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!