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 < ACB intersects the circle at point M. Find CM.
This has been asked a couple times, specifiically here: http://web2.0calc.com/questions/help-please-due-tommorow-incredibly-hard
Although the answer may be correct, they use trig, which was not intended for this problem.
Hints: AB = 10, Use Angle Bisector theorem, and power of a point(aka instersecting chords theorem, as CPhill calls it).
Any help is appreciated! 
+1639 https://web2.0calc.com/questions/question-help_7 (Very clever!!!) Bravo, MaxWong!!!
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BC => a = 6 AC => b = 8 AB => c = 10
∠A = tan-1(6 / 8)
∠ANC = 180 - (45 + ∠A)
By using the Law of Sines we can find side AN of triangle ACN.
AN ≈ 5.714 BN = 10 - AN ≈ 4.286
CN can be calculated by using this formula: CN = sqrt [ab(a + b + c)(a + b - c)] / a + b
CN ≈ 4.849
By using the Intersecting chords theorem we can calculate the length of CM.
MN * CN = AN * BN MN ≈ 5.05
CM = CN + MN ≈ 9.899
+128408 Max does get the correct answer....and BMA is a right triangle.....but I don't see how he arrives at the conclusion that this triangle is isosceles with BA = MA
The only way that would be possible is that if CM were a diameter (which it isn't)
Am I misssing something, jugoslav ????
+128408 Per jugoslav's diagram.....Let N be the intersection of the chords BA and MC
Since BCA is bisected, we have the following relationship
BC /CA = BN /AN
6/8 = BN /AN = 3/4
So BN = (3/7) 10 = 30/7
And AN = (4/7)10 = 40/7
And AN * BN = 1200 / 49
Without re-inventing the wheel the length of the angle bisector is given by
BC * CA sqrt (2) / ( BC + CA) = *8*6 sqrt (2) / ( 8 + 6) = 48sqrt (2) / 14 = 24sqrt (2) / 7 = CN
So
AN * BN = CN * NM
1200/49 = 24sqrt (2) / 7 * NM
1200/ 49 * 7 / (24 sqrt (2)) = NM = 25sqrt (2) / 7
Therefore CN + MN = CM = (25 + 24) sqrt (2) / 7 = 49sqrt (2) /7 = 7sqrt (2)
Here's the proof of the length of the angle bisector :
