In trapezium ABCD. AB||DC and DC = 2AB. A line segment 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 &EF= 10AB.

Explanation

Since GE||DC
`(BE)/(EC)= (BG)/(GD)` (by B.P theorem ) …(1)
Also GF||AB
`(AF)/(FD)= (BG)/(GD)` (by B.P theorem) ….(2)
from (1) and (2) , we get
`(AF)/(FD) = (BE)/(EC) = 3/4 `
In `triangleBGE and triangleBDC`
`angle1=angle2` (corresponding `angles` as GE||DC)
`angle3=angle3` (common)
`triangleBGE~ triangleBDC` (AA corocllary)
`(GE)/(DC)=(BE)/(BC)` (corresponding sides of similar triangles are proportional)
`(GE)/(2AB)=3/7 ( DC = 2AB) `
`Rightarrow (GE)/(AB)=6/7` ....(4)
Now in ` triangleDFG and triangleDAB`
`angle4=angle5` ( corresponding `angles` as FG||AB)
`angle6=angle6` (common)
`triangleDFG~triangleDAB` (AA corollary)
`(FG)/(AB) = (DF)/(DA)`
`Rightarrow (FG)/(AB)4/7` ( from 3)...(5))
Adding (4) and (5) , we get
`(GE)/(AB)+(GF)/(AB)= 6/7+4/7`
`(GE+GF)/(AB)= 10/7`
`(EF)/(AB)=10/7`
` Rightarrow 7EF= 10AB` Hence proved.