I have a hard time properly understanding the definitions of similarity between two triangles (symbolically denoted as $\triangle ABC \sim \triangle A'B'C'$). My professor defined similarity like this:

$$\text{If } a': a = b' : b = c': c \text{ then } \triangle ABC \sim \triangle A'B'C'.$$

This deduction is obtained via the study of homotetia. For a certain triangle $\triangle ABC$ in Euclidean plane, we can choose a point $O$ and connect it to points $A$, $B$ and $C$ with three rays originating in $O$ – we now have $OA$, $OB$ and $OC$. We then choose a point on one of those rays – thus, we get either $OA'$, $OB'$ or $OC'$. The quotient of the length of these line segments will help us to deduce our coefficient of "stretch", $k$. All in all, we can say that $|OA|:|OA'| = 1 : k$. It follows that $k = \frac{|OA'|}{|OA|} = \frac{|OB'|}{|OB|} = \frac{|OC'|}{|OC|}$, which in simpler notation reaffirms the claim $a': a = b' : b = c': c$, and also proves that all of these quotients are equal to the coefficient $k$ ($a': a = b' : b = c': c = k$). This deduction from my professor was very clear and I quickly grasped it. But then I found the following definition in my textbook:

\begin{align*} &\triangle ABC \sim \triangle A'B'C' \text{ if } a : b : c = a': b ': c'. \\ &\triangle ABC \sim \triangle A'B'C' \text{ if } (\alpha=\alpha')\wedge (\beta=\beta') \wedge (\gamma = \gamma').\\ \end{align*}

I don't have a problem when it comes to the second line of criteria (I understand the correlation between similarity and angles completely). I do however have a big problem with the first line. The textbook dropped it from nowhere, no explanation, no proof, nothing, unlike my professor, who led me through the process of understanding his definition of similarity. What I'd like to do is prove that $a': a = b' : b = c': c$ follows from $a : b : c = a': b ': c'$ and/or vice versa.

So, if possible, I'd prefer if you wrote down proof of the claim $a': a = b' : b = c': c \Longleftrightarrow a : b : c = a': b ': c'$. If not, I'll have to settle with a proof of $a': a = b' : b = c': c \Longrightarrow a : b : c = a': b ': c'$ or $a : b : c = a': b ': c' \Longrightarrow a': a = b' : b = c': c$.

The thing that really confuses me though is the inapplicability of the definition from the textbook in exercises. Here's an example of my claim.

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TRANSLATION: In a square $ABCD$ we fold one of its sides in such a way so that point $B$ "maps" (please, ignore the clumsy terminology) to the midpoint of the line segment $CD$. Show that triangles $HGC$, $EPF$ and $EHD$ are similar and that the lenght of their sides are in relationship $3 : 4 : 5$.

The exercise is not particularly hard. We prove the similarity via proving that the triangles have two (and therefore three) angles of the same size. We get the $x$ relatively quickly by applying the Pythagorean theorem ($x=\frac{5}{8}$). With this value discovered, we can get $a = \frac{3}{8}$, $b = \frac{4}{8}$ and $c = \frac{5}{8}$. After this, we're practically finished. I applied the method my professor suggested and got the following values for the other two triangles:

\begin{align*} &\triangle EHD; \quad a' = \frac{3}{6} \quad b' = \frac{4}{6} \quad c' = \frac{5}{6} \\ &\triangle EPF; \quad a'' = \frac{3}{27} \quad b'' = \frac{4}{27} \quad c'' = \frac{5}{27}. \end{align*}

If we accept the definition and the premise of similarity as written in the textbook, then we need to show that the quotient $3 : 4 : 5$ is true for only one of the three similar triangles and we automatically know that the other two also have the quotient $3 : 4 : 5$. At least that is the implication of the dubious definition. But I wanted to make sure. This is my stab at it:

\begin{align*} a : b : c &= \frac{3}{8} : \frac{4}{8} : \frac{5}{8} = \frac{6}{5}, \\ a' : b' : c' &= \frac{3}{6} : \frac{4}{6} : \frac{5}{6} = \frac{9}{10}. \end{align*}

Proposition states: If two triangles are similar, then $a : b : c = a' : b' : c'$. But

$$\frac{6}{5} \neq \frac{9}{10} \Longrightarrow a : b : c \neq a' : b' : c'.$$

The same goes for other combinations of triangles (e.g. $a'' : b'' : c'' = a' : b' : c'$); no equality, only inequalities.

My second wish: please tell me why I get this result. It seems like the definition of similarity in my textbook is false.

There are two possibilities: either the proposition in my textbook is false or my understanding of the operand "$:$" (and consequently my calculations) is false. I am certain that the latter one is to blame for my confusion. It appears that I have a deep underlying issues of improper understanding of these definitions, and in the end, improper understanding of the concept of similarity as a whole.

Please help me. Explain everything in great detail, so I may finally truly understand. I am currently in high school and intend on studying pure mathematics in college. I want to get to the bottom of things, understand, not just pick up some pattern that I was taught and spew it on the test sheet. Without this sense of true understanding, all of my efforts are meaningless.

  • $\begingroup$ Yes, you're definitely misunderstanding something. Did you go to school in a country where $:$ is a division symbol? $\endgroup$ – Henning Makholm Nov 17 '17 at 16:15
  • $\begingroup$ Yes he is, they use it in the picture, see? $\endgroup$ – Aqua Nov 17 '17 at 16:17
  • $\begingroup$ @HenningMakholm Yes. For example, $a : b$ is $\frac{a}{b}$. This shouldn't be a problem in the context of my textbook, as it cites the "additional" criteria (e.g. $a' : a = b' : b = c' : c$) also with "$:$" right alongside this "problematic" definition, and it works (my professor's definition, that is)! So I am truly confused... $\endgroup$ – Gregor Perčič Nov 17 '17 at 16:20

Your first question: Show that $a:a'=b:b'\implies a:b=a':b'$.

This is Euclid's proposition V.16. Your actual request is for a version of this involving $a,b$ and $c$, but it can be handled two at a time. We assume, with Euclid, that all quantities being put into ratios are strictly greater than $0$. With proportions, we have that the product of the means equals the product of the extremes, thus:

$$a:a'=b:b' \Leftrightarrow ab'=a'b \Leftrightarrow a:b=a':b'.$$

In more conventional notation,

$$\frac{a}{a'}=\frac{b}{b'} \Leftrightarrow ab'=a'b \Leftrightarrow \frac{a}{b}=\frac{a'}{b'},$$

where we can simply think of the moves as multiplication and division.

Now, that answer will only extend to the three-ratio case if we clear up what's going on with the notation $a:b:c$. When we write $a:b:c=d:e:f$, this is shorthand for the two proportions $a:b=d:e$ and $b:c=e:f$, which together imply that $a:c=d:f$. It does not represent the equation $(a/b)/c=(d/e)/f$. I think this is why that notation for ratios is not very common anymore.

Thus, when you calculate a numeric value for the expression $\frac38:\frac48:\frac58$, it is not meaningful. That expression is not intended to simplify to a single number.

Answering your initial question, with this notational clarification in mind, we can proceed as follows:

$\begin{align} a:a'=b:b'=c:c' &\implies a:a'=b:b' \text{ and } b:b'=c:c' \text { (separating the compound equality)}\\ &\implies a:b=a':b' \text{ and } b:c=b':c' \text{ (alternating proportions by V.16)}\\ &\implies a:b:c=a':b':c' \text{ (definition of triplicate proportion)} \end{align}$

  • $\begingroup$ Thank you for your answer! Two things though: 1. I still do not understand how we can construct $a : b : c$ with "handling two at a time". 2. Could you please Show me how to get $a : b : c = d : e : f$ from $a : b = d : e$ and $b : c = e : f$. $\endgroup$ – Gregor Perčič Nov 17 '17 at 16:33
  • $\begingroup$ That's a definition. $a:b:c=d:e:f$ literally just means $a:b=d:e$ and $b:c=e:f$. $\endgroup$ – G Tony Jacobs Nov 17 '17 at 16:34
  • $\begingroup$ Wow. Could we represent this definition as well-ordered triples in classical number theory (as @HenningMakholm mentioned)? $\endgroup$ – Gregor Perčič Nov 17 '17 at 16:36
  • $\begingroup$ @GTonyJacobs: Not complextly exactly, since your proposed unfolding would imply that $1:0:1=1:0:2$. $\endgroup$ – Henning Makholm Nov 17 '17 at 16:37
  • 1
    $\begingroup$ @GregorPerčič: Except for the question of whether some of the quantities cam be zero (or negative), the concept Tony is explaining is the same one I am explaining, just from a slightly different focus. A minor nitpick though: you're talking about ordered triples here, not "well-ordered" which has a different (and completely unrelated), meaning. $\endgroup$ – Henning Makholm Nov 17 '17 at 16:45

You're misunderstanding the notation $$ x:y:z = p:q:r $$ You've been taught that $:$ means division, so you think that this is an equality asserting that the results of the divisions $\frac{x/y}z$ and $\frac{p/q}r$ are the same. But that is not what it means!

Something like $x:y:z = p:q:r$ is a (nowadays not extremely common) notation for the claim that the triples $(x,y,z)$ and $(p,q,r)$ are "in proportion" to each other -- or less fancily, that the relation between $x$, $y$ and $z$ is the same as the relation between $p$, $q$ and $r$, in a particular sense.

There are various ways to make precise what this means; one that fits a modern number-centric understanding well is

By $x:y:z=p:q:r$ we mean that there is some number $k\ne 0$ such that $p=k\cdot x$ and $q=k\cdot y$ and $r=k\cdot z$.

It should be clear how this generalizes to different number of operands such as $x:y:z:w=p:q:r:s$, or indeed just $x:y=p:q$.

In particular if $x:y=p:q$ means that there is $k\ne 0$ such that $p=kx$ and $q=ky$, it is a simple matter of algebra to see that this is the case exactly when $\frac xy = \frac pq$ -- where the fractions are now truly dvisions -- except when $y$ and/or $q$ is zero. So in that case this usage is mostly compatible with writing $:$ for division. But this connection does not generalize neatly to more than two elements on each site.

The above defines $x:y:z=p:q:r$ as a single, monolithic relation between six quantities, notated with the compound symbol "$::=::$". Some people prefer to write it as $x:y:z\sim p:q:r$ instead, to emphasize that it shouldn't be read as an equality between two things notated $x:y:z$ and $p:q:r$.

But if we really want it to be an actual equality between $x:y:z$ and $p:q:r$, we could achieve it by defining

The notation $x:y:z$ denotes the set $\{(kx,ky,kz)\mid k\ne 0\}$.

And we can then prove that these sets are identical exactly if the monolithic definition of $x:y:s=p:q:r$ above is satisfied.

(But I don't think this definition of $x:y:z$ as a set is common. I have seen it a few times in the context of projective geometry, but not in elementary geometry or arithmetic).

  • $\begingroup$ I think it was a common notation a long time ago, when everyone still learned geometry from Euclid's Elements. $\endgroup$ – G Tony Jacobs Nov 17 '17 at 16:46

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