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

First, draw the diagonal AC.

In ∆ABC, AB is the shortest side.

∴AB < BC

=> ∠2<∠1 …(i)

[∴angle opposite to the shortest side is smaller]

In ∆ACD, CD is the longest side.

AD < CD

=> ∠4 <∠3 … (ii)

[∴angle opposite to the longest side is longer]

Now, draw the diagonal BD in ∆ABD, AB is the shortest side.