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Let AB be a diameter of a circle, and let C be a point on the circle such that AC = 8 and BC = 4 The angle bisector of angle ACB intersects the circle at point M. Find CM.
[asy]
unitsize(0.5 cm);

pair A, B, C, D, P;

A = (4,0);
B = (0,3);
C = (0,0);
D = (12/7,12/7);
P = intersectionpoint(D--(D + 3*(D - C)), circumcircle(A,B,C));

draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
draw(C--P);

label("$A$", A, SE);
label("$B$", B, NW);
label("$C$", C, SW);
dot("$M$", P, NE);
[/asy]

 Aug 1, 2022
 #1
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0

By power of a point, CM = 5*sqrt(3).

 Aug 1, 2022
 #2
avatar+124676 
+1

ACB forms a right triangle in the circle

 

We have something like this : 

 

 

By the Theory of Intersecting Chords

 

AD * BD  =  CD * MD

 

5.9 * 3.04  = 3.88 * MD

 

(5.9)(3.04) / 3.88  = MD  ≈  4.62

 

So

 

CM =  CD + MD  =  3.88 + 4.62  ≈   8.5

 

 

cool cool cool

 Aug 2, 2022

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