Q12 of 31 Page 65

The two chords AB and AC of a circle are equal. I prove that, the bisector of ∠BAC passes through the centre.


Given, AC = AB, BAF = CAF


To prove: FB = FC


In ABF and CAF


AC = AB (Given)


BAF = CAF (Given)


AF = AF (Common)


ABF CAF


BFA = CFA (CPCT)


FB = FC (CPCT)


AE is a perpendicular bisector of chord BC, So, It has to pass through the center of the circle.


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