Question

In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that
(i) $\mathrm{ar}(∆OAB)+\mathrm{ar}(∆OCD)=\frac{1}{2}\mathrm{ar}\left(\right||\mathrm{gm}ABCD),$
(ii) $\mathrm{ar}(∆OAD)+\mathrm{ar}(∆OBC)=\frac{1}{2}\mathrm{ar}\left(\right||\mathrm{gm}ABCD).$

Answer 1

Constructions: Draw EOF || AB and GOH || AD.
Proof: EOF || AB and DA cuts them.
∴ ∆OAB and parallelogram EABF being on the same base and between the same parallels AB and EF, we have:
ar(∆OAB) = $\frac{1}{2}$ ar(parallelogram EABF) ...(i)
Similarly, ar(∆OCD) = $\frac{1}{2}$ ar(parallelogram EFCD) ...(ii)
On adding (i) and (ii), we get:
ar(∆OAB) + ar(∆OCD) = $\frac{1}{2}$ ⨯ ar(parallelogram EABF)​ + $\frac{1}{2}$ ar(parallelogram EFCD)​
⇒ ar(∆OAB) + ar(∆OCD) = $\frac{1}{2}$ ⨯ ​ar(parallelogram ABCD)

(ii)
∴ ∆OAD and parallelogram AHGD being on the same base and between the same parallels AD and GH, we have:
ar(∆OAD) =$\frac{1}{2}$ar(parallelogram AHGD) ...(iii)
Similarly, ar(∆OBC) = $\frac{1}{2}$ar(parallelogram BCGH) ...(iv)

On adding (iii) and (iv), we get:
ar(∆OAD) + ar(∆OBC) =$\frac{1}{2}$⨯ ar(parallelogram AHGD)​ + $\frac{1}{2}$ar(parallelogram BCGH)​
⇒ ar(∆OAD) + ar(∆OBC) = $\frac{1}{2}$​​ ⨯​ ar(parallelogram ABCD)​