In the figure (iii) given below, ABCD is a parallelogram. E is a point on AB, CE intersects the diagonal BD at O and EF || BC. If AE : EB = 2 : 3, find

(i) EF : AD

(ii) area of △BEF : area of △ABD

(iii) area of △ABD : area of trap. AEFD

(iv) area of △FEO : area of △OBC.

(i) Considering △ADB and △EFB,

∠ B = ∠ B (Common angles)
∠ DAB = ∠ FEB (Corresponding angles are equal)

Hence, by AA axiom △ADB ~ △EFB.

We know that when triangles are similar the ratio of the corresponding sides are equal,

$\therefore \dfrac{AB}{BE} = \dfrac{AD}{EF} \\[1em] \Rightarrow \dfrac{EF}{AD} = \dfrac{EB}{AB} \\[1em]$

Given, $\dfrac{AE}{EB} = \dfrac{2}{3}$.

$\Rightarrow \dfrac{AB- EB}{EB} = \dfrac{2}{3} \\[1em] \Rightarrow 3(AB - EB) = 2EB \\[1em] \Rightarrow 3AB - 3EB = 2EB \\[1em] \Rightarrow 3AB = 2EB + 3EB \\[1em] \Rightarrow 3AB = 5EB \\[1em] \Rightarrow \dfrac{EB}{AB} = \dfrac{3}{5}.$

Since, $\Rightarrow \dfrac{EF}{AD} = \dfrac{EB}{AB} = \dfrac{3}{5}.$

Hence, EF : AD = 3 : 5.

(ii) Considering △ABD and △BEF,

∠ B = ∠ B (Common angles)
∠ DAB = ∠ FEB (Corresponding angles are equal)

Hence, by AA axiom △ADB ~ △BEF.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △BEF}}{\text{Area of △ABD}} = \dfrac{EF^2}{AD^2} \\[1em] \Rightarrow \dfrac{\text{Area of △BEF}}{\text{Area of △ABD}} = \dfrac{3^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △BEF}}{\text{Area of △ABD}} = \dfrac{9}{25}.$

Hence, area of △BEF : area of △ABD = 9 : 25.

(iii) From (ii) we get,

$\therefore \dfrac{\text{Area of △ABD}}{\text{Area of △BEF}} = \dfrac{25}{9}$

⇒ 9 x Area of △ABD = 25 x Area of △BEF
⇒ 9 x Area of △ABD = 25 x (Area of △ABD - Area of trapezium AEFD)
⇒ 9 x Area of △ABD = 25 x Area of △ABD - 25 x Area of trapezium AEFD
⇒ 16 x Area of △ABD = 25 x Area of trapezium AEFD

$\therefore \dfrac{\text{Area of △ABD}}{\text{Area of trapezium AEFD}} = \dfrac{25}{16}$

Hence, area of △ABD : area of trapezium AEFD = 25 : 16.

(iv) Considering △FEO and △OBC,

∠ FOE = ∠ BOC (Vertically opposite angles are equal)
∠ FEO = ∠ OCB (Alternate angles are equal)

Hence, by AA axiom △FEO ~ △OBC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \Big(\dfrac{EF}{BC}\Big)^2$

Since, in parallelogram opposite sides are equal so, BC = AD.

$\Rightarrow \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \Big(\dfrac{EF}{AD}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \dfrac{3^2}{5^2} \\[1em] \Rightarrow \dfrac{\text{Area of △FEO}}{\text{Area of △OBC}} = \dfrac{9}{25}.$

Hence, the ratio of the area of △FEO : area of △OBC = 9 : 25.