In the figure,$\angle B < \angle A$ and $\angle C < \angle D$. Show that $AD < BC$.
(N/A) Given: $\angle B < \angle A$ and $\angle C < \angle D$.
In $\triangle AOB$,since $\angle B < \angle A$,the side opposite to the greater angle is longer. Therefore,$OA < OB$ (side opposite to $\angle B$ is $OA$ and side opposite to $\angle A$ is $OB$) ...... $(1)$
In $\triangle COD$,since $\angle C < \angle D$,the side opposite to the greater angle is longer. Therefore,$OD < OC$ (side opposite to $\angle C$ is $OD$ and side opposite to $\angle D$ is $OC$) ...... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$OA + OD < OB + OC$
Since $OA + OD = AD$ and $OB + OC = BC$,we have:
$AD < BC$.
In $\triangle AOB$,since $\angle B < \angle A$,the side opposite to the greater angle is longer. Therefore,$OA < OB$ (side opposite to $\angle B$ is $OA$ and side opposite to $\angle A$ is $OB$) ...... $(1)$
In $\triangle COD$,since $\angle C < \angle D$,the side opposite to the greater angle is longer. Therefore,$OD < OC$ (side opposite to $\angle C$ is $OD$ and side opposite to $\angle D$ is $OC$) ...... $(2)$
Adding equations $(1)$ and $(2)$,we get:
$OA + OD < OB + OC$
Since $OA + OD = AD$ and $OB + OC = BC$,we have:
$AD < BC$.
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