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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

enter image description here

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:

enter image description here

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 $(*)$:

enter image description here

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!

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2  
Is this what you want? en.wikipedia.org/wiki/Intercept_theorem –  David Mitra Dec 20 '11 at 11:51
    
The proof they use is on p.62 here ncca.ie/en/Curriculum_and_Assessment/Post-Primary_Education/… –  Jp McCarthy Dec 20 '11 at 11:52
    
@David That's the one thank you very much. Might throw my own proof in here later. –  Jp McCarthy Dec 20 '11 at 11:54
    
I can't see how the very first line of the proof is true... –  Jp McCarthy Dec 20 '11 at 12:02
1  
I find the wiki proof somewhat confusing too. My answer below is essentialy the same as the one at PlanetMath (google.com/…), but with added detail. –  David Mitra Dec 20 '11 at 12:19
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1 Answer

up vote 1 down vote accepted

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: $$\tag{2} {AD\over AC}={DE\over BC}$$

enter image description here

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 $E$.

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I followed this a lot easier than the Wikipedia effort thank you. –  Jp McCarthy Dec 20 '11 at 12:24
    
I don't know why the powers that be restrict to the commensurable case --- this proof is more fundamental than their's and proves more. Odd that. –  Jp McCarthy Dec 20 '11 at 12:32
1  
@Jp McCarthy I agree. I also don't understand why the (lovely) Intercept Theorem is never mentioned in the standard curriculum here in the US. Similar triangles get all the credit... –  David Mitra Dec 20 '11 at 12:38
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