Question

A perpendicular bisectors of the sides of a triangle ABC meet at I . prove that IA=IB=IC​

Answer 1

Given : In Δ ABC; ID, IE, IF are 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