



 
Call D and E the points where the tangents BC and BA touch the circle. Then OD = OE because they are radii, and BD = BE because they are tangents from the point B. That tells us that the quadrilateral OEBD is symmetric about the line OB; in other words, OB bisects angle CBA. (Draw an accurate figure!) Now use the theorem that tells us that the bisector of angle B divides the segment CA in the same ratio as the sides BC and BA, namely BC/BA = CO/OA. Combine this with the given information that CO + OA = CA = 25. Chris  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 