# need help prove Medians of a triangle coincident

• May 13th 2009, 03:01 PM
Reignrx
need help prove Medians of a triangle coincident
show that the medians of a triangle ABC are coincident
• May 13th 2009, 04:37 PM
Plato
Quote:

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.
• May 13th 2009, 05:00 PM
Krizalid
• May 13th 2009, 05:58 PM
Reignrx
thanks
• May 14th 2009, 09:34 AM
pankaj
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 $\frac{\overrightarrow a+\overrightarrow b+\overrightarrow c}{3}$