Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the angle bisector of ZBAC.
GIVEN Two equal chords AB and AC of a circle C (0, r).
.
TO PROVE Centre 0 lies on the bisector of LBAC, which intersects the chord BC in M.
CONSTRUCTION Join BC. Draw bisector AD of LBAC
PROOF In triangles BAM and CAM, we have
LBAM = ZCAM
and, perpendilar AM=AM
So, by SAS-criterion of congruence, we have
AB = AC
Δ BAM Δ CAM
BM = CM and ZBMA LCMA
BM = CM and ZBMA = ZMA 90’
AM is the perpendicular bisector of the chord BC.
AM passes through the centre 0.
Hence, the centre of the circle lies on the angle bisector of perpendilar BAC.