We know that the perpendicular from the centre of the circle to the chord bisects the
chord. That is in the figure if the length of chord AB is 8 cm, then AM-4 cm. If the diameter
of the circle is 10 cm, then OA =
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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 ΔOAX and ΔOBX
∠OXA=∠OXB [both are 90 ]
OA=OB (Both are radius of circle )
OX=OX (common side )
ΔOAX≅ΔOBX
AX=BX (by property of congruent triangles )
hence proved.
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