The perpendicular bisectors of the sides of a △ABC meet at I. Prove that : IA = IB = IC.

From figure,

AL, CN and BM are perpendicular bisectors of sides BC, AB and AC respectively.

In △BIL and △CIL,

⇒ BL = LC (Since, AL is the perpendicular bisector of BC)

⇒ ∠BLI = ∠CLI (Both equal to 90°)

⇒ LI = LI (Common side)

∴ △BIL ≅ △CIL (By S.A.S axiom)

⇒ IB = IC …..(1) (Corresponding parts of congruent triangles are equal)

In △CIM and △AIM,

⇒ CM = AM (Since, BM is the perpendicular bisector of AC)

⇒ ∠CMI = ∠AMI (Both equal to 90°)

⇒ MI = MI (Common side)

∴ △CIM ≅ △AIM (By S.A.S axiom)

⇒ IC = IA …..(2) (Corresponding parts of congruent triangles are equal)

From eq.(1) and (2), we get:

⇒ IA = IB = IC.

Hence, proved that IA = IB = IC.