Question

prove that the perpendicular from the centre of the circle to the chord bisects the chord​

Answer 1

Answer:

C is a circle with a centre at O

AB is a chord such that OX AB

To prove: OX bisect chord AB ie. AX = BX

In ∆OA × and ∆OBX

<OXA = <OXB (Both 90° given)

OA = OB (Both radius)

OX = OX (Common)

So, ∆OAX = ∆OB × (RHS rule)

AX = BX (CPCT)