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Question

In the given figure, the bisectors of ∠B and ∠C of ∆ABC meet at I. If IP ⊥ BC, IQ ⊥ CA and IR ⊥ AB, prove that (i) IP = IQ = IR, (ii) IA bisects ∠A.

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Solution

In IPC and IQC, we have:IPC=IQC ...(90° each)

PCI=QCI (CI is the bisector of angle C) IC=IC (Common side)Thus, IPCIQC (AAScriterion)

IP=IQ (Corresponding parts of congruent triangles) ......1

Similarly, in IPB and IRB, we have:IPB=IRB (90° each)

PBI=RBI (BI is the bisector of angle B) IB=IB (Common side)Thus, IPBIRB (AAScriterion)

IP=IR (Corresponding parts of congruent triangles) .....2

From (1) and (2), we have,

IP = IQ = IR

Now consider the right triangles, ARI and AQI.

IR=IQ (Proved above) AI = AI (Common side)ARI=AQI (90° each)ARIAQI (RHS criterion)RAI=QAI (Corresponding parts of congruent triangles)

Thus, IA bisects angle A.

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