In figure, \(\angle{B}\) < \(\angle{A}\) and \(\angle{C}\) < \(\angle{D}\). Show that AD < BC.

We have, \(\angle{B}\) < \(\angle{A}\) ...(given)
Also, \(\angle{C}\) < \(\angle{D}\) ...(given)
Since, side opposite to greater angle is longer
we get, AO < BO ...(i)
similarly, OD < OC ...(ii)
On adding, Equation (i) and (ii), we have,
AO + OD < BO + OC
\(\therefore \) AD < BC
Hence, proved.