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

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

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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) (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

.
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