In the given figure, AB and AC we two equal chords of a circle with centre O. Show that O lies on the bisectors of ∠BAC.
Given: AB = AC
Construction: join OA, OB and OC
Proof:
Consider ΔAOB and ΔAOC
Here,
OC = OB (radius)
OA = OA (common)
AB = AC (given)
∴ By SSS congruency
ΔAOB ΔAOC
∴ ∠OAC = ∠OAB (by C.P.C.T)
Hence, we can say that OA is the bisector of ∠BAC, that is O lies on the bisector of ∠BAC.