Given AB and AC are two equal chords whose centre is O.
To prove Centre O lies on the bisector of `angleBAC`.
Construction Join BC, draw bisector AD of `angleBAC`.
Proof In `DeltaBAM and DeltaCAM`,
AB=AC [given]
`angleBAM=angleCAM ` [given]
and AM=AM [common side]
`:. DeltaBAM cong DeltaCAM` [by SAS congruence rule]
`rArr BM=CM` [by CPCT]
and `angleBMA=angleCMA` [by CPCT]
So, `BM=CM and angleBMA=angleCMA=90^(@)`
So, AM is the perpendicular bisector of the chord BC.
Hence, bisector of `angleBAC` i.e., AM passes through the centre O.
