Mathematics
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that : IA = IB = IC.
Triangles
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Answer
From figure,

AD, BE and CF are perpendicular bisectors of sides BC, AC and AB, respectively.
In △ BID and △ CID,
⇒ BD = CD (Since, AD is the perpendicular bisector of BC)
⇒ ∠BDI = ∠CDI (Both equal to 90°)
⇒ DI = DI (Common side)
∴ △ BID ≅ △ CID (By S.A.S. axiom)
We know that,
Corresponding parts of congruent triangles are equal.
By C.P.C.T.C.,
⇒ IB = IC ……..(1)
In △ CIE and △ AIE,
⇒ CE = AE (Since, BE is the perpendicular bisector of AC)
⇒ ∠CEI = ∠AEI (Both equal to 90°)
⇒ IE = IE (Common side)
∴ △ CIE ≅ △ AIE (By S.A.S. axiom)
Since, △ CIE and △ AIE are congruent.
By C.P.C.T.C.,
⇒ IC = IA ………(2)
From equation (1) and (2), we get :
⇒ IA = IB = IC.
Hence, proved that IA = IB = IC.
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