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.


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