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.

 

Solution:

In ΔABD, we see that

AB < AD < BD

So, ∠ADB < ∠ABD — (i) (Since angle opposite to longer side is always larger)

Now, in ΔBCD,

BC < DC < BD

Hence, it can be concluded that

∠BDC < ∠CBD — (ii)

Now, by adding equation (i) and equation (ii) we get,

∠ADB + ∠BDC < ∠ABD + ∠CBD

∠ADC < ∠ABC

∠B > ∠D

Similarly, In triangle ABC,

∠ACB < ∠BAC — (iii) (Since the angle opposite to the longer side is always larger)

Now, In ΔADC,

∠DCA < ∠DAC — (iv)

By adding equation (iii) and equation (iv) we get,

∠ACB + ∠DCA < ∠BAC+∠DAC

⇒ ∠BCD < ∠BAD

∴ ∠A > ∠C