In the figure ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Prove that $\frac{AE}{ED} = \frac{BF}{FC}$.
In the above figure, if $AD = 2.0$ cm, $AE = 1.8$ cm, $EC = 3.6$ cm and $DE \parallel BC$ the measure of $BD$ will be :
