AB and CD are respectively, the smallest and the longest sides of a quadrilateral ABCD

AB and CD are respectively, the smallest and the longest sides of a quadrilateral ABCD. Show that ∠A >∠C and ∠B >∠D.
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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]
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Now, draw the diagonal BD in ∆ABD, AB is the shortest side.
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