Given: IP ⊥BC, IQ ⊥CA and IR ⊥ AB and the bisectors of ∠B and ∠C of ∆ABC meet at I

To prove: IP=IQ=IR and IA bisects ∠A

Proof:

In ∆BIP and ∆BIR we have,

∠PBI = ∠RBI …given

∠IRB = ∠IPB = 90° …Given

IB = IB …common side

Thus by AAS property of congruence,

∆BIP ≅ ∆BIR

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ IP = IR

Similarly,

IP = IQ

Hence, IP = IQ = IR

Now in ∆AIR and ∆AIQ

IR = IQ …proved above

IA = IA … Common side

∠IRA = ∠IQA = 90°

Thus by SAS property of congruence,

∆AIR ≅ ∆AIQ

Hence, we know that, corresponding parts of the congruent triangles are equal

∴ ∠IAR = ∠IAQ

This means that IA bisects ∠A