Constructing quadrilaterals under projection with straightedge $I$ is the image of $A$ under a projection from some center. I'm supposed to construct the images of $B,C,$ and $D$ using only a straightedge. I believe the question is more generally to represent the tiling under a projection, but I have not made any progress and keep getting lost. I think I know that I'm supposed to find the point of projection but looking at the image of $A$ has me confused. I also believe I'm supposed to designate a horizon but again the image of $A$ has me confused. Any help would be much appreciated.

 A: I don't believe there's a unique solution for the images of D and B.  One possible solution is illustrated in the copy of your diagram below (Diagram $1$). If the corners of the square A and its projection are labelled as in that diagram, then there are eight possible ways in which the corners of the square A can be matched to those of its projection:
\begin{align}
&\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\alpha}&{\color{red}\beta}&{\color{red}\gamma}&{\color{red}\delta}}\hspace{1em}
\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\delta}&{\color{red}\alpha}&{\color{red}\beta}&{\color{red}\gamma}}\hspace{1em}
\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\gamma}&{\color{red}\delta}&{\color{red}\alpha}&{\color{red}\beta}}\\
\ \\
&\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\beta}&{\color{red}\gamma}&{\color{red}\delta}&{\color{red}\alpha}}\hspace{1em}
\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\delta}&{\color{red}\gamma}&{\color{red}\beta}&{\color{red}\alpha}}\hspace{1em}
\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\alpha}&{\color{red}\delta}&{\color{red}\gamma}&{\color{red}\beta}}\\
\ \\
&\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\beta}&{\color{red}\alpha}&{\color{red}\delta}&{\color{red}\gamma}}\hspace{1em}
\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\gamma}&{\color{red}\beta}&{\color{red}\alpha}&{\color{red}\delta}}\ ,
\end{align}
giving rise to eight potential solutions.  One definite solution, illustrated in the diagram, can be constructed from the first of these matchings
The line segments ${\color{red}{\alpha\beta}}$ and ${\color{red}{\delta\gamma}}$ in the diagram are the projections of the parallel line segments $\alpha\beta$ and $\delta\gamma$, so the point $\color{red}\kappa$ where their extensions intersect must lie on the horizon. Likewise, the line segments $\color{red}{\alpha\delta}$ and $\color{red}{\beta\gamma}$ are projections of the parallel line segments $\alpha\delta$ and $\beta\gamma$, so the intersection $\color{red}\mu$ of their extensions also lies on the horizon. Thus, the horizon must be the line passing through $\color{red}\kappa$ and $\color{red}\mu$.
Now the diagonals $\alpha\gamma$, $\beta\zeta$ and $\delta\theta$ are parallel, so their projections must all meet at the same point on the horizon.  Since the line segment $\color{red}{\alpha\gamma}$ is the projection of $\alpha\gamma$, the intersection $\color{red}\lambda$ of its extension with the horizon must be the point of intersection of these three projections. Thus, the line segment $\color{red}{\beta\lambda}$ must be the projection of an extension of the diagonal $\beta\zeta$, and its intersection $\color{red}\zeta$ with the extension of the line ${\color{red}{\delta\gamma}}$ must be the projection of the point $\zeta$ where the diagonal $\beta\zeta$ and the extension of $\delta\gamma$ intersect. Likewise, the point $\color{red}\theta$ where the line $\color{red}{\delta\lambda}$ (the projection of an extension of the diagonal $\delta\theta$) intersects the extension of $\color{red}{\beta\gamma}$ must be the projection of the point $\theta$.
Since the line segment $\epsilon\zeta\eta$ is parallel to the segment $\alpha\delta\iota$, an extension of its projection must pass through the points $\color{red}\mu$ and $\color{red}\zeta$, the point where the extension of $\color{red}{\alpha\delta}$ intersects the horizon, and the point which we already know to be the projection of $\zeta$. Thus, the intersection $\color{red}{\eta}$ of the line segment $\color{red}{\mu\zeta}$ with the line segment $\color{red}{\alpha\gamma\lambda}$ (the projection of an extension of the diagonal $\alpha\gamma$) must be the projection of the point $\eta$, and the  intersection $\color{red}\epsilon$ of its extension with the line $\color{red}{\alpha\beta\kappa}$ must be the projection of the point $\epsilon$. Likewise, the intersection $\color{red}\iota$ of the extension of the line segment $\color{red}{\kappa\theta}$ with the line segment $\color{red}{\alpha\delta\mu}$ must be the projection of the point $\iota$.
This completes the construction, the projections of B, C and D being the rectangles $\color{red}{\delta\gamma\theta\iota}$, $\color{red}{\gamma\zeta\eta\theta}$ and $\color{red}{\beta\epsilon\zeta\gamma}$, respectively.

Diagram $\mathbf1$
This is not the only solution, however.  A similar solution can be constructed from the sixth of the above-listed matchings, $\pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\alpha}&{\color{red}\delta}&{\color{red}\gamma}&{\color{red}\beta}}\ $. The diagram of the projection will look identical to the one in the solution given above, but the images of B and D will be interchanged—that is, the rectangle $\color{red}{\delta\gamma\theta\iota}$, will be the projection of D and $\color{red}{\beta\epsilon\zeta\gamma}$ will be the projection of B. Unless you were given some restriction on the viewpoint of the projection, I don't see how either of these solutions could be eliminated as being illegitimate.
These are the only two possibilities, since none of the remaining matchings give rise to legitimate projections. If the projection of the common corner, $\gamma$, of A and C were the point $\ \color{red}\alpha$, then the projection would have to correspond to one of the matchings $\ \pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\gamma}&{\color{red}\beta}&{\color{red}\alpha}&{\color{red}\delta}}\ $ or $\ \pmatrix{\alpha&\beta&\gamma&\delta\\
{\color{red}\gamma}&{\color{red}\delta}&{\color{red}\alpha}&{\color{red}\beta}}\ $. For the first of these, the images of B,C,and D would have to be as in Diagram 2 below,  the horizon and the projection of the diagonal $\alpha\gamma$ being the same as for the solutions already obtained.  However the the extension of the projection of the diagonal $\beta\zeta$ must pass through the projection $\color{red}\beta$ of $\beta$, and the point $\color{red}\lambda$ where the extension $\color{red}{\alpha\gamma\lambda}$ of the projection of the diagonal $\alpha\gamma$ intersects the horizon. But the extension of the segment $\color{red}{\lambda\beta}$ doesn't intersect the extension of the segment $\color{red}{\delta\alpha}$ anywhere above the horizon, and the point below the horizon where those lines do intersect cannot be the projection of the point $\zeta$, which it has to be for the projection to be legitimate.  The second of these two matchings can be eliminated by essentially the same argument.

Diagram $\mathbf2$
If the projection of the common corner $\ \gamma\ $ between A and C were either of the two points $\color{red}\beta$ or $\color{red}\delta$, then one of the line segments $\color{red}{\alpha\beta}$ would have to be the projection of the common side of A and either B or D. Any such projection can be eliminated as a legitimate possibility by an argument similar to the one above.
