Similar Triangle Theorem in the Incommensurable Case The following is a geometry theorem whose proof is examinable in the Irish 'High School' Exam.
Let $\Delta ABC$ be a triangle. If a line $L$ is parallel to $BC$ and cuts $[AB]$ in the ratio $s:t$, then it also cuts $[AC]$ in the same ratio.
The proof required is the commensurable case and appeals to a previous theorem: If three parallel lines cut off equal segments on some transversal line, then they will cut off equal segments on any other transversal.
I have been struggling to find the 'standard' proof in the incommensurable case (i.e. without using continuity). I have a proof that I want to compare with the 'standard' (I use an area argument to show that similar triangles are enlargements of each other and work from there). I have tried and failed to find it online, the closest I've come is p. 30 of 
http://www.gutenberg.org/files/17384/17384-pdf.pdf --- but note that I want to define two triangles as similar if they have equal angle measures and this assumes something different.
Any help would be greatly appreciated.

EDIT: This is my own proof:
Theorem
Let $\Delta ABC$ be a triangle and $DE$ be a line segment such that $DE\parallel BC$ as shown

Then 
$$\frac{|AD|}{|BD|}=\frac{|AE|}{|CE|}.$$ 
Clearly we can reduce to the case of a right-angled triangle. 
Lemma 1: Suppose $\Delta ABC$ is similar to triangle $\Delta A'B'C'$ and suppose further that both of a right-angle at $B$. Then if $|A'B'|=k|AB|$ then $|B'C'|=k|BC|$; i.e. similar triangles are equivalent modulo enlargement.
Proof of L1: Suppose without loss of generality that $|AB|<|A'B'|$ so $k>1$. Hence we can place $\Delta ABC$ inside $\Delta A'B'C'$ as shown:

Suppose $|B'C'|=\alpha|BC|$. Now we compute the area of $\Delta A'B'C'$ in two different ways:
$$\text{area}(\Delta A'B'C')=\frac1{2}|A'B'||B'C'|=\frac1{2}k\alpha |AB||BC|.$$
Also,
$$\text{area}(\Delta A'B'C')=\text{area}(\Delta ABC)+\text{area}(\Delta CEC')+\text{area}(BCDB')$$
$$=\frac1{2}|AB||BC|+\frac{1}{2}|CD||DC'|+|BC||BB'|.$$
Note that 
$$|CD|=|BB'|=|A'B'|-|AB|=(k-1)|AB|\text{ , and}$$
$$|DC'|=|B'C'|-|BC|=(\alpha-1)|BC|.$$
Hence
$$\text{area}(\Delta A'B'C')=\frac1{2}|AB||BC|+\frac1{2}(k-1)|AB|(\alpha-1)|BC|+(k-1)|AB||BC|.$$
Now set these two equal:
$$\frac1{2}k\alpha |AB||BC|=\frac1{2}|AB||BC|+\frac1{2}(k-1)|AB|(\alpha-1)|BC|+(k-1)|AB||BC|,$$
$$\Rightarrow k\alpha =1+(k-1)(\alpha-1)+2(k-1),$$
$$\Rightarrow k\alpha=1+\alpha k-k-\alpha+1+2k-2,$$
$$\Rightarrow \alpha=k\,\,\,\bullet$$
Lemma 2: Suppose $a,\,b>0$. Then the only positive solutions to
$$\frac{a+x}{b+y}=\frac{a}{b}$$
occur when $x/y=a/b$
Proof: 
$$\frac{a+x}{b+y}=\frac{a}{b}\Leftrightarrow ab+bx=ab+ay\Leftrightarrow \frac{x}{y}=\frac{a}{b}\,\,\,\bullet$$
Proof of Theorem: Take $\Delta ABC$ to be a right-angled triangle and translate $CE$ to $D$ to form $DF$ as shown $(*)$:

Now $\Delta ADE\sim \Delta BDF$ so by Lemma 1
$$|AD|=k|BD|\text{ and }|DE|=k|BF|,$$
$$\Rightarrow \frac{|AD|}{|BD|}=\frac{|DE|}{|BF|}\,\,(**).$$
Now by Pythagoras:
$$|AE|^2=|AD|^2+|DE|^2\text{ and }|CE|^2\underset{(*)}{=}|BD|^2+|BF|^2,$$
$$\Rightarrow \frac{|AE|^2}{|CE|^2}=\frac{|AD|^2+|DE|^2}{|BD|^2+|BF|^2}\underset{(**)\text{ and Lemma 2}}{=}\frac{|AD|^2}{|BD|^2}\,\,\bullet$$
My happiness at finding this proof is singularly dampened by seeing how much slicker the 'standard' proof is!
 A: 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 $E$. 
