In trapezium ABCD, AB || DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such that 4 BE = 3EC. Diagonal DB intersects FE at point G. Prove that : 7EF = 10AB.

Given,

⇒ 4BE = 3EC

⇒ $\dfrac{BE}{EC} = \dfrac{3}{4}$

In △DFG and △DAB,

⇒ ∠DFG = ∠DAB [As corresponding angles are equal]

⇒ ∠FDG = ∠ADB (Common)

⇒ △DFG ~ △DAB [By AA postulate]

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$\dfrac{DF}{DA} = \dfrac{FG}{AB}$ ……….(1)

In trapezium ABCD, we have

EF || AB || DC

$\therefore \dfrac{AF}{DF} = \dfrac{BE}{EC} \\[1em] \Rightarrow \dfrac{AF}{DF} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{AF}{DF} + 1 = \dfrac{3}{4} + 1 \\[1em] \Rightarrow \dfrac{AF + DF}{DF} = \dfrac{3 + 4}{4} \\[1em] \Rightarrow \dfrac{AD}{DF} = \dfrac{7}{4} \\[1em] \Rightarrow \dfrac{DF}{AD} = \dfrac{4}{7} \space ……….(2)$

From (1) and (2), we get :

$\dfrac{FG}{AB} = \dfrac{4}{7}$

FG = $\dfrac{4}{7}$AB ………(3)

In △BEG and △BCD,

⇒ ∠BEG = ∠BCD [As corresponding angles are equal]

⇒ ∠GBE = ∠DBC (Common angle)

⇒ △BEG ~ △BCD [By AA postulate]

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$\dfrac{BE}{BC} = \dfrac{EG}{CD}$ …………..(4)

Given,

$\Rightarrow \dfrac{BE}{EC} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{EC}{BE} = \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{EC}{BE} + 1 = \dfrac{4}{3} + 1 \\[1em] \Rightarrow \dfrac{EC + BE}{BE} = \dfrac{4 + 3}{3} \\[1em] \Rightarrow \dfrac{BC}{BE} = \dfrac{7}{3} \\[1em] \Rightarrow \dfrac{BE}{BC} = \dfrac{3}{7} \\[1em] \Rightarrow \dfrac{EG}{CD} = \dfrac{3}{7} \space [\text{From (4)}] \\[1em] \Rightarrow EG = \dfrac{3}{7}CD \\[1em] \Rightarrow EG = \dfrac{3}{7} \times 2AB \space [\because DC = 2AB] \\[1em] \Rightarrow EG = \dfrac{6}{7}AB \space ……….(5)$

Adding equation (3) and (5),

$\Rightarrow FG + EG = \dfrac{4}{7}AB + \dfrac{6}{7}AB \\[1em] \Rightarrow EF = \dfrac{10}{7}AB \\[1em] \Rightarrow 7EF = 10AB.$

Hence, proved that 7EF = 10AB.