The sides AB and AC of △ABC are produced to D and E respectively and the bisectors of ∠CBD and ∠BCE meet at O. If AB > AC, prove that OC > OB.

In △ABC,

AB > AC

⇒ ∠ACB > ∠ABC

∠ACB + ∠BCE = 180° …..(1) [Linear pair]

∠ABC + ∠CBD = 180° …..(2) [Linear pair]

Adding eq.(1) and (2), we have:

∠ACB + ∠BCE = ∠ABC + ∠CBD

Since, ∠ACB > ∠ABC

⇒ ∠BCE < ∠CBD

⇒ $\dfrac{1}{2}$ ∠BCE < $\dfrac{1}{2}$ ∠CBD

⇒ ∠BCO < ∠CBO

⇒ ∠CBO > ∠BCO

∴ OC > OB.

Hence, proved that OC > OB.