Question

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that $\angle A>\angle C$

Answer 1

$AB$ and $CD$ are smallest and longest side of quadrilateral ABCD respectively

Join $AC,$ In $△ABC$

Here, $BC>AB$

$\therefore \angle 1>\angle 2\to \left(1\right)$ [angle opposite to the longer side is greater].

In $△ADC$

$CD>AD$

$\therefore \angle 3>\angle 4\to \left(2\right)$[angle opposite to the longer side is greater].

Adding (1) and (2) , we get

$\angle 1+\angle 3>\angle 2+\angle 4\Rightarrow \angle A>\angle C$