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in a triangle ABC the perpendicular bisectors of the triangle meet at I.Prove that IA=IB=IC.

 Given : In ? ABC; ID, IE, IF are perpendicular bisector of sides BC, AC and AB respectively.
and I is the circumcentre of ? ABC
Now in ? IBD and ? ICD
BD = DC (ID is the bisector of BC)
?IDB = ? IDC = 90° (ID is the perpendicular to BC)
ID = ID (Common)
Thus ? IBD  ? ICD (by SAS congruence criterion)
? IB = IC  ...... (1)
Similarly
? AIE ? CIE
? AI = IC  ......... (2)

 from (1) and (2)
IA = IB = IC


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