Question

AB and CD are respectively the smallest and longest sides of a

quadrilateral ABCD. Show that ∠ A > ∠ C and ∠ B > ∠ D.​

Answer 1

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.