In ΔABC, the bisectors of ∠ABC and ∠ACB meet at I. If AB > AC then let’s prove that IB > IC.

Given: ΔABC, the bisectors of ∠ABC and ∠ACB meet at I and AB > AC

To prove: IB > IC

The figure for the given criteria is as shown below,

Now given BD is bisector of ∠ABC, so

∠ABD = ∠DBC

Or ∠ABC = 2∠DBC…………(i)

Now given CE is bisector of ∠ACB, so

∠ACE = ∠ECB

Or ∠ACB = 2∠ECB…………(ii)

Given AB > AC

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒ ∠ACB > ∠ABC

Now substituting values from equation (i) and (ii) in above equation, we get

⇒ 2∠ECB > 2∠DBC

⇒ ∠ECB > ∠DBC…….(iii)

Now consider ΔIBC,

From equation (iii), we have

∠ICB > ∠IBC

But we know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

Hence IB > IC

Hence proved