$ABC$ equilateral triangle
$M$ point inside the triangle , let $D, E, F$ be the vertical projection (orthogonal) of M on $AB, BC, AC$ respectively, where $MD=1,ME=2,MF=3$
Then evaluate the side of a triangle ?
I'm thinking calculated $MA,MB,MC$ then esay i use rotational $R(A,\frac{π}{3})$ But how I find it ? I'm try Connect $D,E$ but $BED$ not Isosceles
So I have already to see your solution
As shown above, join $M$ to each of the equilateral triangle vertices of $A$, $B$ and $C$. The line segments $MD$, $ME$ and $MF$ are the heights of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$.
Note the area of the surrounding equilateral triangle must equal the sum of the areas of these $3$ smaller triangles. Let the length of the equilateral triangle side be $s$. The height of $\triangle ABC$ would be $\frac{\sqrt{3}}{2}\left(s\right)$ (you can see this by using the side length ratios of $1:\sqrt{3}:2$ of a $30^{o}\text{ - }60^{o}\text{ - }90^{o}$ triangle). Using that the area of a triangle is $\frac{1}{2}\left(bh\right)$ formula, where $b$ is the base length and $h$ is the height, you then get
$$\begin{equation}\begin{aligned} \frac{1}{2}\left(s\right)\left(\frac{\sqrt{3}}{2}\left(s\right)\right) & = \frac{1}{2}\left(s\right)\left(1 + 2 + 3\right) \\ \frac{\sqrt{3}}{2}s & = 6 \\ s & = 4\sqrt{3} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$