In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

Solution:

In $\triangle \mathrm{AOB}$

$\angle B<\angle A$

$\Rightarrow \mathrm{AO}<\mathrm{BO}$ (Side opposite to smaller angle is smaller) ...(1)

In $\triangle \mathrm{COD}$

$\angle C<\angle D$

$\Rightarrow O D

$ (Side opposite to smaller angle is smaller) ... (2)

On adding equations (1) and (2), we obtain

$\mathrm{AO}+\mathrm{OD}<\mathrm{BO}+\mathrm{OC}$

$\mathrm{AD}<\mathrm{BC}$