Question

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

Answer 1

It is given that

$\angle B<\angle A$ and $\angle C<\angle D$

We know that side opposite to larger angle is larger

$OD<OC$ .... (i)

$AO<BO$ .... (ii)

Adding eq(i) and eqn (ii), we get

$AO+OD<BO+OC$

$\Rightarrow AD<BC$ $\left[henceproved\right]$