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$ABCD$ is a tetrahedron. $E$ is a point on segment $AD$. $Δ$ is a line of plane $(BCD)$ parallel to $(BC)$.

Line $Δ$ intersects $(BD)$ at $I$ and $(CD)$ at $J$.

Plane $(ABC)$ intersects line $(EI)$ at $M$ and line $(EJ)$ at $N$.

Drawing

Show that lines $(MN)$ and $(BC)$ are parallel.

I am 15 years old (classe de seconde in France ≈ 10th grade) and I am stuck. We get that

  • $M$ is on line $(AB)$ and $N$ on line $(AC)$
  • $\frac{DB}{DI}=\frac{DC}{DJ}=\frac{BC}{IJ}$ since $(BC)$ and $(IJ)$ are parallel (per Thales)
  • we'd get the desired result by showing $\frac{AM}{AB}=\frac{AN}{AC}$ or either term is $\frac{MN}{BC}$ (using reciprocal of Thales)
  • we'd get the desired result by showing that lines $(MN)$ and $(IJ)$ are parallel (since two lines parallel to a third one are parallel)
  • we'd get the above by showing $\frac{EM}{EI}=\frac{EN}{EJ}$ or either term is $\frac{MN}{IJ}$ (using reciprocal of Thales)
Jean Marie
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fgrieu
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  • @Jean Marie: thanks for the fixes and simplification. I'm really the father ! See here if in doubt. – fgrieu Nov 18 '18 at 17:49
  • I understand. I have been in the same situation some years ago with my son... – Jean Marie Nov 18 '18 at 17:57
  • I have no solution, but maybe one could turn this 3D question into a seemingly equivalent 2D issue by proving the fact that the projected line segment B'C' onto the base plane BCD is parallel to IJ ? – Jean Marie Nov 18 '18 at 18:30

1 Answers1

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You are right in using the recipeocal of Thales. To continue the proof, you need to invoke another Greek guy’s result.

Consider $\frac{AM}{MB}$ and $\frac{AN}{NC}$. We’d like to show that the two are equal. Where could we get these values from? Well, N is the intersection of a line with another line in a triangle so Menelaus’ Theorem is your friend here.

Applying Menelaus in triangle ABD with the line E-M-I we obtain that $\frac{AM}{MB}\frac{EA}{AD}\frac{IB}{BD} = 1$ and we obtain a value for $\frac{AM}{MB}$ and similarly for $\frac{AN}{NC}$ applying Menelaus in triangle ACD with the line E-N-J you get that $\frac{AN}{NC}\frac{EA}{AD}\frac{JC}{CD} = 1$. Since $\frac{JC}{CD} =\frac{IB}{BD}$ you have all the ingredients to show that $\frac{AM}{MB}$ and $\frac{AN}{NC}$ are equal

fgrieu
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Sorin Tirc
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  • That's helpful. It would be perfectly fine if Menelaus’ Theorem was part of the teaching to a 15yo (classe de seconde in France ≈ 10th grade). It's not, and we'll have to adjust that reasoning to use Thales; it does seems within reach. – fgrieu Nov 18 '18 at 14:48
  • The proof of Menelaus does involve “just” Thales but also some cleverly constructed parallels so I am not quite sure whether such a solution, proving Menelaus, would be “easier” in any way, than just learning Menelaus... – Sorin Tirc Nov 18 '18 at 15:48