The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P. Prove that if AB > AC, then PC > PB.

In △ ABC,

⇒ AB > AC (let)

∴ ∠ACB > ∠ABC (If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.)

⇒ 180° - ∠ACB < 180° - ∠ABC

⇒ ∠BCE < ∠CBD

⇒ $\dfrac{∠BCE}{2} \lt \dfrac{∠CBD}{2}$

⇒ ∠BCP < ∠CBP (Since, BP and CP are bisectors of external angles at B and C.)

In △ BCP,

Since, ∠BCP < ∠CBP

∴ PB < PC (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)

⇒ PC > PB.

Hence, proved that if AB > AC, then PC > PB.