AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD. Show that angle A is greater than angle C and angle B is greater than angle D.

In ΔABC, we have

BC > AB       

⇒ ∠BAC > ∠BCA  ...(1)

Similarly, in  ΔACD, we have

CD > AD       

⇒ ∠CAD > ∠ACD  ...(2)

On adding (1) and (2), we get

∠BAC + ∠CAD > ∠BCA + ∠ACD

⇒∠BAD > ∠BCD

⇒ ∠A > ∠C

Now, in Δ ABD, we have

AD > AB

⇒ ∠ABD > ∠ADB  ...(3)      [ AB is the smallest side]

Similarly, in Δ BCD, we have

CD > BC  ...(4)  [CD is the largest side]

⇒ ∠DBC > ∠BDC

On adding (3) and (4), we get

∠ABD + ∠DBC > ∠ADB + ∠BDC

∠ABC > ∠ADC

⇒∠B > ∠D

Hence, it is proved that ∠A > ∠C and ∠B > ∠D