The perpendicular bisectors of the sides of a triangle ABC meet at I.

Prove that : IA = IB = IC.

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.