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


AB < AD < BD


Therefore,


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


Now,


In


BC < Dc < BD


Therefore,


BDC < CBD (ii)


Adding (i) and (ii), we get


ADB + BDC < ABD + CBD


ADC < ABC


B > D


Similarly,


In


ACB < BAC (iii) (Angle opposite to longer side is larger)


Now,


In


DCA < DAC (iv)


Adding (iii) and (iv), we get


ACB + DCA < BAC + DAC


BCD < BAD


A > C


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