Given a parallelogram ABCD where X and Y are the mid-points of the sides BC and CD respectively. Prove that :

ar (△ AXY) = $\dfrac{3}{8} \times$ ar (//gm ABCD)

Given: ABCD is a parallelogram where X and Y are the mid-points of the sides BC and CD respectively.

To prove: ar (△ AXY) = $\dfrac{3}{8}$ x ar (//gm ABCD)

Construction: Join BD and AC. Join AY, AX and XY.

Proof: We know that the diagonal of a parallelogram divides it into two triangles of equal areas.

Ar.(Δ ACD) = Ar.(Δ ABC) = Ar.(Δ BCD) = Ar.(Δ ABD) = $\dfrac{1}{2}$ Ar.(∥gm ABCD)

In Δ ACD, Y is the mid-point of DC. So, AY is the median.

Median of a triangle divides it into two triangles of equal areas.

∴ Ar.(Δ AYD) = $\dfrac{1}{2}$ Ar.(Δ ADC)

= $\dfrac{1}{2} \times \dfrac{1}{2}$ Ar.(//gm ABCD)

= $\dfrac{1}{4}$ Ar.(//gm ABCD)

In Δ BCD, X and Y are the mid-points of sides BC and CD respectively.

⇒ CY = $\dfrac{1}{2}$ CD

⇒ XY = $\dfrac{1}{2}$ BD

So, sides of Δ CXY are half of the sides of the Δ CBD.

Ar.(Δ CXY) = $\dfrac{1}{4}$ Ar.(Δ CBD)

= $\dfrac{1}{4} \times \dfrac{1}{2}$ Ar.(//gm ABCD)

= $\dfrac{1}{8}$ Ar.(//gm ABCD)

Now, area of Δ AXY = area of //gm ABCD - [area of Δ ADY + area of Δ ABX + area of Δ CXY]

= Ar.(//gm ABCD) - [$\dfrac{1}{4}$ Ar.(//gm ABCD) + $\dfrac{1}{4}$ Ar.(//gm ABCD) + $\dfrac{1}{8}$ Ar.(//gm ABCD)]

= Ar.(//gm ABCD) - $\dfrac{(2 + 2 + 1)}{8}$ Ar.(//gm ABCD)

= Ar.(//gm ABCD) - $\dfrac{5}{8}$ Ar.(//gm ABCD)

= $\dfrac{(8 - 5)}{8}$ Ar.(//gm ABCD)

= $\dfrac{3}{8}$ Ar.(//gm ABCD)

Hence, ar (△AXY) = $\dfrac{3}{8}$ x ar (//gm ABCD).