# Using vectors prove that the medians of a triangle are concurrent?

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#### Explanation

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#### Explanation:

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Feb 23, 2018

#### Explanation:

Let us consider a segment $P Q$ (shown below in Fig.1), which is divided by a point $R$ in the ratio of $l : m$. Then vector representing $R$ is given by $\frac{m \vec{p} + l \vec{q}}{l + m}$

It is apparent that mid point is represented by $\frac{\vec{p} + \vec{q}}{2}$.

Now let us consider the $\Delta A B C$, where $A , B$ and $C$ are reprsented by $\vec{A} , \vec{B}$ and $\vec{C}$ respectively. $D , E$ and $F$ are the midpoints of $B C , A C$ and $A B$ respectively. Let the vectors representing $D , E$ and $F$ be $\vec{d} . \vec{e}$ and $\vec{f}$. $A D , B E$ and $C F$ are joined to intersect at $G$ represented by $\vec{g}$.

It is aparent that $\vec{d} = \frac{\vec{b} + \vec{c}}{2}$, $\vec{e} = \frac{\vec{a} + \vec{c}}{2}$ and $\vec{f} = \frac{\vec{a} + \vec{b}}{2}$.

Let us also find the $\vec{g}$. As $G$ divides $A D$ in the ratio of $2 : 1$, we have

$\vec{g} = \frac{2 \vec{d} + \vec{a}}{2 + 1} = \frac{2 \vec{d} + \vec{a}}{2 + 1} = \frac{2 \left(\frac{\vec{b} + \vec{c}}{2}\right) + \vec{a}}{3} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$

Observe that it is symmetric w.r.t. the median chosen and had we divided $B E$ or $C F$ in theratio of $2 : 1$, the result would have been same. And $G$ is the centroid of $\Delta A B C$.

Hence, medians of a triangle are concurrent.

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