Krdytkyu

$ABCD$ 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 $N$.

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)

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