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For instance, let consider $A$ and $B$ two points of the plane. We want to determine the set of points $M$ such that : $\vert \vert \vec{MA}+\vec{MB}\vert \vert=\vert\vert \vec{MA} \vert \vert$.

I know that the classical method uses the scalar product and also barycenter but it's an analytic method. I wonder how to see "what does this set look like" just by drawing ?

If we just look at the expression of the equation we first draw (with vectors) a parallelogram from $M$ and we want that the diagonal ($Md=MA+MB$) has the same length as $MA$. Maybe the $\triangle AMd$ must be isoscele or equilateral.

Thanks in advance !

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By the parallelogram law of vector addition $\overrightarrow{MA}+\overrightarrow{MB}= 2 \overrightarrow{MO}$ where $O$ is the midpoint of segment $AB$. Then the condition for $M$ can be written as: $$\frac{MO}{MA}=\frac{1}{2}$$

The locus of points with constant ratio between the distances to $2$ fixed points is a circle of Appollonius. To construct the circle "just by drawing", let $M_1, M_2$ be the points on line $AB$ that divide the segment $OA$ in the ratio $1 : 2$ internally and, respectively, externally. Notice that $M_2 \equiv B$ since $BO / BA = 1/2$ . The locus of $M$ is the circle having $M_1M_2 \equiv M_1B$ as a diameter.

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    Why do we have $M_2\equiv B$ ? Is it because for $M_2$ we must have $2OA=AB$ ? – 2017-01-16
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    @Maman $M_2$ is defined as the point on line $AB$ *outside* the segment $OA$ which divides $OA$ (externally) in the ratio $1 : 2$ i.e. $M_2O/M_2A = 1/2$. Such a point is known to exist and be unique, and since $B$ does in fact satisfy $BO/BA=1/2$ it follows that $M_2 \equiv B$. – 2017-01-16
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    Ok I understand, but to draw this set you used a barycenter. That's a good idea but if we just have the expression at the beginning of the claim could we have a clue of what the set would be directly ? Seems that indeed we build a parallelogram at first ! – 2017-01-16
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    @Maman Whether you call $O$ a "*barycenter*" or simply the midpoint of $A,B$ the property that $\overrightarrow{MA}+\overrightarrow{MB}= 2 \overrightarrow{MO}$ is pretty basic, and well known in and by itself, so there is no need for any additional construction. And once you translate the condition into $MO / MA = 1/2$ all the rest follows geometrically. Don't know how much more "*direct*" than that it can get. – 2017-01-16