. Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the angle bisector of <BAC.
I think it's based on congruency of two triangles formed by joining B and C to the centre, but I'm not sure...
Look at the picture :
I've drawn the segment bisectors of AB and AC. Any point that is on a segment bisector is at equal distance of the extremes of the segment.
D is the intersection point.
There is only one circle passing through 3 given distinct points. So D is the center of the circle.
Now, consider triangles AED and AFD.
E and F are the midpoints of AB and AC.
Because AB=AC, AE=AF.
Moreover, the two triangles have a common side : AD.
Plus, given the property of bisection, DE is perpendicular to AE and DF is perpendicular to AF. So :
angle DEA=angle DFA=90°
Are the triangles congruent ?
Say O is the center of the circle.
Draw radii OB, OA and OC.
Since AB = AC, then (minor circular arc AB) = (minor circular arc AC)
Then, (central angle BOA) = (cental angle (COA)
OA = OA
OB = OC .......both are radii.
Therefore, triangle BOA is congruent to triangle COA ....SAS
Therefore angle OAB = angle OAC
Therefore, OA is the bisector of angle BAC. And radius OA passes through the center O.