1.

Prove that the perpendicular bisector of a chord of a circle always passesthrough the centre.4.

Answer»

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 = DBCD = CD (Using reflexive property)Thereforetriangle CDAand triangle CDBare congruent triangles.Then, CA = CBSince 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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