Proof of $ASA$ criterium I am studying affine geometry and I found this problem in which I am having some trouble solving it.
Let $\mathcal{A}$ be an affine space over an Euclidian space $E$ of dimension $2$ and let $A,A’,B,B’,C,C’ \in \mathcal{A}$ be points. Consider the triangles $\triangle ABC$ and $\triangle A’B’C’$ in $\mathcal{A}$. If $d(A,B) = d(A’,B’)$, $\measuredangle\{\overrightarrow{BA},\overrightarrow{BC}\} = \measuredangle\{\overrightarrow{B’A’},\overrightarrow{B’C’}\}$ and $\measuredangle\{\overrightarrow{AB},\overrightarrow{AC}\} = \measuredangle\{\overrightarrow{A’B’},\overrightarrow{A’C’}\}$, then
$$d(B,C) = d(B’,C’) \qquad \ d(A,C) = d(A’,C’) \qquad \measuredangle\{\overrightarrow{CB},\overrightarrow{CA}\} = \measuredangle\{\overrightarrow{C’B’},\overrightarrow{C’A’}\}.$$
My attempt.
Since, $\measuredangle\{\overrightarrow{BA},\overrightarrow{BC}\} = \measuredangle\{\overrightarrow{B’A’},\overrightarrow{B’C’}\}$, we know that $\cos\measuredangle\{\overrightarrow{BA},\overrightarrow{BC}\} = \cos\measuredangle\{\overrightarrow{B’A’},\overrightarrow{B’C’}\}$. Hence, it follows that $\frac{\overrightarrow{BA}\mid\overrightarrow{BC}}{d(B,C)} = \frac{\overrightarrow{B’A’}\mid\overrightarrow{A’C’}}{d(B’,C’)}$. Using the same reason, we get from the fact $\measuredangle\{\overrightarrow{AB},\overrightarrow{AC}\} = \measuredangle\{\overrightarrow{A’B’},\overrightarrow{A’C’}\}$ that $\frac{\overrightarrow{AB}\mid\overrightarrow{AC}}{d(A,C)} = \frac{\overrightarrow{A’B’}\mid\overrightarrow{A’C’}}{d(A’,C’)}$. By applying the Chasles relation, one can obtain that $\frac{d^2(A,B)-\overrightarrow{AB}\mid\overrightarrow{AC}}{d(B,C)} = \frac{d^2(A’,B’)-\overrightarrow{A’B’}\mid\overrightarrow{A’C’}}{d(B’,C’)}$. If I apply the law of cosines, I get that $\frac{d^2(B,C) - d^2(A,C)}{d(B,C)} = \frac{d^2(B’,C’) - d^2(A’,C’)}{d(B’,C’)}$. Although I can’t deduce from here that $d(B,C) = d(B’,C’)$ or that $d(A,C) = d(A’,C’)$.
What am I missing? Can someone give me a hint of what should I do?
Thank you for your attention!
 A: Preliminaries
A ray in $E$ is a set of the form $\{\lambda a\; : \;\lambda\geq0\}$ for some $a\in E\setminus\{0\}$. An angle in $E$ is an ordered pair $(\Delta_1, \Delta_2)^\wedge$ of rays. Two angles $(\Delta_1, \Delta_2)^\wedge$ and $(\Delta'_1, \Delta'_2)^\wedge$ are equal by definition if there exists a rotation in $E$, i.e. an element $r$ of the group $\mathbb{SO}(E)$, such that
$$ r(\Delta_1)=\Delta_2\qquad \text{and}\qquad r(\Delta'_1)=\Delta'_2. \tag1 $$
In this way the set $\mathcal U$ of equivalence classes of equal angles is in a bijection with the group $\mathbb{SO}(E)$, so is itself a group, called the group of angles of $E$. These groups are therefore isomorphic, and are denoted additively the first and multiplicatively the second. They are also commutative because $\dim(E)=2$. The isomorphism
$$ \rho \;:\; \mathcal U \;\to\; \mathbb{SO}(E) \tag2 $$
associates to each class $[\vartheta]$ of equal angles the corresponding rotation, which we will often write simply as $\rho_\vartheta$.
We observe that, in general, we have $\;\rho_\vartheta^{-1} = \rho_{-\vartheta}$.
An angle $(\Delta_1, \Delta_2)^\wedge$ will often be written, for simplicity, as $(a_1, a_2)^\wedge$, where $a_1\in\Delta_1\setminus\{0\}$ and $a_2\in\Delta_2\setminus\{0\}$, since there is no ambiguity, as it is easy to show.
The ASA Criterion
Let $ABC$ and $A'B'C'$ triangles in $\mathcal A(E)$ such that
$$ \|B-A\| = \|B'-A'\| $$
$$ \alpha := (B-A, C-A)^\wedge = (B'-A', C'-A')^\wedge =:\alpha' \tag3 $$
$$ \beta := (A-B, C-B)^\wedge = (A'-B', C'-B')^\wedge =: \beta'. $$
We will show that $ABC = A'B'C'$, i.e. there exists an isometry of $\mathcal A(E)$ that transform $ABC$ into $A'B'C'$.
Let
$$ \theta := (B-A, B'-A')^\wedge \tag4 $$
and consider the function $f\; : \;\mathcal A(E) \to \mathcal A(E)$ defined by
$$ \boxed{f(P) := A' + \rho_\theta(P-A).} \tag5 $$
We have:
$\quad$1.$\quad$ $f$ is a bijective isometry: for all $P,Q\in\mathcal A(E)$, one has
\begin{align}
\|f(P)-f(Q)\| &= \|A'+\rho_\theta(P-A)-(A'+\rho_\theta(Q-A))\| =\\[1ex]
&= \|\rho_\theta(P-A-(Q-A))\| = \tag6 \\[1ex]
&= \|\rho_\theta(P-Q)\| = \|P-Q\|
\end{align}
being $\rho_\theta$ an isometry. This also proves that $f$ is injective:
$$ f(P)=f(Q) \;\implies\; 0 = \|f(P)-f(Q)\| = \|P-Q\| \;\implies\; P = Q. $$
But it is simpler to prove injectivity and suriectivity simultaneously by showing that $f$ is invertible. In fact it is immediately seen that
$$ f^{-1}(P') = A + \rho_\theta^{-1}(P'-A') = A + \rho_{-\theta}(P'-A'). \tag7 $$
$\quad$2.$\quad$ $f(A) = A'+\rho_\theta(A-A) = A'+\rho_\theta(\vec 0) = A'.$
$\quad$3.$\quad$ $f(B) = A'+\rho_\theta(B-A) = A'+\lambda(B'-A')$ for some $\lambda\geq 0$. Now, from $\|\rho_\theta(B-A)\| = \|\lambda(B'-A'\| = |\lambda|\|B'-A'\|$ it follows $\lambda=1$ by the hypothesis $\|B-A\|=\|B'-A'\|$, and $\lambda\geq0$. Hence $f(B)=B'$, and also $\rho_\theta(B-A)=B'-A'$.
$\quad$4.$\quad$ Indicating in general with $P\rightsquigarrow Q$ the affine ray having origin in $P$ and containing $Q$, i.e. the set $\{P+\lambda(Q-P)\;:\;\lambda\geq0\}$, then:
$$ f(A\rightsquigarrow C) = A'\rightsquigarrow C'\qquad \text{and} \qquad f(B\rightsquigarrow C) = B'\rightsquigarrow C'. \tag8 $$
In fact, we van write
\begin{align}
f(A+\lambda(C-A)) &= A'+\rho_\theta(\lambda(C-A)) =\\[1ex]
&= A'+\lambda\,\rho_\theta(C-A) =\\[1ex]
&= A'+\lambda\,\rho_\theta\,\rho_\alpha(\rho_{-\alpha}(C-A)) = \tag9 \\[1ex]
&= A'+\lambda\,\rho_\theta\,\rho_\alpha(\mu(B-A)) = \qquad\qquad\text{[for some }\mu\geq0]\\[1ex]
&= A'+\lambda\mu\,\rho_\alpha\,\rho_\theta(B-A) = \qquad\qquad[\mathbb{SO}(E)\text{ is commutative}]\\[1ex]
&= A'+\lambda\mu\,\rho_\alpha(B'-A') =\ldots
\end{align}
Now we observe that by the hypothesis $\alpha$ and $\alpha'$ are equal in amplitude: this means that $\alpha=\alpha'\;$ OR $\;\alpha=-\alpha'$.
IF$\;\;\boldsymbol{\alpha=\alpha'}$, $(9)$ proceeds easily:
\begin{align}
\ldots &= A'+\lambda\mu\,\rho_{\alpha'}(B'-A') =\\[1ex]
&= A'+\lambda\mu(\nu(C'-A')) =\qquad\qquad\qquad\quad\text{[for some }\nu\geq0] \tag9\\[1ex]
&= A'+\lambda\mu\nu(C'-A') \in A'\rightsquigarrow C'.
\end{align}
This proves the inclusion $f(A\rightsquigarrow C) \subseteq A'\rightsquigarrow C'$. The opposite inclusion goes in a similar way by appropriately using $f^{-1}$ instead of $f$.
The second formula in (8) is demonstrated in a perfectly analogous way $\;$IF $\boldsymbol{\;\beta=\beta'}$.
Assuming $\alpha=\alpha'$ and $\beta=\beta'$, therefore, we have
$$ \{f(C)\} = f\Big((A\rightsquigarrow C)\cap(B\rightsquigarrow C)\Big) = f(A\rightsquigarrow C)\cap f(B\rightsquigarrow C) = (A'\rightsquigarrow C')\cap(B'\rightsquigarrow C') = \{C'\} $$
so $f(C) = C'$. Our criterium follows immediately if we remember that any affine function preserves the barycentric combinations, with the same coefficients: if $\;\lambda+\mu+\nu=1$, then
$$ f(\lambda A + \mu B + \nu C) = \lambda f(A) + \mu f(B) + \nu f(C) = \lambda A' + \mu B' + \nu C' \tag{10}, $$
and if also $\lambda, \mu, \nu\geq0$, the combinations $\lambda A+\mu B+\nu C\;$ and $\;\lambda A'+\mu B'+\nu C'$ describes respectively our triangles $ABC$ and $A'B'C'$. For example, if $\nu=0$, the combination $\lambda A+\mu B$ describes, for $\lambda+\mu=1$, the side $AB$, etc.
It is interesting in itself to show directly that, always under the same assumptions already made (that is $\alpha=\alpha', \;\beta=\beta'$), the angles
$$ \gamma := (A-C, B-C)^\wedge \qquad\text{and}\qquad \gamma' :=(A'-C', B'-C')^\wedge $$
have the same amplitude, or rather they are equal.
In fact,
\begin{align}
A'-C' &= f(A)-f(C) = A'+\rho_\theta(A-A)-\big[A'+\rho_\theta(C-A)\big] =\\[1ex]
&= \rho_\theta(\vec0)+\rho_\theta(A-C) = \rho_\theta(A-C)
\end{align}
and similarly
\begin{align}
B'-C' &= f(B)-f(C) = A'+\rho_\theta(B-A)-\big[A'+\rho_\theta(C-A)\big] =\\[1ex]
&= \rho_\theta(B-A+(A-C)) = \rho_\theta(B-C)
\end{align}
imply
$$ (A-C, A'-C')^\wedge = \theta = (B-C, B'-C')^\wedge. $$
The conclusion then follows from the group identity
$$ (A-C, B-C)^\wedge = (A-C, A'-C')+(A'-C', B'-C')^\wedge+(B'-C', B-C)^\wedge, $$
which gives
$$ (A-C, B-C)^\wedge-(A'-C', B'-C')^\wedge = (A-C, A'-C')^\wedge-(B-C, B'-C')^\wedge = \theta-\theta=0. $$
$\quad$5.$\quad$ Greater care is needed for the case $\alpha=-\alpha'$. First a Lemma:
$$ \alpha=\alpha' \iff \beta=\beta'. \tag{11}$$
$\big[$ hence $\alpha = -\alpha' \iff \beta = -\beta'. \big]$
Let $\;L\;:\;\mathcal A(E) \to \mathbb R$ be an affine form whose kernel consists of the points on the line $AB$, that is
$$ L(P)=0 \iff P\in AB. \tag{12} $$
For example, $L(P)=(P-A\big|z)$, where $z$ is any fixed non zero vector orthogonal to $AB$. We recall that while $L$ describes the line $AB$, its linear part $dL$ describes the direction of $AB$ (a vector subspace of $E$).
It follows that
$$ L' := L\circ f^{-1} \tag{13} $$
is an affine form whose kernel is the set of the points of the line $A'B'$, as it is easy to check. And it is also easy to check that the linear part of $L'$ is given by
$$ dL' = dL\circ df^{-1} = dL\circ\rho_\theta^{-1}. \tag{14} $$
Let us therefore assume that $\alpha=\alpha'$ and $\beta = -\beta'$. We will obtain a contradiction by calculating $dL'(C'-B')$ in two different ways.
We have, on the one hand,
\begin{align}
dL'(C'-B') &= dL'(C'-A'+(A'-B')) = dL'(C'-A') =\\[1ex]
&= dL\rho_\theta^{-1}(C'-A') = dL\rho_\theta^{-1}\rho_{\alpha'}(\lambda(B'-A')) = \qquad\quad\big[\text{for some }\lambda>0\big]\\[1ex]
&= \lambda\,dL\rho_\alpha\rho_\theta^{-1}(B'-A') = \lambda\,dL\rho_\alpha(B-A) = \tag{15}\\[1ex]
&= \lambda\,dL(\mu(C-A))= \lambda\mu\,dL(C-A) =\qquad\qquad\qquad\big[\text{for some }\mu>0\big]\\[1ex]
&= \lambda\mu\,dL(C-B+(B-A)) = \lambda\mu\,dL(C-B) =\\[1ex]
&= \lambda\mu\,dL(v+\nu z) =\lambda\mu\nu\,dL(z)\qquad\qquad\qquad\qquad\qquad\big[\nu\neq 0\big]
\end{align}
where $z$ is any non zero vector orthogonal and $v$ is a vector parallel to the direction of $AB$.
On the other hand,
\begin{align}
dL'(C'-B') &= dL\rho_\theta^{-1}(C'-B') = dL\rho_\theta^{-1}\rho_{\beta'}(\lambda'(A'-B'))=\qquad\quad\big[\text{for some }\lambda'>0\big] \\[1ex]
&= \lambda'\,dL\rho_{\beta'}\rho_\theta^{-1}(-(B'-A')) = -\lambda'\,dL\rho_{\beta'}(B-A) = \tag{16}\\[1ex]
&= \lambda'\,dL\rho_{-\beta}(A-B) =\ldots\\
\end{align}
Here the problem of calculating $\;\rho_{-\beta}(A-B)\;$ arises. The idea is to consider the orthogonal symmetry $s$ of $\mathcal A(E)$ with axis the line $\;AB$. By a well known property of symmetries [see Note below], we have
$$ \beta = (A-B, C-B)^\wedge = (s(C)-B, A-B)^\wedge \tag{*}$$
hence
$$ \hskip5cm\rho_\beta(s(C)-B) = \mu'(A-B) \qquad\qquad\qquad\big[\text{for some }\mu'>0 \big]$$
hence
$$ \rho_{-\beta}(A-B) = \mu'^{\,-1}(s(C)-B) = \mu'^{\,-1}ds(C-B). $$
So (16) then proceeds as follows:
\begin{align}
\ldots&= \lambda'\mu'^{\,-1} dL\,ds(C-B) =\\[1ex]
&= \lambda'\mu'^{\,-1}dL\left(C-B-2\,\frac{(C-B\big|z)}{(z\big|z)}\,z\right)=\qquad\qquad\big[z \text{ as in (15)},\quad \text{see Note below}\big]\\[1ex]
&= \lambda'\mu'^{\,-1}dL\left(\nu z-2\,\frac{(\nu z\big|z)}{(z\big|z)}\,z\right) =\qquad\qquad\qquad\quad\big[C-B \text{ as in (15)}\big]\tag{16}\\[1ex]
&= \lambda'\mu'^{\,-1}dL\left(\nu z-2\nu z\right) =\\[1ex]
&= -\lambda'\mu'^{\,-1}\nu\,dL(z),
\end{align}
and we obtain a contradiction as the signs of (15) and (16) are opposite. The Lemma is proved.
In conclusion, if $\alpha=-\alpha'$, then also $\beta=-\beta'$. In this case, the symmetry $s$, which is an isometry, transforms $ABC$ into a new triangle $AB\overline C$, where $\overline C=s(C)$, to which we can apply the result of n. $\boldsymbol 4$, obtaining that $ABC$ is (inversely) equal to $A'B'C'$. This concludes the proof of the ASA criterion.
$\quad$
Note.$\quad$ The (orthogonal) symmetry in $\mathcal A(E)$ with axis the line $AB$ is given (easy calculation) by
$$ s\;:\; P \;\longmapsto \;P-2\,\frac{(P-A\big|z)}{(z\big|z)}\,z\qquad\qquad \left[\,ds\;:\;v \;\longmapsto\; v-2\,\frac{(v\big|z)}{(z\big|z)}\,z\,\right] \tag{17}
$$
where $z$ is any non zero vector orthogonal to the line $AB$.
The (*) is an immediate consequence of the following property of symmetries:
Two angles $(\Delta_1, \Delta_2)^\wedge$ and $(\Delta'_1, \Delta'_2)^\wedge$ of $E$ are equal if and only if there exists a vector symmetry $s$ such that
$$ s(\Delta_1)=\Delta'_2 \qquad\text{and}\qquad s(\Delta_2)=\Delta'_1. $$
The symmetry $s$ is unique. [See Dieudonné, Algèbre Linéaire et Géométrie élémentaire, Hermann; (5.4.7)].
