Given two distinct parallel lines and two distinct fixed points on one of the lines and a point that can vary on the other line. Then the areas of all the triangles formed by those 3 points are all equal. – Does this theorem have a name? I am not asking for an explanation of the mathematics, but simply an elevator-pitch name, something that rolls off the tongue more smoothly than "Do you know the area = one half base times height theorem?"
[Math] Does this triangle-area theorem have a name
euclidean-geometry
Related Solutions
You want the:
Intercept Theorem:
The parallel lines $\color{maroon}{\ell_1}$ and $\color{maroon}{\ell_2}$ split the sides of the angle $\angle BAC$ into equal proportions: $$ \tag{1} {AD\over DC}={AE\over EB},\quad{\rm and}\quad{AD\over AC}={AE\over AB} $$ Also: $$\tag{2} {AD\over AC}={DE\over BC}$$
Proof: Observe:
1) $\color{darkgreen}{\triangle ADE}$ and $\color{darkblue}{\triangle BDE}$ have the same height, $\color{gray}{h_1}$, with respect to the baseline $AB$. Therefore $${AE\over EB} = {{\rm area}(\color{darkgreen}{\triangle ADE}) \over{\rm area}(\color{darkblue}{\triangle BDE})}.$$
2) The triangles $\color{darkblue}{\triangle BDE}$ and $\color{pink}{\triangle DEC}$ have a common base $DE$. Moreover, since $\color{maroon}{\ell_1}$ is parallel to $\color{maroon}{\ell_2}$, the heights of these triangles, $\color{darkblue}{h_3}$ and $\color{pink}{h_4}$, with respect to the base $DE$ are equal. Thus, $${\rm area}(\color{darkblue}{\triangle BDE})= {\rm area}(\color{pink}{\triangle DEC}).$$
3) $\color{darkgreen}{\triangle AED}$ and $\color{pink}{\triangle DEC}$ have the same height, $\color{gray}{h_2}$, with respect to the baseline $AC$. Therefore $${{\rm area}(\color{darkgreen}{\triangle AED}) \over {\rm area}(\color{pink}{\triangle DEC})}={AD\over DC }.$$
The first part of $(1)$ follows immediately from these observations.
The second part of $(1)$ can now be established by using the first part and a bit of algebra.
$(2)$ can be established by applying $(1)$ to the angle $ \angle ABC$, the line $AC$, and the line parallel to $AC$ passing through $E$.
You do not need the parallel axiom to show the statement that if $\alpha+\beta=180^\circ$ then the two lines do not meet. It's a theorem of so-called neutral geometry (geometry with the usual axioms except the parallel axiom), and the proof is as given in your post, and as you say they do not use the parallel axiom in that proof.
Best Answer
This is a particular case of Euclid's Proposition 38 from Book I of the Elements:
[ EDIT ] Credit goes to @Micah's comment for pointing out that this particular case makes in fact the object of Proposition I.37: