Question

In quadrilateral ABCD, AB is the longest side CD is the shortest side. Prove that $\angle A>\angle C$.

Answer 1

In $\Delta ADC,DC<AD$

$\angle 2<\angle 4$ (angles opposite to greater side is greater)

$\Delta ABC,AB>BC$

$\angle 3>\angle 1$ (angles opposite to greater side is greater)

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

Hence, proved.