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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