Q. 124.8( 5 Votes )

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

Answer :

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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