Constructing a quadrilateral with $3$ equal sides given their midpoints

Question

Construct a quadrilateral with three equal sides, given the midpoint of each of these three sides. Also it is given that which of these three midpoints is for the middle side (a side between the two other sides).

Here is what I've done to solve the problem,

Assume the three points $E, G$ and $F$ are given and the point $G$ is for the middle side. Now consider the problem is solved,

I noticed that the vertex $A$ lie on the perpendicular bisector of $GE$ and similarly, $B$ lies on the perpendicular bisector of $GF$. So starting with three points,

Now it is enough to find the vertices $A$ and $B$ on each side of the green angle so that the segment $AB$ (which passes through $G$) make $AG= BG$. But from here I'm not sure how to continue.

Answer 1

You have $E,G,F$ (non-collinear).

Draw three lines:

• take the perpendicular bisector of $EG$ and call the line $t$
• reflect $t$ through $G$ and call this parallel line $r$
• take the perpendicular bisector of $GF$ and call the line $s$ (not parallel to $t$ and $r$ as the points are not collinear)

Then set the points:

• $B$ on the intersection of $r$ and $s$
• $A$ as the reflection of $B$ through $G$ (so lies on $t$)
• $D$ as the reflection of $A$ through $E$
• $C$ as the reflection of $B$ through $F$

and the reflections and perpendicular bisectors ensure all the desired distances are equal.

Answer 2

The red lines connect in your modified diagram, connect the given midpoints in order, $EG$ , then $GF$, where it is understood that $G$ is the midpoint of the middle side. Since $A$ is to be equidistant from E and G, then it lies on the perpendicular bisector of EG. Similarly point $B$ lies on the perpendicular bisector of $GF$. These bisector are drawn in blue in the diagram above.

Using analytic geometry, and assuming the coordinates of $E , G, F$ are known, then the blue lines have known equations. Let the left one be

$ A(t) = \dfrac{E + G}{2} + t U_1\tag{1} $

where $U_1$ is a unit vector perpendicular to $EG$ , and let the right blue line be

$ B(s) = \dfrac{G + F}{2} + s U_2 \tag{2}$

where $U_2$ is a unit vector perpendicular to $GF$.

The implicit algebraic equation of this last line is

$ GF \cdot ( (x, y) - \dfrac{G + F}{2} ) = 0 \tag{3} $

To find $A$ and $B$ we proceed as follows.

Starting with equation $(1)$, reflect $A(t)$ about point $G$.

Let this point be $A'$, then

$ A' - G = G - A $

i.e.

$ A' = 2 G - A = \dfrac{ 3 G - E }{2} - t U_1 $

Now substitute $A'$ in equation $(3)$ and solve for $t$.

This will give

$ GF \cdot ( \dfrac{2 G - E - F}{2} - t U_1 ) = 0 $

From which

$ t = \dfrac{1}{2} \dfrac{ GF \cdot (2 G - E - F) }{ GF \cdot U_1 } $

With this $t$, we have point $A(t)$ , and $A'(t) = B(s) $

Now

$ C - F = F - B$, so $C = 2 F - B $

and

$ D - E = E - A$, so $D = 2 E - A $

So now we have all the four vertices.