Question 940968
For this solution, I set the radius of the circle with center O to be smaller than the radius of the circle with center O'. Also, C (on the line tangent to circle O' at A) can only lie on one side of A, so I picked C to be the intersection of the line with the circle with center O. However if the radius of the circle with center O is larger, a similar argument should hold.


If we let M be the midpoint of segment AC (where C is on the circle with center O), we have *[tex \large \angle O'AM = 90^{\circ}] and *[tex \large OMA = 90^{\circ}] so segments O'A and OM are parallel. 


Let *[tex \large \angle AOO' = \alpha]. Because *[tex \large \triangle AOO'] is isosceles, *[tex \large \angle OAO' = \alpha] and also *[tex \large AOM = \alpha] by parallel lines. It follows that *[tex \large \triangle AOM] is congruent to *[tex \large \triangle AOO'] since their angles and hypotenuse are equal.


Then *[tex \large \angle OAC = \angle OAB], so OA bisects *[tex \large \angle BAC].