document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #573544 by richard1234(7193) $\"\"$ $\"About$

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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.\r
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\n" ); document.write( "\n" ); document.write( "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. \r
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\n" ); document.write( "\n" ); document.write( "Let

. Because

is isosceles,

and also

by parallel lines. It follows that

is congruent to

since their angles and hypotenuse are equal.\r
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\n" ); document.write( "\n" ); document.write( "Then

, so OA bisects

. \n" ); document.write( "

\n" );