ABCD is a parallelogram. P and Q are the mid-points of sides AB and AD respectively. Prove that area of triangle APQ = $\dfrac{1}{8}$ of the area of parallelogram ABCD.

From figure,

ABCD is the parallelogram and BD is the diagonal of the parallelogram.

∴ BD divides || gm ABCD into two triangles of equal area.

∴ Area of Δ ABD = Area of Δ DBC = $\dfrac{1}{2}$ Area of || gm ABCD ……….(1)

Since, P is the mid-point of AB,

∴ DP is the median of Δ ABD.

∴ Area of Δ APD = Area of Δ DPB = $\dfrac{1}{2}$ Area of Δ ABD (Median divides triangle into two triangles of equal area)

⇒ Area of Δ APD = $\dfrac{1}{2}$ Area of Δ ABD

⇒ Area of Δ APD = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of || gm ABCD [From equation (1)]

⇒ Area of Δ APD = $\dfrac{1}{4}$ Area of || gm ABCD …….(2)

In Δ APD,

Q is the mid-point of AD.

∴ PQ is the median.

∴ Area of Δ APQ = Area of Δ DPQ = $\dfrac{1}{2}$ Area of Δ APD

⇒ Area of Δ APQ = $\dfrac{1}{2}$ Area of Δ APD

⇒ Area of Δ APQ = $\dfrac{1}{2} \times \dfrac{1}{4}$ Area of || gm ABCD [From equation (2)]

⇒ Area of Δ APQ = $\dfrac{1}{8}$ Area of || gm ABCD.

Hence, proved that area of Δ APQ = $\dfrac{1}{8}$ area of || gm ABCD.