Let AB be a diameter of a circle, and let C be a point on the circle such that AC = 8 and BC = 8. The angle bisector of angle ACB intersects the circle at point M. Find CM.
if AB is a diameter then angle ACB is 90 degrees. The bisector of angle ACB is also a diameter. Hence a right angled triangle and we just need to find the hypotenuse of the isosceles 90 degree triangle whose side is length 8. sqr of 8^2 + 8^2 = sqr128
