According to our development, congruence has been a relationship between segments or a relationship between angles. In Geometry we are accustomed to seeing congruence as a relationship between triangles. We can make this so by a definition.
Definition: Two triangles are congruent if there exists a one-to-one correspondence between their vertices sot that the corresponding sides and corresponding angles are congruent.
Proposition 7.1: If in 
we have 
, then 
.
Figure 8.1: Isosceles triangles
Proof: This simple proof is due to Pappus. Consider the correspondence
of vertices 
, 
, and 
. Under this correspondence, two sides and the included angle of
are congruent respectively to the corresponding sides and
included angle of 
. Hence, by SAS the triangles are congruent.
Therefore, the corresponding angles are congruent and .
Proposition 7.2:[Segment Subtraction]
If 
, 
, 
, and 
, then
.
Figure 8.2: Segment subtraction
Proof:
From what we are given, let us assume that 
. By Axiom
C-1, there exists a unique point G on ray 
so that 
.
By our hypothesis, we have that 
. Since and ,
by Axiom C-3 
. But then by Axiom C-2 we have that
. It follows from Axiom C-1 that F=G. We have reached a
contradiction. Thus, it must be that , and we are done.
Proposition 7.3: Given , then for any point B
between A and C, there exists a unique point E, ,
such that .
Proof: The proof of this proposition is left to the reader.
We can use this result to help us establish a partial ordering on the line segments in the plane.
Definition: AB < CD (or CD > AB) means that there exists a point E
between C and D such that 
.
Proposition 7.4:[Segment Ordering]
, then AB < EF.
, then AB > EF.
Proof: We will prove the first item only. The proofs of the remaining three items will be left to you as homework.
We are given segments AB and CD. If 
, then we are done. So,
let us assume that 
. By Axiom C-1 there is a unique
point 
so that . 
, else 
which is impossible under our assumption. Thus, we must have that
or 
from Axiom B-2. If
, then by definition AB < CD, and we are done.
Suppose then that . By Proposition 8.3 there
exists a unique point 
so that 
. In this case by the
definition AB > CD, and we are done.
Proposition 7.5: Supplements of congruent angles are congruent.
Proposition 7.6:
Proposition 7.7: For every line 
and every point P
there exists a line through P perpendicular to .
Figure: Theorem 8.7
Proof:
Either 
or 
. First, let us assume that .
Let 
. Such points exist by Axiom I-2. The ray 
lies on one side of . By Axiom C-4 there exists a ray 
on the opposite side of from P so that 
By Axiom C-1 there is a point 
so that 
.
Since P and P' are on opposite sides of , the segment PP'
intersects the line . Let 
.
If Q=A, then 
. Thus,
.
If 
, then 
by SAS. Thus, from
the definition of congruent triangles, 
, and
.
Now, if , there is an 
. From this point X apply the
previous technique to construct a perpendicular line to through X. By
Axiom C-4 we can copy this angle on one side of at P. From the
second part of Proposition 8.6 the other side of this angle is part
of a line through P perpendicular to .
Proposition 7.8:[ASA] Given and 
with 
, 
, and .
Then 
.
Proposition 7.9: If in we have that , then and is isosceles.
Proposition 7.10:[Angle Addition]
Given 
between
and 
, 
between 
and 
,
and 
. Then 
.
Proposition 7.11:[Angle Subtraction]
Given between
and , between and ,
, and , then .
Proof: Since lies between and , we may
apply the Crossbar Theorem to find that intersects AC.
Without loss of generality, we may assume that this point of intersection is,
in fact, the point G. Then, we have that 
. Assume that
the points D, F, and H are chosen so that 
, 
,
and . This is nothing but a relabeling of the points.
Let us assume then that 
. Then, there exists a
unique ray, 
, on the same side of 
as so
that
By our assumption, 
, so
by Proposition 8.10, 
. Since , the uniqueness of implies that
, a contradiction. Thus, 
.
As with line segments, there is a natural method for defining an ordering on angles.
Definition: 
means that there exists a ray
between and so that 
.
This gives us the following results, which are completely analogous to those for segments.
Proposition 7.12:[Ordering of Angles]
and 
, then 
.
and , then 
.
and 
, then .
Proposition 7.13:[SSS]
Given triangles and
. If , , and , then
.
Proposition 7.14:[Euclid's Fourth Postulate] All right angles are congruent to each other.
