I need help plz
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.
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Triangle ABC is a right triangle because it has it's hypotenuse on a diameter of a circle. (point C is backwards on my graph, but that doesnt affect the result because its a reflection of the original image)
Finding AB with pythagorean theorem, we get \(AB = \sqrt{4^2 + 8^2} = 4\sqrt{5}\).
We also Get ACM=MCB = 45 degrees
Because interior angles of a circle are congruent, we get AMB = BAM = 45. Now we see that ABM is a right isoceles triangle. Also, CAMB is a cyclic quadrilateral. We can calculate AM and BM too. \(AM = BM = \frac{4\sqrt{5}}{\sqrt{2}} = 2\sqrt{10}\)
Because ACBM is a cyclic, Ptolemy's theorem can be applied.
Set the length of CM as X
then \(4*2\sqrt{10}+8*2\sqrt{10}=x*4\sqrt{5}\) and solving the equation we get x = \(x = 6\sqrt{2}\)
