In this section, our factorization of isometries into osometries and translations pays off handsomely. It "factors" the difficulty of proving that isometries preserve displacements, angles, barycentric coordinates, lines and triangles, etc. into separate (and easier) proofs for osometries and translations.
(The next theorem is a correction of the one given in Tondeur’s textbook, Theorem 4.1. There it is tacitly assumed that the origin is at the vertex of the angle.)
Proof. Before beginning the proof, let’s review a formula from vector calculus. For any vectors `X` and `Y`, the dot product is related to the angle between them by
`X Y` = 
`XY\ cos\ /_XY`. 
(1) 
Substituting, we have
$ (P^\alpha  Q^\alpha)(R^\alpha  Q^\alpha)\ =$ 
$  P^\alpha  Q^\alpha   R^\alpha  Q^\alpha \ cos\angle P^\alpha Q^\alpha R^\alpha $ 

$=_{bb{1}}$ 
$  P  Q   R  Q \ cos \angle P^\alpha Q^\alpha R^\alpha $ 
(2) 
where equality $=_{bb{1}}$ follows from the definition of isometry. So we have reduced the problem of showing that isometries preserve dot products, to showing that they preserve angles.
The idea now is to take advantage of the factorization of `alpha` by proving the theorem for translations and osometries, and then proving it for compositions of isometries.
Step 1: Proof for $\alpha=\tau_D$.
Assuming $\alpha$ is a translation by $D$, we have
$P^\alphaQ^\alpha\ $ = 
$P^\tauQ^\tau$ 
= 
$(P+D)(Q+D)$ 
= 
$PQ$. 
Thus,
$ (P^tauQ^tau)(R^tauQ^tau)\ =_{bb{1}}$ 
$ (PQ)(RQ) $ 
$=_{bb{2}}$ 
$ PQ RQ \ cos \angle PQR $ 
where $ =_{bb{1}} $ is by substitution, and $ =_{bb{2}} $ follows from equation (1) above.
Combining this with equation (2) yields
$PQRQ\ cos \angle P^\tau Q^\tau R^\tau\ =\ PQRQ\ cos \anglePQR$. 
And so,
$cos\ \angle P^\tau Q^\tau R^\tau\ =\ cos\ \angle PQR$. 
This completes Step 1. The next step is to consider osometries:
Step 2: Proof for $\alpha=\beta$, where $\beta$ is an osometry.
$(P^\betaQ^\beta) (R^\betaQ^\beta)\ $ = 
$(PQ)^\beta (RQ)^\beta$ 
(by O3, O4) 

= 
$(PQ) (RQ)$ 
(by O2) 

= 
$PQ RQ \ cos \angle PQR$. 
(by (1)) 
Then combining this with (2), we get our result
$cos\ \angle P^\beta Q^\beta R^\beta\ =\ cos\ \angle PQR$. 
We now have enough to prove the theorem for any isometry:
Step 3: Proof for general $\alpha$.
By the Osometry Factorization Theorem (OFT), we can factor $\alpha$ as $\ \alpha=\tau_D\ \beta.
Then
$cos\ \angle P^\alpha Q^\alpha R^\alpha\ $ = 
$cos\ \angle\tau\beta(P)\ \tau\beta(Q)\ \tau\beta(R)$ 

= 
$cos\ \angle\tau(P^\beta)\ \tau(Q^\beta)\ \tau(R^\beta)$ 

= 
$cos\ \angle P^\beta Q^\beta R^\beta$ 
(by Step 1) 

= 
$cos\ \angle PQR$. 
(by Step 2) 
And we have proven the theorem.
`square`
To prove this theorem, we’ll employ a method similar to the one used in the previous proof.
Proof. As in the previous proof, we will prove the theorem for translations and osometries separately, and then prove it for the composition.
Out first case are translations:
Step 1: Proof for $\alpha=\tau_D$.
`aA^tau+bB^tau+cC^tau` = 
`a(A+D)+b(B+D)+c(C+D)` 
(by definition) 

= 
`aA+bB+cC+(a+b+c)D` 

= 
`aA+bB+cC+D` 
(`a+b+c=1`) 

= 
`(aA+bB+cC)^tau`. 
(by definition) 
Next, we consider ososmetries:
Step 2: Proof for $ \alpha=\beta $, where $ \beta $ is an osometry.
$ aA^\beta+bB^\beta+cC^\beta $ = 
$ (aA)^\beta+(bB)^\beta+(cC)^\beta $ 
(by O3) 

= 
$ (aA+bB+cC)^\beta $. 
(by O4) 
Finally, for general isometries:
Step 3: Proof for general $\alpha$.
By the Osometry Factorization Theorem (OFT), factor $\alpha=\tau_D\ beta$.
`aA^alpha+bB^alpha+cC^alpha` = 
`a(tau \ beta(A))+b(tau \ beta(B))+c(tau \ beta(C))` 

= 
`a(tau(A^beta))+b(tau(B^beta))+c(tau(C^beta))` 

= 
`tau(aA^beta+bB^beta+cC^beta)` 
(by step 1) 

= 
`tau((aA+bB+cC)^beta)` 
(by step 2) 

= 
`(aA+bB+cC)^alpha`. 
This concludes the proof of the Barytheorem.
`square`
If you only read the first and last line in each of the first two parts of the above proof, they seem to say the same thing. Both express the "linearity" of a transformation, but translations are not linear isometries (osometries). What gives?
You need to pay attention to the crucial role that $ a + b + c = 1 $ plays for the first case. We might say (but shouldn’t) that translations are "linear" for some linear combinations of points, namely those that express the barycentric coordinates of a point.
Recall that we required a point transformation, or trafo, to be a onetoone and onto (bijective) transformation of the plane. Although we included this property in the definition of an isometry, it doesn’t have to be. It is, in fact, a consequence of what we have learned in this lesson.
Proof. Let $ \alpha $ be an isometry.
To show $ \alpha $ is onetoone, suppose $ \alpha(X)=\alpha(Y) $ or equivalently, $ X^\alphaY^\alpha\ =\ 0 $. Then
$ X^\alphaY^\alpha\ $ = 
$0$. 
But since $ \alpha $ is an isometry,
$ X^\alphaY^\alpha\ $ = 
$ XY $. 
Therefore, $ XY=0 $ and by property L2 above, we have $ X=Y $.
To show $ \alpha $ is onto, we apply the factorization procedure of the previous two proofs.
Step 1: Proof for $ \alpha=\tau_D $.
Let $ Y $ be any point. We must find an $ X $ such that $ X^\tau=Y $.
Indeed, writing out the equation $ X+D=Y $ implies that we should choose $ X\ :=\ YD $.
Then a simple calculation shows
$ \tau(X) $ = 
$ \tau(YD) $ 
= 
$YD+D$ 
= 
$Y$. 
Next, consider osometries.
Step 2: Proof for $ \alpha=\beta $, where $ \beta $ is an osometry.
This case follows from the previous theorem. Let $ A,B,C $ be three noncollinear points.
Step 3: Proof for general `alpha`.
By the Osometry Factorization Theorem (OFT), we factor $ \alpha$ as $ \ \alpha=\tau_D\ \beta $.
Now given $ Y $, there is a point $ Z $ such that $ Z^\tau=Y $ by Step 1.
And by Step 2, there is a point $ X $ such that $ X^\beta=Z $.
Hence, $ \alpha(X)\ =\ \tau\beta(X)\ =\ \tau(Z)\ =\ Y $.
$\square$
We close this lesson with some immediate consequences of the theorem that isometries preserve barycentric coordinates. Because it provides an easy solution to many problems about isometries, We shall affectionately refer to this theorem as the Barytheorem. It is important to understand just what the Barytheorem actually says. Let $ \triangle ABC $ define one system of barycentric coordinates and $ \alpha $ be an isometry.
Why is $ \triangle A^\alpha B^\alpha C^\alpha $ a triangle?
It’s easiest way to prove this is by contradiction. So suppose $ (A^\alpha B^\alpha C^\alpha) $ are collinear. We exclude the possibility that two of these image points are the same because isometries are 1:1 transformations. By the Main Collinearity Lemma (MCL) of Affine Geometry there exist nonzero real numbers $ a + b = 1 $ for which $ A^\alpha = b B^\alpha + c C^\alpha $. Applying the inverse $ \alpha ^{1} $ to both sides of the equation yields $ A = bB + cC$, and $ b + c = 1 $ still. This says that $ (ABC) $ are collinear. And that contradicts the hypothesis.
So, where did we use the Barytheorem? We used two facts. One that the inverse is also an isometry. Then we applied the Barytheorem to $ bB^\alpha + cC^\alpha = 0A^\alpha + bB^\alpha + c C^\alpha $ since $ 0 + b + c = 1 $ still.
How does the Barytheorem show that isometries map lines to lines?
Recall that $ \ell_{AB} = \{ X : (\exists a+b=1)(X = aA + bB) \}$. Note that relative to any point $ C $ not on the line, $ X = aA + bB + 0C $. Applying isometry $ \alpha $ to each point in the set we get that
$ \alpha( \ell_{AB} ) = \{ a A^\alpha + b B^\alpha \} = \ell_{A^\alpha B^\alpha } $ 
Do isometries transform parallels into parallels? Hint: Factor the answer into two cases.
This is not a trick question. Two lines $ \ell = (AB) , k = (XY) $ are parallel if the displacement vectors $ (BA) = t (YX) $ are proportional.
For a translation $ \tau_D $, calculate $ A^\tau  B^\tau = A  B $, and you can continue $ = t(YX) = t (Y^\tau  Y^\tau) $. Are you surprised that the proportionality factor $ t $ doesn’t change under the translation?
For an osometry $ \beta $ use linearity thus.
$ A^\beta  B^\beta = (A  B)^\beta =(tX tY)^\beta =tX^\beta  tY^\beta= t(X^\beta  Y^\beta) $ 
Since isometries are such versatile transformations, wouldn’t it be nice to know all the different kinds of isometries? In one sense we already do. Every isometry is the composition of two unique isometries, a translation and an osometry (linear isometry). As we will see later, osometries are composed of only two kinds, rigid rotations about the origin, and reflections in lines through the origin. So, why isn’t that enough? And it is for many purposes.
As "scientists" of geometry we want two things further, discover the decomposition of an isometry into utterly simple factors, and partition the set of nontrivial isometries into mutually exclusive sets. The former goal is to identify the "atomic" isometries that generate all isometries under composition. The second is the classification of isometries by their composition which then determines what they do to points.
Our exposition could diverge here from the geometrical approach taken by Tondeur. We could discover that every linear isometry can be represented by a matrix, which takes every point to the product of the matrix by the coordinates of the point. In fact, we could, by going one dimension higher, concoct a scheme whereby translation becomes a linear transformation, but in the projective plane. This is the approach taken by computer graphics.
We could diverge yet another way, namely by imposing the complex numbers and their algebra as the numerical description of the Euclidean plane. That is the approach taken by Michael Hvidsten, the author of the textbook for MA 402, and also a PhD student of Prof. Tondeur. This approach lets one study the Euclidean and nonEuclidean plane using identical mathematical methods.
Recall that the concept of isometry is the correct physical interpretation of Euclid’s concept of congruence. Each different way of studying the isometry group lends new insight in what profound contribution to human knowledge Euclid and his Greek colleagues made some 2300 years ago. Geometry, along with the theory of numbers, astronomy and music (the quadrivium or graduate curriculum of the GraecoRoman world) is truly foundational, and well worth our study of it.