AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠ A > ∠ C and ∠ B > ∠ D.

Let ABCD be a quadrilateral such that AB is it's smallest side and CD is it largest side.
Now join AC and BD 
In ΔABC ,we have 

BC>AB 

⇒∠8>∠3 ...(1) {∵ angle opposite to longer side is greater.}

Since CD is the longest side of quadrilateral ABCD
In ΔACD, we have 

CD>AD

⇒∠7>∠4 ...(2)
Adding (1) and(2) we get
∠8+∠7>∠3+∠4

⇒∠A>∠C

Now in ΔABD, we have 

AD>AB {∵AB is the shortest side}

⇒∠1>∠6 ...(3)

In ΔBCD, we have

CD>BC

⇒∠2>∠5 ....(4)

Adding (3) and (4)
∠1+∠2>∠5+∠6

⇒∠B>∠D

Hence proved.