PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: Very high Entry average rating: No information on entry rating
congruence axioms (Axiom)

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 $ \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 $ \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
$\displaystyle \overline{ab}\cong\overline{cd}$    iff $\displaystyle \qquad (a,b)\cong (c,d),$
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$: $ \triangle abc\cong \triangle 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 $ \triangle 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 $ \triangle abc\cong\triangle 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: $ \angle abc$ is congruent to $ \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 $ \triangle a_1bc_1\cong \triangle p_1qr_1$.

It is customary to also write $ \angle abc\cong \angle pqr$ to mean that $ \angle abc$ is congruent to $ \angle pqr$. Clearly for any points $ x\in\overrightarrow{ba}$ and $ y\in\overrightarrow{bc}$, we have $ \angle xby\cong \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.

Bibliography

1
D. Hilbert, Foundations of Geometry, Open Court Publishing Co. (1971)
2
K. Borsuk and W. Szmielew, Foundations of Geometry, North-Holland Publishing Co. Amsterdam (1960)
3
M. J. Greenberg, Euclidean and Non-Euclidean Geometries, Development and History, W. H. Freeman and Company, San Francisco (1974)



"congruence axioms" is owned by CWoo. [ full author list (2) ]
(view preamble)

View style:

Other names:  axioms of congruence
Also defines:  congruence relation
Log in to rate this entry.
(view current ratings)

Cross-references: properties, mean, axiom, congruent, sides, obvious, endpoints, closed line segments, angles, triangles, line segments, boundary, closed half plane, line, intersect, lying on, point, ray, strict betweenness relation, ordered geometry, equivalence relation, transitive, imply, symmetric, Reflexive, relations
There are 10 references to this entry.

This is version 13 of congruence axioms, born on 2005-10-08, modified 2007-06-26.
Object id is 7426, canonical name is CongruenceAxioms.
Accessed 3843 times total.

Classification:
AMS MSC: 51F20 (Geometry :: Metric geometry :: Congruence and orthogonality)
Pending Errata and Addenda
None.
[ View all 2 ]
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)