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General

Easy

Question

# The perpendicular bisectors of the sides of a triangle ABC meet at point I. prove that IA=IB=IC

Hint:

### Find the congruence rules used and thus find the equal sides.

## The correct answer is: IA = IB = IC

In ΔABC, in which AD is perpendicular bisector of BC

BE is perpendicular bisector of CA

CF is perpendicular bisector of AB

AD, BE and CF meet at I

To prove IA=IB=IC

In ΔBID and ΔCID

BD = DC (given)

(AD is perpendicular bisector of BC)

ID = ID (common)

Hence by SAS congruence rule, we have

ΔBID ⩭ ΔCID

So, IB = IC (Corresponding part of congruent triangle)

Similarly, In ΔCIE and ΔAIE

CE = AE (given)

(BE is perpendicular bisector of AC)

IE = IE (common)

By SAS congruence rule, ΔCIE ⩭ ΔAIE

IC = IA (Corresponding part of congruent triangle)

Thus, IA = IB = IC

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