In $\displaystyle AB$ of the triangle $\displaystyle \triangle ABC$ is given point$\displaystyle M$ in order that $\displaystyle AM:MB=AC:CB$
prove that line $\displaystyle CM$ is simetral(line that shares the angle) of$\displaystyle \angle ACB$
In $\displaystyle AB$ of the triangle $\displaystyle \triangle ABC$ is given point$\displaystyle M$ in order that $\displaystyle AM:MB=AC:CB$
prove that line $\displaystyle CM$ is simetral(line that shares the angle) of$\displaystyle \angle ACB$
Would "angle bisector" be equivalent to "simetral"?
Thank you!
If h denotes the perpendicular distance from C to AB, then $\displaystyle \frac{\text{area}(\triangle AMC)}{\text{area}(\triangle BMC)} = \frac{\frac12h*AM}{\frac12h*BM} = \frac{AM}{BM} = \frac{AC}{BC}$.
By the sine rule, $\displaystyle \frac{\text{area}(\triangle AMC)}{\text{area}(\triangle BMC)} = \frac{\frac12AC*MC\sin(\angle ACM)}{\frac12BC*MC\sin(\angle BCM)} = \frac{AC\sin(\angle ACM)}{BC\sin(\angle BCM)}$.
Compare the two equations to see that $\displaystyle \angle ACM = \angle BCM$.