Question

Prove that: The perpendicular bisector of any chord of a circle passes through the centre of that circle.
From the diagram shown on the right, prove that the line JK consists of points equidistant from G and I and one of these points must be the centre of the circle. Help please~ please give a full working out ty.

Answer 1

prove the opposite

Answer 2

Using RHS rite..? both triangle are 90 degrees, both ave 1 similar side OH, and both hypotenuse are equal.. but by just doing this then did I proved that all the points on the line works as well? what if i pick the point which is not the center of the circle?

Answer 3

OG=OI...both radius
OH common
GH=IH given