In figure, $\angle B < \angle A$and $\angle C < \angle D$. Show that $AD < BC$.

Hint: Solve the problem by using the theorem of a triangle that is the opposite side of the higher angle is always greater, and the side opposite to the lowest angle is always the shortest.

Complete answer:
Given:
In $\Delta OAB$, $\angle B < \angle A$ and in $\Delta OCD$, $\angle C < \angle D$.
From $\Delta OAB$ by the theorem of side opposite to larger angle is longer, opposite side of $\angle B$ is less than opposite side of $\angle A$.
$\Rightarrow OA < OB$…...(1)
From $\Delta OCD$, the opposite side of $\angle C$ is less than the opposite side of $\angle D$.
$\Rightarrow OD < OC$.…..(2)
Add both equations (1) and (2).
$\begin{array}{c}
OA + OD < OB + OC\\
AD < BC
\end{array}$
Therefore, side AD is shorter than the side BC.

Note: While applying the theorem always remember to choose the opposite side of the larger angle, not adjacent side. This is a common error that is usually done by students so to eliminate this error remembers the correct concept of the theorem.