Maths-
General
Easy

Question

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

hintHint:

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)
    straight angle B D I equals straight angle C D I equals 90 to the power of ring operator(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)
    straight angle C E I equals straight angle A E I equals 90 to the power of ring operator(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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