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I have try to solve this problem I cannot find the solution. Thank you for all help.

Let $ABC$ be an equilateral triangle with its centroid $O$. Let $M$ be any point inside $ABC$. Let $P,Q,R$ be the reflective point of $M$ through $AB, BC, CA$ respectively. Prove that $O$ is the centroid of $DEF$.

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    $\begingroup$ Please this is not a site where we solve your homework, you must show your attempts so we could guide you. $\endgroup$ – neonpokharkar Sep 12 '17 at 8:20
  • $\begingroup$ You have just posted something similar and as the above comment shows, we are not here to solve your homework. $\endgroup$ – complexmanifold Sep 12 '17 at 8:36
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Using vectors, it's straightforward . . .

Place triangle $ABC$ in $\mathbb{R}^3$ with $A=(1,0,0),\;B=(0,1,0),\;C=(0,0,1)$.

Let $M=(x,y,z)$, where $x + y + z = 1$. Then \begin{align*} \vec{AM} &=\vec{M}-\vec{A}=\langle{x-1,y,z}\rangle\\[6pt] \text{proj}_{\vec{AB}}(\vec{AM}) &= \left({\small{\frac{\vec{AM}\cdot \vec{AB}}{\vec{AB}\cdot \vec{AB}}}}\right)\vec{AB} ={\small{\frac{1}{2}}}\langle{x-y-1,y-x+1,0}\rangle\\[4pt] \vec{D}&=\vec{M}+2\left(\left(\text{proj}_{\vec{AB}}(\vec{AM})\right)-\vec{AM}\right) =\langle{1-y,1-x,-z}\rangle\\[8pt] &\text{and by analogous computations,}\\[8pt] \vec{E}&=\vec{M}+2\left(\left(\text{proj}_{\vec{BC}}(\vec{BM})\right)-\vec{BM}\right) =\langle{-x,1-z,1-y}\rangle\\[4pt] \vec{F}&=\vec{M}+2\left(\left(\text{proj}_{\vec{CA}}(\vec{CM})\right)-\vec{CM}\right) =\langle{1-z,-y,1-x}\rangle\\[4pt] \end{align*} Then the centroid of triangle $DEF$ is \begin{align*} {\small{\frac{1}{3}}}(\vec{D}+\vec{E}+\vec{F}) &= {\small{\frac{1}{3}}} \langle{2-(x+y+z),2-(x+y+z),2-(x+y+z)}\rangle\\[4pt] &= {\small{\frac{1}{3}}}\langle{1,1,1}\rangle\\[4pt] &= \langle{{\small{\frac{1}{3}}},{\small{\frac{1}{3}}},{\small{\frac{1}{3}}}}\rangle\\[4pt] \end{align*} which is the same as the centroid of triangle $ABC$.

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The technique here is similar to the [other problem you posted]

Let the centroid be at $\vec{0}$ and the vertices at $\vec{a}, \vec{b},\vec{c}$

Let the perp distances from $M$ to $BC, AC, AB$ be $d_1, d_2, d_3$ resp.

Then $$\vec{m} = \frac{d_1\vec{a}+d_2\vec{b}+d_3\vec{c} }{d_1+d_2+d_3}$$

Reflecting $M$ across $AB$ results in $D$ whose perp distances from $AB, BC, CA$ are now $d_3, (d_2+d_3), (d_1+d_3)$ and hence we have

$$\vec{d} = \frac{-d_3\vec{c}+(d_2+d_3) \vec{b}+(d_1+d_3)\vec{a} }{d_1+d_2+d_3}$$

Recall that $\vec{a}+ \vec{b}+\vec{c} = \vec{0}$

Hence $$\vec{d} = \frac{-2d_3\vec{c}+d_2 \vec{b}+d_1\vec{a} }{d_1+d_2+d_3}$$

Its now easy to see that $\vec{d}+\vec{e}+\vec{f} = \vec{0}$ and hence the centroid of $\triangle DEF$ coincides with that of $\triangle ABC$

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