**Question:**

Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

**Solution:**

Let $\triangle \mathrm{OAB}$ be any triangle such that $\mathrm{O}$ is the origin and the other co-ordinates are $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) . \mathrm{P}$ and $\mathrm{R}$ are the mid-points of the sides $\mathrm{OA}$ and $\mathrm{OB}$ respectively.

We have to prove that line joining the mid-point of any two sides of a triangle is equal to half of the third side which means,

$\mathrm{PR}=\frac{1}{2}(\mathrm{AB})$

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

So,

Co-ordinates of P is,

$\mathrm{P}\left(\frac{x_{1}}{2}, \frac{y_{1}}{2}\right)$

Similarly, co-ordinates of R is,

$\mathrm{R}\left(\frac{x_{2}}{2}, \frac{y_{2}}{2}\right)$

In general, the distance between $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is given by,

$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Similarly,

$\mathrm{PR}=\sqrt{\left(\frac{x_{2}}{2}-\frac{x_{1}}{2}\right)^{2}+\left(\frac{y_{2}}{2}-\frac{y_{1}}{2}\right)^{2}}$']

$=\frac{1}{2} \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$=\frac{1}{2}(\mathrm{AB})$

Hence,

$\mathrm{PR}=\frac{1}{2}(\mathrm{AB})$

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