In a triangle $A B C, D, E, F$ are the mid-points of the sides $B C, C A$ and $A B$ respectively then prove that, $\to A D = - (\to B E + \to C F)$ .

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Solution

In $△ C F B$
$\Rightarrow \to C F + \to F B + \to B C = 0$ ................. I

In $△ C B E$
$\Rightarrow \to B E + \to E C + \to C B = 0$ ............... II

In $△ A D C$
$\Rightarrow \to A D + \to D C + \to C A = 0$ .............. III

In $△ A D B$
$\Rightarrow \to A D + \to D B + \to B A = 0$ ............... IV

Adding equation III and IV, we get
$\Rightarrow 2 \to A D + \to C A + \to B A = 0$ $[∵ \to D C a n d \to D B$ are equal and opposite]................. V

Now, $\to C A = 2 \to C E$ and $\to B A = 2 \to B F$ ........... VI

Using equations V and VI, we get
$\Rightarrow \to A D = - (\to C E + \to B F)$ .................. VII

Adding equation I and II, we get
$\Rightarrow \to B E + \to C F = \to C E + \to B F$ $[∵ \to B C = - \to C B, \to F B = - \to B F, \to E C = - \to C E]$ ........... VIII

Putting equation VIII in equation VII, we get
$\Rightarrow \to A D = - (\to B E + \to C F)$

Hence, proved.

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