the perpendicular bisectors of the sides of a triangle abcd meet at I. prove that IA=IB=IC
Answers
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Step-by-step explanation:
Given ; A ΔABC in which AD is the perpendicular bisector of BC
BE is the perpendicular bisector of CA
CF is the perpendicular bisector of AB
AC , BE and CF meet at I
We need to prove that
IA = IB = IC
In ΔBIC and ΔCID
BD + DC [Given]
∠BDI=∠CDI=90
o
[ AD is the perpendicular bisector of BC ]
BC = BC [ common]
By Side - Angle - Side criterion of congruence
ΔBID≅ΔCID
The corresponding parts of the congruent triangle are congruent
IB = IC [ c. p . c. t ]
Similarly , in ΔCIE and ΔAIE
CE = AE [ Given ]
∠CEI=∠AEI=90
o
[ AD is the perpendicular bisector of BC ]
IE = IE [ common ]
By Side - Angle - Side criterion of congruence
ΔCIE≅ΔAIE
The corresponding parts of the congruent triangles are congruent
IC = IA [ c.p. c. t ]
Thus , IA = IB = IC