1. ## triangle

In $AB$ of the triangle $\triangle ABC$ is given point $M$ in order that $AM:MB=AC:CB$
prove that line $CM$ is simetral(line that shares the angle) of $\angle ACB$

2. What is a "simetral" of an angle? I can't find any definition online. Sorry!

When you reply, please include a clear listing of what you've tried so far. Thank you!

3. this line on the picture that shares the angles in two parts is the "simetral"

4. Would "angle bisector" be equivalent to "simetral"?

Thank you!

5. Originally Posted by beq!x
In $AB$ of the triangle $\triangle ABC$ is given point $M$ in order that $AM:MB=AC:CB$
prove that line $CM$ is simetral [the English word is bisector] (line that shares the angle) of $\angle ACB$
If h denotes the perpendicular distance from C to AB, then $\frac{\text{area}(\triangle AMC)}{\text{area}(\triangle BMC)} = \frac{\frac12h*AM}{\frac12h*BM} = \frac{AM}{BM} = \frac{AC}{BC}$.

By the sine rule, $\frac{\text{area}(\triangle AMC)}{\text{area}(\triangle BMC)} = \frac{\frac12AC*MC\sin(\angle ACM)}{\frac12BC*MC\sin(\angle BCM)} = \frac{AC\sin(\angle ACM)}{BC\sin(\angle BCM)}$.

Compare the two equations to see that $\angle ACM = \angle BCM$.

6. I didn't understand this step can you explain me a little bit please
Originally Posted by Opalg

By the sine rule, $\frac{\text{area}(\triangle AMC)}{\text{area}(\triangle BMC)} = \frac{\frac12AC*MC\sin(\angle ACM)}{\frac12BC*MC\sin(\angle BCM)} = \frac{AC\sin(\angle ACM)}{BC\sin(\angle BCM)}$.

Compare the two equations to see that $\angle ACM = \angle BCM$.