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I have try to solve this problem but it is indeed difficult. Thank you for all kind help and support.

Let $ABC$ be an equilateral triangle and $M$ be any point inside this triangle. Let $D,E,F$ be the projection of $M$ on $AB, BC, CA$ respectively. Prove that $$ \overrightarrow{MD}+\overrightarrow{ME}+\overrightarrow{MF}=\frac{3}{2}\overrightarrow{MO}, $$ where $O$ is the centroid of triangle $ABC$. enter image description here

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    $\begingroup$ Which center of the triangle? $\endgroup$
    – Narasimham
    Sep 12 '17 at 7:51
  • $\begingroup$ @Kevin Try to think carefully your comments $\endgroup$
    – Blind
    Sep 12 '17 at 8:02
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The solution I give below mirrors my solution to your problem

$\qquad$Centroid of equilateral triangle

Only minor modifications are needed.

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

Then $O = ({\large{\frac{1}{3}}},{\large{\frac{1}{3}}},{\large{\frac{1}{3}}})$.

Let $M=(x,y,z)$, where $x + y + z = 1$.

Hence ${\large{\frac{3}{2}}}\vec{MO} = {\large{\frac{3}{2}}}\left(\vec{O}-\vec{M}\right)={\large{\frac{1}{2}}}\langle{1-3x,1-3y,1-3z}\rangle$. \begin{align*} \text{Then}\;\;\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}+\left(\left(\text{proj}_{\vec{AB}}(\vec{AM})\right)-\vec{AM}\right) ={\small{\frac{1}{2}}}\langle{x-y+1,y-x+1,0}\rangle\\[4pt] \vec{MD}&=\vec{D}-\vec{M}={\small{\frac{1}{2}}}\langle{1-x-y,1-x-y,-2z}\rangle\\[8pt] &\text{and by analogous computations,}\\[8pt] \vec{ME}&={\small{\frac{1}{2}}}\langle{-2x,1-y-z,1-y-z}\rangle\\[4pt] \vec{MF}&={\small{\frac{1}{2}}}\langle{1-x-z,-2y,1-x-z}\rangle\\[8pt] &\text{hence}\\[8pt] \vec{MD}+\vec{ME}+\vec{MF} &= {\small{\frac{1}{2}}} \langle{\bigl(2-(x+y+z)\bigr)-3x,2-(x+y+z)\bigr)-3y,2-(x+y+z)\bigr)-3z}\rangle\\[4pt] &={\small{\frac{1}{2}}}\langle{1-3x,1-3y,1-3z}\rangle\\[4pt] &={\small{\frac{3}{2}}}\vec{MO}\\[4pt] \end{align*}

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  • $\begingroup$ Thanks for your solution. Could you give a solution without using coordinate system? $\endgroup$
    – Blind
    Sep 13 '17 at 6:30
  • $\begingroup$ Maybe, but as I see it, using coordinates and vectors is the most straightforward approach, and as you can see, the work is not too complicated. $\endgroup$
    – quasi
    Sep 13 '17 at 6:54
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Let the centroid be at $\vec{0}$ and let the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c} $ resp. Since, $\triangle ABC$ is equilateral, $\vec{a}+\vec{b}+\vec{c} = \vec{0}$

Let $ME = d_1, MF = d_2, MD = d_3$. Since trilinear coordinates and barycentric coordinates coincide for an equilateral triangle, we have the position vector of $M$ is $$ \vec{m} = \frac{d_1\vec{a}+d_2\vec{b}+d_3\vec{c} }{d_1+d_2+d_3}$$

Using elementary geometry, the perpendicular distances of $D$ from $AC, BC$ are resp. $\displaystyle d_2+\frac{d_3}{2}, d_1+\frac{d_3}{2}$ and hence the position vector of $D$ is given by $$ \vec{d} = \frac{\left(\displaystyle d_2+\frac{d_3}{2}\right)\vec{b}+ \left(\displaystyle d_1+\frac{d_3}{2}\right)\vec{a}}{d_1+d_2+d_3}$$

Thus we get $$\vec{DM} =\vec{m}-\vec{d} =\frac{d_3 \left(\vec{c} -\displaystyle \frac{\vec{b}}{2} - \frac{\vec{a}}{2} \right)}{d_1+d_2+d_3} $$

$\vec{EM}, \vec{FM}$ are similarly obtained.

Now, $$\vec{DM}+\vec{EM}+\vec{FM} = \displaystyle \frac{\displaystyle \sum \left(d_1 - \frac{d_2}{2}-\frac{d_3}{2}\right) \vec{a}}{d_1+d_2+d_3}$$

Now, if the side of the triangle is $s$, from the sum of the areas of triangles formed by the vertices at $M$ we have $d_1+d_2+d_3 = \frac{\sqrt{3}}{2} s$

Hence $$\vec{DM}+\vec{EM}+\vec{FM} = \displaystyle \frac{\displaystyle \sum \left(\frac{3d_1}{2}- \frac{\sqrt{3}}{2} s\right) \vec{a}}{d_1+d_2+d_3}$$

$=\frac{3}{2}\vec{OM}$ since $\displaystyle \sum \frac{\sqrt{3}}{2} s \vec{a}=0$

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