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Math Help - need help prove Medians of a triangle coincident

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    need help prove Medians of a triangle coincident

    show that the medians of a triangle ABC are coincident
    Last edited by Reignrx; May 13th 2009 at 05:32 PM.
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    Quote Originally Posted by Reignrx View Post
    How can i "show that the medians of a triangle ABC are coincident (at point X). and "What is the value of A'X/AX"?
    This is a completely standard theorem. Its proof is in any standard textbook.
    There is a very straightforward proof using vectors.
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    thanks
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    Senior Member pankaj's Avatar
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    Let \overrightarrow a,\overrightarrow b,\overrightarrow c, denote the vertices A,B,C respectively and also assume D(\overrightarrow d),E(\overrightarrow e),F(\overrightarrow f) denote the mid-points of AB,BC,CA respectively.Therefore,

    \overrightarrow d=\frac{\overrightarrow b+\overrightarrow c}{2}

    \overrightarrow e=\frac{\overrightarrow c+\overrightarrow a}{2}

    \overrightarrow f=\frac{\overrightarrow a+\overrightarrow b}{2}

     <br />
2\overrightarrow d=\overrightarrow b+\overrightarrow c;<br />
2\overrightarrow e=\overrightarrow c+\overrightarrow a;<br />
2\overrightarrow f=\overrightarrow a+\overrightarrow b<br />

    2\overrightarrow d+\overrightarrow a=2\overrightarrow e+\overrightarrow b=2\overrightarrow f+\overrightarrow c=\overrightarrow a+\overrightarrow b+\overrightarrow c

     <br />
\frac{2\overrightarrow d+\overrightarrow a}{3}=<br />
\frac{2\overrightarrow e+\overrightarrow b}{3}=<br />
\frac{2\overrightarrow f+\overrightarrow c}{3}=\frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3}<br />

    Now, \frac{2\overrightarrow d+\overrightarrow a}{3} is the position vector of the point on median AD which divides AD in the ratio 2:1.Simlarly, \frac{2\overrightarrow e+\overrightarrow b}{3} is the position vector of the point on median BE which divides BE in the ratio 2:1 and \frac{2\overrightarrow f+\overrightarrow c}{3} is the position vector of the point on median CF which divides CF in the ratio 2:1.

    Since these position vectors have turned out to be equal,obviously AD,BE\\,and\\,CF are concurrent and the point of concurrency is \frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3}
    Last edited by pankaj; May 15th 2009 at 06:45 PM.
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