SOLUTION: 2 circles with centers O and O' are drawn to intersect each other at points A and B. Center O of one circle lies on the circumference of the other circle. CD is drawn tangent to th

Algebra ->  Circles -> SOLUTION: 2 circles with centers O and O' are drawn to intersect each other at points A and B. Center O of one circle lies on the circumference of the other circle. CD is drawn tangent to th      Log On


   

Question 940968: 2 circles with centers O and O' are drawn to intersect each other at points A and B. Center O of one circle lies on the circumference of the other circle. CD is drawn tangent to the circle with center O' at A. Prove that OA bisects angle BAC.
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
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 and so segments O'A and OM are parallel.

Let . Because is isosceles, and also by parallel lines. It follows that is congruent to since their angles and hypotenuse are equal.

Then , so OA bisects .