Let \(\overline{AB}\) be a diameter of a circle, and let C be a point on the circle such that AC = 8 and BC = 6. The angle bisector of\( \angle ACB\) intersects the circle at point M. Find CM in simpliest radical form.
I tried solving the sides of the triangle then dividing length BA by 2 and soving for the midpoint of CM which i called O. this did not work... I have found other answers such as \(3\sqrt{13}\), \(2\sqrt{11}\), and a really long dicimal number.
Please help me
AB is a diameter so angle ACB = 90 deg, so angle ACM = 45 deg and so angle ABM is also 45 deg.
Now use a part of the sine rule which is often forgotten, the bit that says that the ratios are equal to twice the radius of the circumscribing circle.
So, from the triangle CBM, CM / sin(/_CBM) = 10.
Angle CBM is made up of the angle CBA (and you can get the sine and cosine of that from the triangle ABC), and the 45 deg angle ABM.
Expand sin(/_CBM) = sin(/_CBA + /_ABM), etc..
+9519 \(BM = MA\) (why?)
\(BA = \sqrt{6^2 + 8^2} = 10\)
\(\angle BMA = 90^\circ\text{ and } BM = MA \Longleftrightarrow \triangle BMA\text{ is a right isosceles triangle (with angles }90^\circ,45^\circ , 45^\circ )\)
\(BM^2 + MA^2 = BA^2\\ 2BM^2 = 10^2\\ BM = MA = 5\sqrt 2\)
\(BM \cdot CA + MA \cdot BC = BA \cdot CM\\ 5\sqrt 2\cdot 8 + 5\sqrt 2 \cdot6 = 10 \cdot CM\\ CM = 7 \sqrt 2\)
