In given figure, AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

ABCD is a quadrilateral such that AB is its smallest side and CD is its largest side. Join AC and BD. Since AB is the smallest side of quadrilateral ABCD.
∴In △ABC, we have BC > AB ⇒ ∠8 > ∠3 (1) [Since angle opposite to longer side is greater]
Since CD is the longest side of quadrilateral ABCD. Now, In △ACD, we have CD > AD ⇒ ∠7 > ∠4 (2) [Since angle opposite to longer side is greater]
Adding (1) and (2), we get ∠8 + ∠7 > ∠3 + ∠4 ⇒ ∠A > ∠C
Again, in △ABD, we have AD > AB [Since AB is the shortest side] ⇒ ∠1 > ∠6 (3)
In △BCD, we have CD > BC [Since CD is the longest side] ⇒ ∠2 > ∠5 (4)
Adding (3) and (4), we get ∠1 + ∠2 > ∠5 + ∠6 ⇒ ∠B > ∠D
Thus, ∠A > ∠C and ∠B > ∠D. [hence proved]