Let 
be any triangle in the plane. This triangle gives us not
just three segments, but in fact three lines.
Definition: An angle supplementary to an angle of a triangle is called an exterior angle of the triangle. The two angles of the triangle not adjacent to this exterior angle are called the remote interior angles .
Theorem 9.2:[Exterior Angle Theorem] An exterior angle of a triangle is greater than either remote interior angle. See Figure 10.2
Figure 10.2: Exterior angle theorem
Proof: We shall show that 
. In a like manner,
you can show that 
. Then by using the same techniques,
you can prove the same for the other two exterior angles.
By Proposition 8.12 part i) either:
If 
, then by the Alternate Interior
Angle Theorem, 
and 
are non-intersecting. This is
impossible, since they both contain B.
Assume, then, that 
. By the definition of this ordering
on angles, there exists a ray 
between 
and 
so
that
By the Crossbar Theorem,
intersects BC in a point G. Again by the Alternate Interior Angle
Theorem and are non-intersecting. This is a
contradiction.
Thus, 
.
Proposition 9.1:[SAA Congruence]
In triangles and 
given that 
, 
, and
, then 
.
Proof: If 
, we are done by Angle-Side-Angle. Thus, let
us assume that 
. Then, by Proposition 8.4 we must
have that either AB<DE or AB>DE.
If AB<DE, then there is a point 
so that 
. Then by the
SAS Axiom 
. Thus, 
. But 
is exterior to 
, so by the Exterior
Angle Theorem 
. Thus, 
by Proposition 8.12, and we have a contradiction. Therefore, AB is
not less than DE. By a similar argument, we can show that assuming that
AB>DE leads to a similar contradiction.
Thus, our hypothesis that cannot be valid. Thus,
and by ASA.
Proposition 9.2: Two right triangles are congruent if the hypotenuse and a leg of one are congruent respectively to the hypotenuse and a leg of the other.
Proposition 9.3: Every segment has a unique midpoint.
Proof: Let AB be any segment in the plane, and let C be any point
not on . Such a point exists by Axiom I-3. There exists a
unique ray 
on the opposite side of from P such that
, by Axiom C-4. There is a unique point 
so that 
, by Axiom C-1. Q is on the opposite side of
from P by Homework Problem 9, Chapter 3 and Axiom B-4.
Since P and Q are on opposite sides of ,
. Let M denote this point of intersection.
By Axiom B-2, either
We want to show that 
, so let us assume 
.
Since 
, by construction, we have from the
Alternate Interior Angle Theorem that 
and 
are
non-intersecting. If M=A then 
and
which intersects . We can dispose of
the case M=B similarly.
Thus, assume that 
. This will mean that the line 
will intersect side MB of 
at a point between M and B.
Thus, by Pasch's Theorem it must intersect either MQ or BQ. It
cannot intersect side BQ as and are non-intersecting. If intersects MQ then it must contain MQ for
P, Q, and M are collinear. Thus, M=A which we have already shown is
impossible. Thus, we have shown that is not possible.
In the same manner, we can show that 
is impossible. Thus, we
have that . This means that 
since
they are vertical angles. By Angle-Angle-Side we have that 
. Thus, 
and M is the midpoint of AB.
Proposition 9.4:
Proposition 9.5: In a triangle the greater
angle lies opposite the greater side and the greater side lies opposite the
greater angle; i.e., AB>BC if and only if 
.
Proposition 9.6: Given and 
, if 
and 
, then 
if
and only if AC<A'C'.
