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Question

The perpendicular from the centre of a circle to a chord bisects the chord.


Solution

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 )
$$ΔOAX\cong ΔOBX$$
$$AX = BX$$  (by property of congruent triangles )
hence proved.

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Mathematics

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