In right triangle, ABC, the length of side AC is 8 the length of side BC is6 and C is 90 degrees The circumcircle of triangle ABC is drawn. The angle bisector of ACB meets the circumcircle at point M Find the length CM
+1490 AC = 8
BC = 6
AB = ? AB = sqrt (6² + 8²) AB = 10 (AB is a diameter of a circle)
∠ACM = 45°
∠BAC = ? tan(BAC) = 6/8 BAC = 36.87°
Let the intersection point of CM and AB be an N.
∠ANC = 98.13°
Connecting M with Q which is exactly on the opposite side of the circle.
Now we have a right triangle CMQ
∠CMQ = 8.13°
And finally... cos(CMQ) = CM / MQ CM = 9.899497463 
CPhill: Doesn't the "Intrsecting Chord Theorem" apply to this question? If yes, then the diameter AB =10, which is intersected by CM, say at point "O", has lengths of AO =5.714 and BO =4.286 and and CO=4.849. Let MO =x, then we have: 5.714 * 4.286 =x * 4.849 and x =5.051.
Therefore, the length of CM =4.849 + 5.051 =9.9 unit
