Question

The perpendicular bisectors of the sides of a triangle ABC meet at I.

Prove that: IA = IB = IC.

Answer 1

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
AD, BE and CF meet at I

WE need to prove that
IA = IB= IC
Proof:
In  ΔBID and  ΔCID
BD = DC                 ...[ Given ]
∠BDI = ∠CDI = 90°...[ AD is the perpendicular bisector of BC]
DI = DI               ...[ Common ]
∴ By the Side-Angle-Side criterion of congruence,
 Δ BID ≅ Δ CID
The corresponding parts of the congruent triangles are congruent.
∴ IB = IC               ...[ c.p.c.t ]

Similarly, in  Δ CIE and  Δ AIE
CE = AE              ...[ Given ]
∠CEI = ∠AEI = 90° ...[ AD is the perpendicular bisector of BC ]
IE = IE                ...[ Common ]
∴ By Side-Angel-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