Maths-
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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