Question

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

Answer 1

Given,

AB bisects ∠A (∠DAB = ∠DAC)

∠C > ∠B

In ΔADB

∠ADB + ∠DAB + ∠B = 180°(i)

In ΔADC

∠ADC + ∠DAC + ∠C = 180°(ii)

From (i) and (ii), we get

∠ADB + ∠DAB + ∠B = ∠ADC + ∠DAC + ∠C

∠ADB > ∠ADC (Therefore, ∠C > ∠B)

Hence, proved