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Question

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

Prove that : IA = IB = IC.

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Solution

Given: In Δ ABC; ID, IE, IF are perpendicular bisector of sides BC, AC and AB respectively.

and I is the circumcentre of ? 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

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