Question

Prove that the perpendicular bisector of the chord of a circle always passes through its centre.

Answer 1

AB is the chord of a circle. Join AC and BD.

CD perpendicular toAB such that .angle CDA=AngleCDB=90°

Also as CD is the perpendicular bisector of AB so AD = DB

CD = CD (Using reflexive property)

Therefore triangle CDA and triangle CDB are congruent triangles.

Then, CA = CB

Since the center of the circle is the only point within the circle that has points on the circumference equal distance from it.

Hence, C is the center of the circle.