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# 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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