Question

Prove that the perpendicular from the centre of a circle to a chord bisects the chord.

Answer 1

To prove that the perpendicular from the centre to a chord bisect the chord.

Consider a circle with centre at $O$ and $AB$ is a chord such that $OX$ perpendicular to $AB$

To prove that $AX=BX$

In $\Delta OAX$ and $\Delta OBX$

$\angle OXA=\angle OXB$ [both are 90 ]

$OA=OB$ (Both are radius of circle )

$OX=OX$ (common side )

$\Delta OAX\cong \Delta OBX$ by RHS rule.

$AX=BX$ (by property of congruent triangles )

hence proved.