need help prove Medians of a triangle coincident

**Reignrx** · May 13th 2009, 02:01 PM

show that the medians of a triangle ABC are coincident

**Plato** · May 13th 2009, 03:37 PM

Originally Posted by Reignrx

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.

**Krizalid** · May 13th 2009, 04:00 PM

Apply

Ceva's theorem - Wikipedia, the free encyclopedia

**Reignrx** · May 13th 2009, 04:58 PM

thanks

**pankaj** · May 14th 2009, 08:34 AM

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,

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

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

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

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

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

$ \frac{2\overrightarrow d+\overrightarrow a}{3}= \frac{2\overrightarrow e+\overrightarrow b}{3}= \frac{2\overrightarrow f+\overrightarrow c}{3}=\frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3} $

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 $\frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3}$

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