2016/017) In a quadrilateral ABCD ab is the smallest and CD is the largest side. Prove that angles -- (1) $A > C$ and $B > D$

Join AC. In triangle ABC $AB < BC$ so $\angle BAC > \angle BCA$
in triangle ADC $CD > AD$ so $\angle CAD > \angle DCA$
adding above 2 we get the result. Similarly the 2nd part