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.
+14917 The angle bisector of ACB intersects the circle at point M Find CM.
Hello Guest!
Point C is on the periphery of the Thales circle of diameter AB.
Triangle ABC is a right triangle. The following applies:
\(\overline{AB}=\sqrt{\overline{AC}\ ^2+\overline{BC}\ ^2}=\sqrt{8^2+4^2}=\sqrt{64+16}=8.944\)
To be continued today.
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