Question

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

Answer 1

In the given ΔABC, AD bisects ∠Aand ∠C >∠B. We need to prove ∠ADB >∠ADC.

 

Let,

 ∠BAD = ∠1

∠DAC = ∠2

∠ADB = ∠3

∠ADC = ∠4

Also, 

As AD bisects ∠A, 

∠1 = ∠2…..(1) 

Now, in ΔABD, using exterior angle theorem, we get,

∠4 = ∠B + ∠1

Similarly,

∠3 = ∠2 + ∠C

∠3 = ∠1 + ∠C  [using (1)]

Further, it is given,

∠C >∠B

Adding ∠1to both the sides

∠C +∠1 >∠B + ∠1

∠3 > ∠4

Thus, ∠3 > ∠4

Hence proved.