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)
Δ AIE ≅ Δ CIE
⇒ AI = IC ......... (2)
from (1) and (2)
IA = IB = ICRegards