Question

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

Answer 1

In the figure it is given that $\angle B<\angle A$ and $\angle C<\angle D$
Consider triangle $AOB$
Since $\angle B<\angle A$
We get
$AO<BO\dots \dots \left(1\right)$
Consider triangle $COD$
Since $\angle C<\angle D$
$DO<CO\dots \dots .\left(2\right)$
By adding both the equations we get
$AO+DO<BO+CO$
So we get
$AD<BC$
Therefore, it is proved that $AD<BC$.