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Given an equilateral triangle and a point $D$ on one of its sides, I need to construct a square $DEFG$ with the vertices $E, F$ lying on the other two sides of the triangle and $G$ somewhere inside it (see picture).

I know if $D$ is the midpoint of the respective side, the problem is easy, but how about the general case? Are there any solutions at all? Actually, my intuition says there should not be if $D$ is not quite close to the middle.

Furthermore, I have tried using analytic geometry but it quickly became messed up....so I wonder also if we can construct such a square with compass and ruler only.

Thanks in advance.

As $DF=DEsqrt 2$ and the angle $angle EDF=45^circ,$ the point $F$ is obtained from $E$ through the rotation composed with the homothety (common center $D$, angle and ratio as above).

Construct in this transformation the image of the side that should contain $E.$ Its intersection (if it exists) with the side that doesn't contain $D$ is $F.$

$;;;$

Suppose triangle $ABC$ is equilateral.

Using coordinates, and then solving algebraically, we get the following result:

If $E$ is on side $CA$, strictly between $C$ and $A$, there is at most one square $DEFG$ such that

• 
  $D$ is on side $BC$, strictly between $B$ and $C$.$\[4pt]$
  


• 
  $F$ is on side $AB$, strictly between $A$ and $B$.$\[4pt]$
  


• 
  $G$ is in the interior of triangle $ABC$.
  

and such a square exists if and only if
$$2sqrt3-3 < fracCA < 4-2sqrt3qquad(mathbf*)$$
Moreover, if $(mathbf*)$ is satisfied, then letting
$$
w=fracCA
qquadqquad;;;;;,
$$
the points $D,F,G$ are uniqely determined by
beginalign*
frac&=-1+w(sqrt3+1)\[4pt]
fracAF&=(1-sqrt3)+w(1+sqrt3)\[4pt]
endalign*
and where $G$ is the reflection of $E$ over the line $DF$.

Let

the vertex of the triangle between $D$ and $E$ be $A$

and

the vertex of the triangle between $E$ and $F$ be $C$

Let the length of the side of the triangle be $a$

length of the side of the square be $u$

$angle ADE = theta$ (therefore, $angle EFC = frac5 pi6 - theta$)

length of $AE$ be $x$, therefore length of $CE$ is $a-x$

Then consider the triangles $ ADE$ and $CEF$ using Sine rule:

$displaystyle fracusin fracpi3 = fracxsin theta = fraca-xsin left( frac5 pi6 - theta right) = fracasin theta + sin left( frac5 pi6 - theta right)$

Hence

$displaystyle u = fraca sin fracpi3sin theta + sin left( frac5 pi6 - theta right)$

$displaystyle x = fraca sin thetasin theta + sin left( frac5 pi6 - theta right)$