Question

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

Answer 1

In ΔABC,

∠A = 60°

∠C = 40°

∴ ∠B = 180° - 60° - 40°

⇒ ∠B = 80°

Now,

BP is the bisector of ∠ABC.

∴ ∠PBC = `"∠ABC"/2`

⇒ ∠PBC = 40°

In ΔPBC,

∠PBC = ∠PCB = 40°

∴ BP = CP             ...[Sides opp. to equal angles are equal.]