Q. 9

In a Δ ABC, AD bisects A and C >B. Prove that ADB >ADC.

Answer :

Given,

AB bisects A (DAB = DAC)


C > B


In ,


ADB + DAB + B = 180o (i)


In ,


ADC + DAC + C = 180o (ii)


From (i) and (ii), we get


ADB + DAB + B = ADC + DAC + C


ADB > ADC (Therefore, C > B)


Hence, proved


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