Let $A, B, C$ and $D$ be four points of a plane, $I$ the midpoint of $[AB]$ and $J$ the midpoint of $[CD]$.
1) Under which condition can we have $I=J$?
2) We assume that $I\neq J$. Determine $\mathcal{E}$, the set of points of the plane such that $$\big|\big|\;\overrightarrow{MA} + \overrightarrow{MB}\;\big|\big|=\big|\big|\;\overrightarrow{MC} + \overrightarrow{MD}\;\big|\big|.$$
3) We assume that $I\neq J$ and $AB=CD$. Determine $\mathcal{F}$, the set of points of the plane such that $$MA^2 + MB^2=MC^2 + MD^2.$$
Here is what I found:
For 1) I do not know how to prove it, but I guess $I=J$, if it is the centroid of the four points?
For 2), I found that $\mathcal{E}$ is the perpendicular bisector of $[IJ]$.
For 3), I also found that $\mathcal{F}$ is the perpendicular bisector of $[IJ]$.
So my questions: -What do I have to write for $1)$,
- and are my descriptions of $\mathcal{E}$ and $\mathcal{F}$ correct?
Thank you in advance!
For 1), think of $A,B,C,$ and $D$ as the corners of a quadrilateral. If you order the points correctly, line segments $\overline{AB}$ and $\overline{CD}$ are then diagonals of a simple quadrilateral that bisect each other. Do you remember what kind of quadrilateral has diagonals that bisect each other? (The parallelogram $ACBD$.)
For 2), $\lVert \overrightarrow{MA} + \overrightarrow{MB} \rVert$ is twice the distance between $M$ and the midpoint of $A$ and $B$, and similarly for $C$ and $D$. So you are correct: $\mathcal{E}$ is the perpendicular bisector of $\overline{IJ}$. (A previous version of this answer overlooked the vector signs over $MA$ etc., so it was wrong.)
For 3), by the initial assumptions we see that $\overline{MI}$ is a median of triangle $\Delta ABM$ and $\overline{MJ}$ is a median of triangle $\Delta CDM$. Apollonius' theorem then gives us an equation for each triangle:
$$MA^2+MB^2=2IA^2+2MI^2$$ $$MC^2+MD^2=2JC^2+2MJ^2$$
The added assumptions for 3) make the left-hand-sides equal to each other, and $IA$ and $JC$ equal to each other. This makes
$$MI^2=MJ^2,\text { and thus}$$ $$MI=MJ$$
This means $M$ is on the perpendicular bisector of $\overline{IJ}$. So you are correct about $\mathcal{F}$.