The perpendicular bisectors of the sides of $△ A B C$ meet at $I$ . Then:

I A = I B \neq I C

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I B = I C \neq I A

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I A = I C \neq I B

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I A = I B = I C

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Solution

The correct option is D $I A = I B = I C$
Given:
$△ A B C$ , $I D$ , $I E$ and $I F$ are perpendicular bisectors.
To Prove:
$I A = I B = I C$
In $Δ I B D$ and $Δ I C D$
$B D = C D$ ( $I D$ bisects $B C$ )
$∠ I D B = ∠ I D C = 90^{\circ}$
$I D = I D$ (Common)
Thus,
$Δ I B D ≅ Δ I C D$ (SAS rule)
Thus,
$I B = I C$ (By CPCT)
Similarly,
$I A = I B = I C$

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