Question

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

Answer 1

Step-by-step explanation:

In △ADB

Sum of all angles of a triangle =180

o

∴∠B+∠ADB+

2

1

∠A=180

∠ADB=180

o

−∠B−

2

1

∠A

Similarly, ∠ADC=180

o

−∠C−

2

1

∠A

Since, $$\angle C > \angle B$$

Hence, $$\angle ADB > \angle ADC$$