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fatimanamra03
@fatimanamra03
July 2021
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Prove that the perpendicular bisector of the chord of a circle always passes through its centre.
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Akkasshhhh
AB is the chord of a circle. Join AC and BD.
CD perpendicular toAB such that .angle CDA=AngleCDB=90°
Also as CD is the perpendicular bisector of AB so AD = DB
CD = CD (Using reflexive property)
Therefore triangle CDA and triangle CDB are congruent triangles.
Then, CA = CB
Since the center of the circle is the only point within the circle that has points on the circumference equal distance from it.
Hence, C is the center of the circle.
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Answers & Comments
AB is the chord of a circle. Join AC and BD.
CD perpendicular toAB such that .angle CDA=AngleCDB=90°
Also as CD is the perpendicular bisector of AB so AD = DB
CD = CD (Using reflexive property)
Therefore triangle CDA and triangle CDB are congruent triangles.
Then, CA = CB
Since the center of the circle is the only point within the circle that has points on the circumference equal distance from it.
Hence, C is the center of the circle.