4. 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
