+218 I just figured out this information for this problem, and I wanted help completing it. The problem is:
Let AB be a diameter of a circle, and let C be a point on the circle such that AC = 8 and BC = 6. The angle bisector of angle ACB intersects the circle at point M.Find CM.
Label the point of intersection of BA and CM as P.
By the angle bisector theorem which states:
"If a ray bisects an angle of a triangle, then it divides the opposite side of the triangle into segments that are proportional to the other two sides."
Then, AP and BP are proportional to 8:6
We can find that AP is 407/, while BP is 30/7.
What now? Thanks!
Noori
+1490 Hi, Noori! I'm gonna use the angle bisector theorem to solve this simple question.
1/ AB = MR = 10 (AB and MR are diameters of a circle O) AB ⊥ MR
2/ AP : PB = 8 : 6 AP = 8( 10 / 14 ) = 5.714285714
3/ PO = AP - AO = 0.714428714
4/ Triangle PMO is the right-angled triangle.
5/ Angle PMO = arctan(PO / OM) = 8.130102354º
6/ And finally, CM = cos(PMO) * MR = 9.899494937 or 7√2 
