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

 In Δ ABC; ID, IE, IF are the perpendicular bisector of sides BC, AC, and AB respectively.
 and I is the circumcentre of Δ ABC Now in Δ IBD and Δ ICD

BD = DC (ID is the bisector of BC)

Δ IDB = Δ IDC = 90° (ID is the perpendicular to BC)

ID = ID (Common)

Thus Δ IBD ≅ Δ ICD (by SAS congruence criterion)

⇒ IB = IC ...... (1)

Similarly

Δ AIE ≅ Δ CIE

⇒ AI = IC ......... (2)

from (1) and (2)

IA = IB = IC