Question

$can \: ya \: all \: please \: help$
In trapezium ABCD, AB||CD=2AB.EF drawn parallel to AB cuts AD in F and BC in E such that BE/EC=3/4.Diagonal DB intersects EF at G.Prove that 7FE=10AB.

Answer 1

Hey ❤,

Well do refer to your own diagram,

✔

ATQ,

In triangles BGE and BDC, GE and DC are parallel,

Therefore,

Angle BGE = Angle BDC

Angle BEG = Angle BCD

So,

Triangles BGE and BDC are similar. 

✔

⤵

= GE/DC

= BE/BC

= 3/7

[ BE : EC = 3 : 4 (given), = BC = BE+EC = 7] 
   
OR

GE = (3/7) x DC

= 2 x (3/7) x AB

as DC = 2 x AB (Given)

    OR GE = (6/7) x AB....(1)

DG/DB = CE/CB = 4/7...(2)

⤴

⭐

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) x AB 

Hence

EF =  GE + FG

= (6/7) x AB + (4/7) x AB

= (10/7) x AB

7EF = 10AB  

⭐

Hence proved....!

Hope this helps you out! ♥