Given,

AB bisects ∠A (∠DAB = ∠DAC)

∠C > ∠B

In ,

∠ADB + ∠DAB + ∠B = 180^o (i)

In ,

∠ADC + ∠DAC + ∠C = 180^o (ii)

From (i) and (ii), we get

∠ADB + ∠DAB + ∠B = ∠ADC + ∠DAC + ∠C

∠ADB > ∠ADC (Therefore, ∠C > ∠B)

Hence, proved