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Question
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
Solution
Given: A ΔABC in which AD is the perpendicular bisector of BC
BE is the perpendicular bisector of CA
CF is the perpendicular bisector of AB
AD, BE and CF meet at I
WE need to prove that
IA = IB= IC
Proof:
In ΔBID and ΔCID
BD = DC ...[ Given ]
∠BDI = ∠CDI = 90°...[ AD is the perpendicular bisector of BC]
DI = DI ...[ Common ]
∴ By the Side-Angle-Side criterion of congruence,
Δ BID ≅ Δ CID
The corresponding parts of the congruent triangles are congruent.
∴ IB = IC ...[ c.p.c.t ]
Similarly, in Δ CIE and Δ AIE
CE = AE ...[ Given ]
∠CEI = ∠AEI = 90° ...[ AD is the perpendicular bisector of BC ]
IE = IE ...[ Common ]
∴ By Side-Angel-Side Criterion of congruence,
ΔCIE ≅ ΔAIE
The corresponding parts of the congruent triangles are congruent.
∴ IC = IA ...[ c.p.c.t ]
Thus, IA = IB = IC
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