$$ABCD$$ is a trapezium. Sides AB and $$CD$$ are parallel to each other. $$AB$$ = 6 cm, $$CD$$ = 18 cm, $$BC$$ = 8 cm and $$AD$$ = 12 cm. $$A$$ line parallel to $$AB$$ divides the trapezium in two parts of equal perimeter. This line cuts $$BC$$ at $$E$$ and $$AD$$ at $$F$$. If $$\frac{BE}{EC} = \frac{AF}{FD}$$, than what is the value of $$\frac{BE}{EC}$$?

Given ,

$$\frac{BE}{EC}=\frac{AF}{FD}\ .$$

So, $$\frac{8-EC}{EC}=\frac{12-FD}{FD}\ .\ \left(given,\ BC=8\ and\ AD=12\right)$$

So, $$\frac{EC}{FD}=\frac{8}{12}=\frac{2}{3}\ .$$

Let say, EC=2k and FD= 3k.

So,

So, AF=(12-3k) and  BE=(8-2k) .

According to question :

Perimeter of ABEF= Perimeter of FECD=(6+8+18+12)/2=22 cm .

So,

$$FE+3k+2k+18=22\ .$$

or, $$FE+12-3k+8-2k+6=22\ .$$

or, $$FE=\left(5k-4\right)\ .$$

Again, 

$$FE+CD+FD+EC=22\ .$$

or, $$3k+2k+18+5k-4=22\ .$$

or, $$10k=8\ .$$

or, $$k=\frac{8}{10}=\frac{4}{5}.$$

So, $$\frac{BE}{EC}=\frac{8-2k}{2k}=\frac{8-\frac{8}{5}}{\frac{8}{5}}=\frac{40-8}{8}=4\ .$$

C is correct choice.