$ABCD$ 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