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.

The diagram:

https://artofproblemsolving.com/texer

The previous posts on this topic were incorrect or had the wrong values. I was also unable to find out the values from their work.

Thanks,

Guest

Guest Jun 24, 2023

#1**0 **

To find CM, we can use the angle bisector theorem. According to the angle bisector theorem, the ratio of the lengths of the segments formed by the angle bisector on a triangle is equal to the ratio of the lengths of the opposite sides.

In triangle ABC, let CM intersect AB at point D. We know that AD + DB = AB (the full length of the diameter). Since AB is the diameter, it is twice the radius of the circle.

Let's assign the radius as r. Therefore, AB = 2r.

Given AC = 8 and BC = 4, we can determine AD and DB.

AD/DB = AC/BC AD/DB = 8/4 AD/DB = 2

Since AD + DB = AB = 2r, we have:

AD = 2/(2 + 1) * AB AD = 2/3 * 2r AD = 4r/3

Similarly,

DB = 1/(2 + 1) * AB DB = 1/3 * 2r DB = 2r/3

Now, we can find the length of CM.

CM = CD + DM

Since CM is the angle bisector, we have:

CD/DB = CM/MB

CD/DB = CM/(AB - CM)

Substituting the values we have:

4r/3 / (2r/3) = CM / (2r - CM)

4r/3 * (3/2r) = CM / (2r - CM)

2 = CM / (2r - CM)

2(2r - CM) = CM

4r - 2CM = CM

4r = 3CM

CM = 4r/3

Now we need to find the value of r. Since AC = 8 and BC = 4, we can use the Pythagorean theorem on triangle ABC:

AC^2 + BC^2 = AB^2

8^2 + 4^2 = (2r)^2

64 + 16 = 4r^2

80 = 4r^2

r^2 = 80/4

r^2 = 20

r = √20

Now we can find CM:

CM = 4r/3

CM = 4(√20)/3

Simplifying the expression:

CM = (4/3) * (2√5)

CM = (8√5)/3

Therefore, CM is equal to (8√5)/3 units.

Guest Jun 25, 2023