In figure (1) given below, ABCD is a parallelogram. Points P and Q on BC trisect BC into three equal parts. Prove that :

area of ∆APQ = area of ∆DPQ = $\dfrac{1}{6}$ (area of ||gm ABCD)

Construct: Through P and Q, draw PR and QS parallel to AB and CD.

Area of ∆APD = Area of ∆AQD [Since ∆APD and ∆AQD lie on the same base AD and between the same parallel lines AD and BC]

Area of ∆APD – Area of ∆AOD = Area of ∆AQD – Area of ∆AOD [On subtracting ar ∆AOD on both sides]

Area of ∆APO = Area of ∆OQD ……. (i)

Area of ∆APO + Area of ∆OPQ = Area of ∆OQD + Area of ∆OPQ [On adding area of ∆OPQ on both sides]

Area of ∆APQ = Area of ∆DPQ ……. (ii)

We know that, ∆APQ and ||gm PQSR are on the same base PQ and between the same parallel lines PQ and AD.

Area of ∆APQ = $\dfrac{1}{2}$ Area of ||gm PQRS ……. (iii)

From figure,

Height of || gm ABCD = Height of || PQRS = AE

Since, P and Q trisect BC so,

⇒ PQ = $\dfrac{BC}{3}$

⇒ BC = 3PQ.

$\Rightarrow \dfrac{\text{Area of || gm ABCD}}{\text{Area of || PQRS}} = \dfrac{BC \times AE}{PQ \times AE} \\[1em] \Rightarrow \dfrac{\text{Area of || gm ABCD}}{\text{Area of || PQRS}} = \dfrac{3PQ \times AE}{PQ \times AE} \\[1em] \Rightarrow \dfrac{\text{Area of || gm ABCD}}{\text{Area of || PQRS}} = 3 \\[1em] \Rightarrow \text{Area of || PQRS} = \dfrac{1}{3} \text{Area of || gm ABCD}.$

Substituting above value in (iii) we get,

⇒ Area of ∆APQ = $\dfrac{1}{2} \times \dfrac{1}{3}$ Area of ||gm ABCD = $\dfrac{1}{6}$ Area of ||gm ABCD.

⇒ Area of ∆APQ = $\dfrac{1}{6}$ Area of ||gm ABCD ……..(iv)

From (ii) and (iv) we get,

area of ∆APQ = area of ∆DPQ = $\dfrac{1}{6}$ (area of ||gm ABCD).

Hence, proved that area of ∆APQ = area of ∆DPQ = $\dfrac{1}{6}$ (area of ||gm ABCD).