Question

In trapezium ABCD,  AB||CD, DC=2AB. EF||AB cuts AD in F and BC in E.$\frac{BE}{EC}= \frac{4}{4}$ . DB intersects EF at G .PROVE THAT 7EF=11AB

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  
Hope This Helps :)

Answer 2

I think it's 10AB...so according to that the answer is given below