Q5 of 47 Page 7

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


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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