$AB$ and $CD$ are the smallest and largest sides of a quadrilateral $ABCD$. Determine which is greater,$\angle B$ or $\angle D$.

(A) Given: $A$ quadrilateral $ABCD$ in which $AB$ is the smallest side and $CD$ is the largest side.
To prove: $\angle B > \angle D$.
Construction: Join $BD$.
Proof: In $\triangle ABD$,since $AB$ is the smallest side of the quadrilateral,we have $AD > AB$. The angle opposite to the longer side is greater,therefore $\angle ABD > \angle ADB$ ... $(1)$.
In $\triangle CBD$,since $CD$ is the largest side of the quadrilateral,we have $CD > BC$. The angle opposite to the longer side is greater,therefore $\angle CBD > \angle BDC$ ... $(2)$.
Adding $(1)$ and $(2)$,we get:
$\angle ABD + \angle CBD > \angle ADB + \angle BDC$
$\Rightarrow \angle ABC > \angle ADC$
$\Rightarrow \angle B > \angle D$.
Hence,it is proved that $\angle B$ is greater than $\angle D$.