In triangle ABC; ∠A = 60°, ∠C = 40° and bisector of angle ABC meets side AC at point P. Show that BP = CP.

In △ ABC,

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 60° + ∠B + 40° = 180°

⇒ ∠B + 100° = 180°

⇒ ∠B = 180° - 100° = 80°.

Since, BP is bisector of ∠B,

∴ ∠PBC = $\dfrac{∠B}{2} = \dfrac{80°}{2}$ = 40°.

∵ ∠PBC = ∠PCB (Both equal to 40°)

∴ CP = BP (Sides opposite to equal angles are equal)

Hence, proved that BP = CP.