ...(AA similarity)

3. From the similarity, corresponding sides are proportional.

So, `(OA)/(OC) = (OB)/(OD)`. 

Hence, OA × OD = OB × OC   ...(1)

4. Angles ∠AOD and ∠BOC are vertically opposite formed by intersection of AC and BD.

So, ∠AOD = ∠BOC.

Hence, sin ∠AOD = sin ∠BOC = sin θ.   ...(2)

5. `"Area" (ΔAOD) = 1/2 xx OA xx OD xx sin ∠AOD`   ...(Using `1/2 xx ab xx sin C` formula).

From (1) and (2), this equals `1/2 xx OB xx OC xx sin ∠BOC = "Area" (ΔBOC)`.

Therefore, Area (ΔAOD) = Area (ΔBOC), as required.