Dear student, AB is the chord of a circle Join AC and BD CD⊥⊥AB such that ∠CDA=∠CDB=90°∠CDA=∠CDB=90° Also as CD is the perpendicular bisector of AB so AD = DB CD = CD (Using reflexive property) Therefore â–³CDAâ–³CDA and â–³CDBâ–³CDB 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 Regards

Please solve q24!

Dear student,

AB is the chord of a circle. Join AC and BD.
CD⊥⊥AB such that ∠CDA=∠CDB=90°∠CDA=∠CDB=90°.
Also as CD is the perpendicular bisector of AB so AD = DB
CD = CD (Using reflexive property)
Therefore △CDA△CDA and △CDB△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.

Regards

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