prove that a perpendicular drawn to a chord, which
is not a diameter of the circle, from the centre
of the circle, bisect the chord
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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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