Question

6. The perpendicular bisectors of the sides of a
triangle ABC meet at I.
Prove that : IA = B = IC.​

Answer 1

well its obvious

As perpendicular bisectors intersect at circumcentre and so lA= lB = lC = radius

of circumscribed circle

As LE is perpendicular bisector

In ALB

LA = LB As L is on perpendicular bisector of AB

Ohk lets prove

In ALE and ELB

AEL= BEL= 90

LE = LE

BE = EA

So they are similar

So LA = LB

Similarly LD is perpendicular bisector of BC

So LB= LC

So LA = LB = LC