Question

In quadrilateral $ABCD$, side $AB$ is the longest and side $DC$ is the shortest. Hence, $\angle D>\angle B$

If true enter 1 else 0.

Answer 1

In $△ABD$
AB > AD

$\angle BDA>\angle DBA$ (Angle opposite to the greater side is greater) ...(1)
In $△BDC$

$BC>CD$

$=>\angle BDC>\angle CBD$ (Angle opposite to the greater side is greater) ...(2)
On adding
equations (1) and (2), we obtain
$\angle BDC+\angle BDA>\angle DBA+\angle CBD$
$=>\angle D>\angle B$