Question

In ΔABC, AB < AC, PB and PC are the bisectors of ∠B and ∠C. Prove that PB < PC.

Answer 1

In ΔABC,

• Given that AB < AC.
• PB and PC are the angle bisectors of ∠B and ∠C.
• Prove that PB < PC.

Step 1: Relation between sides and angles

From the property of triangles:

AB < AC ⟹ ∠C < ∠B (opposite side rule: larger side opposite larger angle).

So we know: 

∠B > ∠C

Step 2: Angle bisector property

PB is the bisector of ∠B.

PC is the bisector of ∠C.

So: 

∠PBC = ∠PBA = `1/2 angleB`

∠PCB = ∠PCA = `1/2 angleC`

Step 3: Compare ∠PBC and ∠PCB

Since ∠B > ∠C, we get:

`1/2 angleB > 1/2 angleC`

= ∠PBC > ∠PCB

Step 4: Use triangle inequality logic in ΔPBC

In ΔPBC:

The larger angle lies opposite the longer side.

∠PBC is larger than ∠PCB.

So, side PC (opposite ∠PBC) is longer than side PB (opposite ∠PCB).

= PB < PC