**General enunciation:** In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. Prove that PB>PC.

**Particular enunciation:** In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. It is required prove that PB>PC.

Proof: PB is the bisector of ∠B.

∴∠PBC=½∠B

PC is the bisector of ∠C.

We know,

In a triangle the angle opposite the greater side is greater than the angle opposite the smaller side since AB>AC

∴ ∠ACB>∠ABC

⟹½∠ACB>½∠ABC

⟹∠PCB>∠PBC

⟹PB>PC

Therefore PB>PC.**(Proved)**