Question

Let ABCD be a quadrilateral in which AB is the smallest side and CD is the largest side. Prove that ∠A > ∠C and ∠B > ∠D. (Hint: Join AC and BD.)

Answer 1

Let us join AC.

In ΔABC:

AB < BC (AB is the smallest side of quadrilateral ABCD)

∴ ∠2 < ∠1 ... (1) (The angle opposite to a shorter side is smaller)

In ΔADC:

AD < CD (CD is the longest side of quadrilateral ABCD)

∴ ∠4 < ∠3 ... (2) (The angle opposite to the shorter side is smaller)

On adding equations (1) and (2), we obtain:

∠2 + ∠4 < ∠1 + ∠3

⇒ ∠C < ∠A

⇒ ∠A > ∠C

Let us join BD.

In ΔABD:

AB < AD (AB is the smallest side of quadrilateral ABCD)

∴ ∠8 < ∠5 ... (3) (The angle opposite to the shorter side is smaller)

In ΔBDC:

BC < CD (CD is the largest side of quadrilateral ABCD)

∴ ∠7 < ∠6 ... (4) (The angle opposite to the shorter side is smaller)

On adding equations (3) and (4), we obtain:

∠8 + ∠7 < ∠5 + ∠6

⇒ ∠D < ∠B

⇒ ∠B > ∠D