AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50). Show that ∠ A >∠ C and∠ B >∠ D

It is given in the question that:

In ΔABD

AB < AD < BD

Therefore,

∠ADB < ABD (i) (Angles opposite to longer side is larger)

Now,

In ΔBCD

BC < DC < BD

Therefore,

∠BDC < ∠CBD (ii)

Adding (i) and (ii), we get

∠ADB + ∠BDC < ∠ABD + ∠CBD

∠ADC < ∠ABC

∠B > D

Similarly,

In ΔABC

∠ACB < ∠BAC

(iii) (Angle opposite to longer side is larger)

Now,

In ADC

∠DCA < ∠DAC (iv)

Adding (iii) and (iv), we get

∠ACB + ∠DCA < ∠BAC + ∠DAC

∠BCD < ∠BAD

∠A > ∠C