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 !
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.