1. ## Circle

. 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...

2. Hello,

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.

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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 ?

3. Originally Posted by ice_syncer
. 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...
Consider the chord BC. The bisector of angle BAC bisects the chord BC (why?) and is therefore perpendicular to BC (why?). Therefore the centre of the circle lies on this bisector (theorem).

4. Originally Posted by ice_syncer
. 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...
You are right.

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.