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

Construction: join AC.
In ΔABC, BC > AB [∵ AB is the shortest side of the quadrilateral ABCD]
Hence, ∠1 > ∠3 …(1) [∵ In a triangle, longer side has greater angle opposite to it]
Similarly,
In ΔADC,
CD > AD [∵ CD is the longest side of the quadrilateral ABCD]
Hence, ∠2>∠3 + ∠4 …(2) [∵ In a triangle, Longer side has greater angle opposite to it]
From the equation (1) and (2), we have
∠1 +∠2 > ∠3 + ∠4
⇒ ∠A> ∠C
Construction: Join BD.
In ABD, AD> AB [∵ AB is the shortest side of the quadrilateral ABCD]
Hence, ∠5 > ∠7 …(3) [∵ In a triangle, longer side has greater angle opposite to it]
Similarly.
In ΔBDC, CD > BC [∵ CD is the longest side of the quadrilateral ABCD]
Hence, ∠6 > ∠8 …(4) [∵ In a triangle, longer side has greater angle opposite to it]
From the equation (3) and (4), we have
∠5 + ∠6 > ∠7 + ∠8 ⇒ ∠B > ∠D.