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 ACB intersects the circle at point M Find CM.
One way -- I hope that it's not the easist way ...
I'm going to look at triangle(ACM) and find CM by using the Law of Cosines.
To do this, I need the length of AC (which is 8), the length of AM (which I have
to calculate), and the size of angle(A) (which I have to calculate).
First, note that since angle(ACB) is inscribed in a semicircle, it is a right angle.
this makes triangle(ACB) a right triangle.
To find the size of angle(A):
Looking at right triangle(ACB): A = tan-1(CB/CA) = tan-1(4/8)
A = 26.565o
To find the length of AM:
First, find the length of AB:
Pythagorean Theorem: AB2 = AC2 + BC = 82 + 42 = 80
AB = sqrt(80)
Now, use the Angle-Bisector Theorem to find the length of AM:
AM / MB = AC / BC = 8 / 4
Let AM = x ---> This makes MB = sqrt(80) - x
AM / MB = x / [ sqrt(80) - x ] = 8 / 4 = 2
AM = x = 2·[ sqrt(80 - x ] = 2sqrt(80) - 2x
x = 2sqrt(80) / 3
Finally, use the Law of Cosines to find CM:
CM2 = AM2 + AC2 - 2 · AM ·AC · cos(A)
We can now finish this by replacement ...