Question

In trapezium ABCD,  AB||CD, DC=2AB. EF||AB cuts AD in F and BC in E.$\frac{BE}{EC}= \frac{3}{4}$ . DB intersects EF at G .PROVE THAT 7EF=10AB

Answer 1

As per figure, in triangles BGE and BDC, GE and DC are parallel, hence angle BGE = angle BDC and angle BEG = angle BCD, hence triangles BGE and BDC are similar. 
(1) GE/DC = BE/BC = 3/7 [ BE : EC = 3 : 4 (given), ==> BC =BE+EC = 7] 
    OR GE = (3/7)xDC = 2 x (3/7) x AB as DC = 2xAB (Given).
     OR GE = (6/7)xAB
(2) DG/DB = CE/CB = 4/7

(3) In triangles DFG and DAB, FG is parallel to AB. Hence triangles DFG and DAB are similar. [For similar reasons as for triangles BGE and BDC]

Therefore FG/AB = DG/DB = 4/7 [from (2) above]
                 OR FG = (4/7)xAB 

Hence EF =  GE + FG = (6/7)xAB + (4/7)xAB = (10/7)xAB
==> 7EF = 10AB  

Answer 2

$<b>Step-by-step explanation:</b>$

$<u>From figure:</u>$

(i)

In ΔDFG and ΔDAB, we have

⇒ ∠FDG = ∠ADB

∴ ΔDFG ~ ΔDAB

⇒ (DF/DA) = (FG/AB)

(ii)

In trapezium ABCD, EF ║ AB ║ DC

⇒ (AF/DF) = (BE/EC)

⇒ (AF/DF) = 4/3

⇒ (AF/DF) + 1 = 4/3 + 1

⇒ (AF + DF)/DF = 7/3

⇒ AD/DF = 7/3

⇒ DF/AD = 3/7  

(iii)

From (i) & (ii), we have

⇒ FG/AB = 3/7

⇒ FG = (3/7) AB

(iv)

From ΔBEG and ΔBCD, we have

⇒ ∠BEG = ∠BCD

∴ ΔBEG ~ ΔBCD

⇒ (BE/BC) = EG/CD

⇒ 4/7 = EG/CD

⇒ EG = (4/7) CD

⇒ EG = (4/7) * 2AB

⇒ EG = (8/7) AB

On solving (iii) & (iv), we get

⇒ FG + EG = (3/7) AB + (8/7) AB

⇒ EF = (11/7)AB

⇒ $\sf{\bold{\large{7\: EF = 11\: AB.}}}$

Hope it helps!