Let $\overline{AB}$ 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]

itsdshen112 Apr 10, 2023

#1**0 **

First, we can use the Pythagorean theorem to find the length of AC. Since AB is a diameter of the circle, we have:

AC^2 = 4^2 + 8^2 = 80

AC = 4*sqrt(5)

Now, let's draw the angle bisector of ∠ACB and label the point where it intersects the circle as M. We can use the angle bisector theorem to find the ratio of AM to MC:

AM/MC = AB/BC AM/MC = 8/4 AM/MC = 2

We also know that AM + MC = AC, so we can substitute in the values we found for AM/MC and AC:

2MC + MC = 4*sqrt(5)

3MC = 4*sqrt(5)

MC = (4/3)*sqrt(5)

Therefore, CM = (4/3)*sqrt(5).

Guest Apr 10, 2023