In the figure (2) given below, O is any point inside a parallelogram ABCD. Prove that

(i) area of ∆OAB + area of ∆OCD = $\dfrac{1}{2}$ area of || gm ABCD.

(ii) area of ∆OBC + area of ∆OAD = $\dfrac{1}{2}$ area of || gm ABCD.

(i) Draw a line PQ || to AB and CD from point O.

AB || PQ and AP || BQ (Since, AD || BC)

ABQP is a || gm

Similarly,

PD || CQ and PQ || DC

PQCD is a || gm

Now, ∆OAB and || gm ABQP are on the same base AB and between same || lines AB and PQ

Area of ∆OAB = $\dfrac{1}{2}$ Area of ||gm ABQP …..(1)

Similarly, ∆OCD and || gm PQCD are on the same base CD and between same || lines CD and PQ

Area of ∆OCD = $\dfrac{1}{2}$ Area of || gm PQCD ….. (2)

Now by adding (1) and (2),

Area of ∆OAB + Area of ∆OCD = $\dfrac{1}{2}$ Area of || gm ABQP + $\dfrac{1}{2}$ Area of || gm PQCD

= $\dfrac{1}{2}$ [Area of || gm ABQP + Area of || gm PQCD]

= $\dfrac{1}{2}$ Area of || gm ABCD

Hence, proved that Area of ∆OAB + Area of ∆OCD = $\dfrac{1}{2}$ Area of || gm ABCD.

(ii) From figure,

⇒ Area of ∆OAB + Area of ∆OBC + Area of ∆OCD + Area of ∆OAD = Area of || gm ABCD

⇒ Area of ∆OAB + Area of ∆OCD + Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD

⇒ $\dfrac{1}{2}$ Area of || gm ABCD + Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD

⇒ Area of ∆OBC + Area of ∆OAD = Area of || gm ABCD - $\dfrac{1}{2}$ Area of || gm ABCD

⇒ Area of ∆OBC + Area of ∆OAD = $\dfrac{1}{2}$ Area of || gm ABCD.

Hence, proved that Area of ∆OBC + Area of ∆OAD = $\dfrac{1}{2}$ Area of || gm ABCD.