Question

5. The perpendicular bisectors of the sides of a
triangle ABC meet at I.
Prove that : IA = IB = IC.

pls do it in a copy!​

Answer 1

$\huge \sf {\orange {\underline {\pink{\underline {A᭄ɴsᴡᴇʀ࿐ :−}}}}}$

Hence Proved....

Answer 2

Proof

in triangle BID and triangle CID

BD = CD............... (given)

<BDI = <CDI = 90° [AD perpendicular bisector of BC]

DI = DI............. (common )

By SAS theoram

triangle BAD and triangle CID are congruent

so IB = IC

similarly in triangle CIE and triangle AIE

CE = AE............. (given)

<CEI = < AEI = 90° [AD perpendicular bisector of AC ]

IE = IE............. (common )

By SAS theoram

triangle CIE and triangle AIE are congruent

The corresponding parts of the congruent triangles are congruent

therefore. IC= IA

Thus. IA = IB = IC