In a trapezium ABCD, AB||DC and DC=2AB. EF||AB, where E and F lie on BC and AD respectively such that . . .

Question : In a trapezium ABCD, AB||DC and DC=2AB. EF||AB, where E and F lie on BC and AD respectively such that BE/EC=4/3. Diagonal DB intersects EF at G. Prove that 7EF=11AB.

Doubt by Muskan 

Solution :

Given : ABCD is a trapezium in which AB||DC.

DC=2AB

EF||AB

BE/EC = 4/3

BD is a diagonal.

To Prove : 7EF=11AB

Proof : 

AB||DC

EF||AB

⇒ EF||DC

∴ AB||EF||DC

BE/EC = BG/GD [By Converse of BPT]

BG/GD = AF/FD [By Converse of BPT]

⇒BE/EC = AF/FD

Also, BE/EC = 4/3 (Given)

∴ BE/EC = AF/FD = 4/3

EC/BE = FD/AF = 3/4

Now 

EC/BE = 3/4

Adding 1 both sides.

(EC/BE)+1 = (3/4)+1

BC/BE = 7/4

BE/BC = 4/7-------------(1)

Also
FD/AF = 3/4
AF/FD = 4/3
Adding 1 both sides

(AF/FD)+1 = (4/3)+1 
AD/FD = 7/3
FD/AD = 3/7
DF/DA = 3/7 ---------------(2)

In ∆BEG and ∆BCG

∠B = ∠B (Common)

∠BEG = ∠BCD (Corresponding Angle)

∆BEG ~ ∆BCG (by AA similarity criteria)

BE/BC = EG/CD (by CPST)

4/7 = EG/CD [From (1)]

EG/CD = 4/7

EG =4CD/7 ---------(3)

Similary ∆DFG~∆DAB

DF/DA=FG/AB (by CPST)

3/7 = FG/AB [(From (1)]

FG = 3AB/7 ---------(4)

Adding eq (3) and (4)

EG+FG = 4CD/7+3AB/7

EF = 4[2AB]/7+3AB/7 [∵CD=2AB]

EF = 8AB/7 + 3AB/7

EF = 11AB/7

7EF = 11AB