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Suppose we are a creative individual, and during our math exam would like to draw a picture of a cube using the vanishing point perspective. Let $A$ and $B$ be two adjacent vertices of a square in the plane, and $V$ the vanishing point. We extend segments from each vertex of the square to $V$.

The question is how far along each segment should we place the remaining vertices of the cube. Let $E$ be the remaining vertex lying on the segment $\overline{BV}$:

Proposition: $E$ is the intersection of the bisector of $\angle VAB$ and the segment $\overline{BV}$, so that $\angle VAE=\angle EAB$.

Proof: In the plane, the bisectors of opposite vertices of any square should be equal. If this fact holds in any projected square, this gives a unique solution and we are done.

Can anyone provide a proof for this claim?

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  • $\begingroup$ Which claim would you like help with? And what thoughts of your own do have about the problem? $\endgroup$ – Rob Arthan Nov 23 '18 at 0:06
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    $\begingroup$ No, this doesn't hold for the projected square. And if I am right, this problem is indeterminate, as this depends on the focal length, which is free. $\endgroup$ – Yves Daoust Nov 23 '18 at 0:07
  • $\begingroup$ @RobArthan I need help to prove the highlighted conjecture that my construction of the point $E$ is correct $\endgroup$ – M. Nestor Nov 23 '18 at 0:08
  • $\begingroup$ @M.Nestor: so make that clear in your question! Don't write "Proposition ..." when you mean "I think ... holds and that would solve the problem (but I don't know how to prove it)". $\endgroup$ – Rob Arthan Nov 23 '18 at 0:11
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    $\begingroup$ As @Yves Daoust has said, this problem is indeterminate. Thus, you have found "a" solution (more or less pleasing to the eye), but you cannot claim that this is "the" solution... A clue : to give unique solutions you need to have 3 points on a same straight line (L), then you can work on the placement of a fourth point on (L) using the so-called "cross-ratio", which is an indispensable tool for doing "projective geometry". $\endgroup$ – Jean Marie Nov 23 '18 at 0:53

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