In trapezium $ABCD$, $AB$ is parallel to $DC$. The diagonals $AC$ and $BD$ intersect at $X$, and $XY$ is constructed parallel to $AB$, intersecting at $X$, and $XY$ is constructed parallel to $AB$, intersecting $BC$ at $Y$.
a) Prove $AB:CD= BY:YC$.
b) In a certain trapezium, the length of $AB$ is $18$ cm. Given that $BY:BC$ = $3:4$, what is the length of the shorter side?
$$\angle XAB=\angle YXC=\angle XCD\;\; (why?)$$
$$\angle ABD =\angle BDC\;\;(why?)$$
from the above, it follows by a.a. that $\;\Delta ABX\sim\Delta CDX\;$ , so
$$\frac{AB}{CD}=\frac{AX}{CX}=\frac{BX}{DX}$$
Now just apply Thales Theorem , aka Intercept Theorem, to $\;\angle DBC\;$ and the intercepts $\;XY\,,\,DC\;$ , to get $\;\frac{BX}{DX}=\frac{BY}{CY}\;$ and we're done with (a) .
About (b): I've no idea what "shorter side" you write about...side of what?