Answer :

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


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