congruence axioms
General Congruence Relations^{}. Let $A$ be a set and $X=A\times A$. A relation on $X$ is said to be a congruence relation on $X$, denoted $\cong $, if the following three conditions are satisfied:

1.
$(a,b)\cong (b,a)$, for all $a,b\in A$,

2.
if $(a,a)\cong (b,c)$, then $b=c$, where $a,b,c\in A$,

3.
if $(a,b)\cong (c,d)$ and $(a,b)\cong (e,f)$, then $(c,d)\cong (e,f)$, for any $a,b,c,d,e,f\in A$.
By applying $(b,a)\cong (a,b)$ twice, we see that $\cong $ is reflexive^{} according to the third condition. From this, it is easy to that $\cong $ is symmetric^{}, since $(a,b)\cong (c,d)$ and $(a,b)\cong (a,b)$ imply $(c,d)\cong (a,b)$. Finally, $\cong $ is transitive^{}, for if $(a,b)\cong (c,d)$ and $(c,d)\cong (e,f)$, then $(c,d)\cong (a,b)$ because $\cong $ is symmetric and so
$(a,b)\cong (e,f)$ by the third condition. Therefore, the
congruence relation is an equivalence relation^{} on pairs of elements
of $A$.
Congruence Axioms in Ordered Geometry^{}. Let $(A,B)$ be an
ordered geometry with strict betweenness relation $B$.
We say that the ordered geometry $(A,B)$ satisfies the congruence
axioms if

1.
there is a congruence relation $\cong $ on $A\times A$;

2.
if $(a,b,c)\in B$ and $(d,e,f)\in B$ with

–
$(a,b)\cong (d,e)$, and

–
$(b,c)\cong (e,f),$
then $(a,c)\cong (d,f)$;

–

3.
given $(a,b)$ and a ray $\rho $ emanating from $p$, there exists a unique point $q$ lying on $\rho $ such that $(p,q)\cong (a,b)$;

4.
given the following:

–
three rays emanating from ${p}_{1}$ such that they intersect with a line ${\mathrm{\ell}}_{1}$ at ${a}_{1},{b}_{1},{c}_{1}$ with $({a}_{1},{b}_{1},{c}_{1})\in B$, and

–
three rays emanating from ${p}_{2}$ such that they intersect with a line ${\mathrm{\ell}}_{2}$ at ${a}_{2},{b}_{2},{c}_{2}$ with $({a}_{2},{b}_{2},{c}_{2})\in B$,

–
$({a}_{1},{b}_{1})\cong ({a}_{2},{b}_{2})$ and $({b}_{1},{c}_{1})\cong ({b}_{2},{c}_{2})$,

–
$({p}_{1},{a}_{1})\cong ({p}_{2},{a}_{2})$ and $({p}_{1},{b}_{1})\cong ({p}_{2},{b}_{2})$,
then $({p}_{1},{c}_{1})\cong ({p}_{2},{c}_{2})$;

–

5.
given three distinct points $a,b,c$ and two distinct points $p,q$ such that $(a,b)\cong (p,q)$. Let $H$ be a closed half plane with boundary $\overleftrightarrow{pq}$. Then there exists a unique point $r$ lying on $H$ such that $(a,c)\cong (p,r)$ and $(b,c)\cong (q,r)$.
Congruence Relations on line segments^{}, triangles^{}, and angles. With the above five congruence axioms, one may define a congruence relation (also denoted by $\cong $ by abuse of notation) on the set $S$ of closed line segments of $A$ by
$$\overline{ab}\cong \overline{cd}\mathit{\hspace{1em}\hspace{1em}}\text{iff}\mathit{\hspace{1em}\hspace{1em}}(a,b)\cong (c,d),$$ 
where $\overline{ab}$ (in this entry) denotes the closed line segment with endpoints^{} $a$ and $b$.
It is obvious that the congruence relation defined on line segments of $A$ is an equivalence relation. Next, one defines a congruence relation on triangles in $A$: $\mathrm{\u25b3}abc\cong \mathrm{\u25b3}pqr$ if their sides are congruent:

1.
$\overline{ab}\cong \overline{pq}$,

2.
$\overline{bc}\cong \overline{qr}$, and

3.
$\overline{ca}\cong \overline{rp}$.
With this definition, Axiom 5 above can be restated as: given a
triangle $\mathrm{\u25b3}abc$, such that $\overline{ab}$ is congruent to
a given line segment $\overline{pq}$. Then there is exactly one
point $r$ on a chosen side of the line $\overleftrightarrow{pq}$ such that
$\mathrm{\u25b3}abc\cong \mathrm{\u25b3}pqr$. Not surprisingly, the congruence
relation on triangles is also an equivalence relation.
The last major congruence relation in an ordered geometry to be
defined is on angles: $\mathrm{\angle}abc$ is congruent to $\mathrm{\angle}pqr$ if there are

1.
a point ${a}_{1}$ on $\overrightarrow{ba}$,

2.
a point ${c}_{1}$ on $\overrightarrow{bc}$,

3.
a point ${p}_{1}$ on $\overrightarrow{qp}$, and

4.
a point ${r}_{1}$ on $\overrightarrow{qr}$
such that $\mathrm{\u25b3}{a}_{1}b{c}_{1}\cong \mathrm{\u25b3}{p}_{1}q{r}_{1}$.
It is customary to also write $\mathrm{\angle}abc\cong \mathrm{\angle}pqr$ to mean
that $\mathrm{\angle}abc$ is congruent to $\mathrm{\angle}pqr$. Clearly for any
points $x\in \overrightarrow{ba}$ and $y\in \overrightarrow{bc}$, we have $\mathrm{\angle}xby\cong \mathrm{\angle}abc$, so that $\cong $ is reflexive. $\cong $ is also
symmetric and transitive (as the properties are inherited from the
congruence relation on triangles). Therefore, the congruence
relation on angles also defines an equivalence relation.
References
 1 D. Hilbert, Foundations of Geometry, Open Court Publishing Co. (1971)
 2 K. Borsuk and W. Szmielew, Foundations of Geometry, NorthHolland Publishing Co. Amsterdam (1960)
 3 M. J. Greenberg, Euclidean and NonEuclidean Geometries, Development and History, W. H. Freeman and Company, San Francisco (1974)
Title  congruence axioms 

Canonical name  CongruenceAxioms 
Date of creation  20130322 15:31:59 
Last modified on  20130322 15:31:59 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  16 
Author  CWoo (3771) 
Entry type  Axiom 
Classification  msc 51F20 
Synonym  axioms of congruence 
Defines  congruence relation 