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Thread: triangle

  1. #1
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    triangle

    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$\displaystyle \angle ACB$
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  2. #2
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    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!
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  3. #3
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    this line on the picture that shares the angles in two parts is the "simetral"
    sorry for my English please
    Attached Thumbnails Attached Thumbnails triangle-simetral.bmp  
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  4. #4
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    Would "angle bisector" be equivalent to "simetral"?

    Thank you!
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  5. #5
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    Quote Originally Posted by beq!x View Post
    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 [the English word is bisector] (line that shares the angle) of$\displaystyle \angle ACB$
    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 $\displaystyle \angle ACM = \angle BCM$.
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  6. #6
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    I didn't understand this step can you explain me a little bit please
    Quote Originally Posted by Opalg View Post

    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 $\displaystyle \angle ACM = \angle BCM$.
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