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.Mathematics

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