i. Let $AD$ meet $BC$ at $E$, $AC$ meet $BD$ at $F$, and let $M$ be the midpoint of $AB$. Since $AB > CD$, the points $D$ and $C$ must lie on the segments $AE$ and $BE$ respectively.
Since $\triangle EDC \sim \triangle EAB$, we have $\frac{EA}{EB} = \frac{DA}{CB} > 1$. This implies $EA > EB$, and hence $\angle CBA > \angle DAB$.

ii. By Steiner's theorem, $E$, $F$, $M$ are collinear. Since $EA > EB$, both $E$ and $F$ lie on the same side as $B$ to the perpendicular bisector of $AB$. This implies $FA > FB$. Now, as $\triangle FAB \sim \triangle FCD$, we have $\frac{AC}{BD} = \frac{FA}{FB} > 1$. This means $AC > BD$.