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Question
If the perpendicular bisector of the sides of a triangle PQR meet at I, then prove that the line joining from P, Q, R to I are equal.
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Solution
Given:
In ΔPQR,
PA is the perpendicular bisector of QR ⇒ QA = RA
RC is the perpendicular bisector of PQ ⇒ PC = QC
QB is the perpendicular bisector of PR ⇒ PR = RB
PA, RC and QB meet at I.
To prove: IP = IQ = IR
Proof:
In ΔQIA and ΔRIA
QA = RA ....[Given]
∠QAI = ∠RAI ....[Each = 90]
IA = IA ....[Common]
∴ By Side-Angle-Side criterion of congruence,
ΔIQ = IR ....(i)
Similarly, in ΔRIB and ΔPIB
RB = PB ...[Given]
∠RBI = ∠PBI ...[Each = 90°]
IB = IB ...[Common]
∴ By Side-Angle-Side criterion of congruence,
ΔRIB ≅ ΔPIB
The corresponding parts of the congruent triangles are congruent.
∴ IR = IP ....(ii)
From (i) and (ii), we have
IP = IQ = IR.
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