In Fig. 7.49, $ \angle \mathrm{B}
"
Given:
$\angle B<\angle A$ and $\angle C<\angle D$.
To do:
We have to show that $AD$
Solution:
Let us consider $\triangle OAB$
We know that,
The side opposite the smaller angle is always smaller.
This implies,
$AO < OB$
In a similar way,
In $\triangle ODC$ we get,
$OD < OC$
By adding equation (a) with (b),
We get,
$AO+OD < BO+OC$
Therefore,
$AD < BC$.
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