Let parallel to $ CD$ through $ X$ cut $ AC, BD$ at $ C_0, D_0$ and let $ X_0 \equiv CD_0 \cap DC_0.$ Since $ CA, DB$ are medians of $ \triangle XCD$ $ \Longrightarrow$ $ \triangle X_0C_0D_0$ is anticomplementary to $ \triangle XCD$ $ \Longrightarrow$ $ X$ is midpoint of $ D_0C_0$ and $ D_0X = XC_0 = CD.$ $ \angle AC_0X = \angle CC_0D_0 = \angle DCC_0 = \angle DCA = \angle DBA = \angle D_0BX\ (\mod \pi)$ $ \angle BD_0X = \angle BD_0C_0 = \angle CDD_0 = \angle CDB = \angle CAB = \angle C_0AX\ (\mod \pi)$ $ \angle C_0XA = \angle D_0XB$ $ \Longrightarrow$ $ \triangle XAC_0 \sim \triangle XD_0B$ are oppositely similar $ \Longrightarrow$ $ AC_0BD_0$ is cyclic $ \Longrightarrow$ $ AX \cdot XB = D_0X \cdot XC_0 = CD^2.$ By AM-GM, $ AB = AX + XB \ge 2 \sqrt {AX \cdot XB} = 2 CD$ or $ \frac {AB}{CD} \ge 2.$

By Menelaus $ \frac{|DE|}{|EB|}+\frac{|EC|}{|EA|}=1$ where $ E$ is the intersection point of the diagonals. $ |AE|=a, \, |BE|=b, \, |CE|=c, \, |DE|=d$ $ ac=bd, \quad \frac{d}{b}+\frac{c}{a}=1$ $ \Longrightarrow \; \frac{|AB|}{|CD|}=\frac{a}{d}=\frac{b}{c}=\frac{c}{d}+\frac{d}{c}=\frac{a}{b}+\frac{b}{a} \geq 2$ Inequality holds when $ AB || CD$

actually,let DX and AC,CX and DB intersect at Y,Z then $DC=2YZ;\frac{AB}{CD}=\frac{AB}{2YZ}$ let AC,BD intersect at K,XK,YZ intersect at L.then $\frac{XK}{LK}=4$.Using Liuchang Lee's theorem(i,e,the length of a triangle's median is greater than that of its conjugate line),since ABK is similar to $ZYK$,then $\frac{AB}{YZ}=\frac{XK}{TK}\ge\frac{XK}{LK}=4$(where TK is the conjugate line of LK) hence $\frac{AB}{CD}\ge 2$.