In the 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
I added Q in order to create the right triangle CMQ.
MQ is a diameter of a circle. MQ = 10
The angle CMQ = 8.13° cos(8.13°) = CM / 10
CM = 9.899 
If this a competition question, assuming AB=MC is one way to get quick approx.
so we know AB=10 ,thus MC=10
well the real value of MC= 9.9
If it is MCQ, choose the value closer to 10 :D
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 units.
Hi Melody: Sorry, I didn't realize that it had been posted twice! But, looking at the page, the original post is just 5 questions below this one. I just copied and pasted my answer under the original one.
+118673 Thanks,
Thanks for your answer too.
My comment was aimed at the asker.
Thje heading was "The help was false"
This meant that the asker was given and answer and was too impolite to continue with the question in the original thread.
Hence my request for the original thread address.
Could you please put a link on both so they are together, thanks.
As yours was not the only answer.
