68 The parallel axiom; Euclidean geometry Ch. 5
A ′
B′
C′
A
B
C
B′′
C′′
Figure 5.7. Similar triangles.textProof. ChooseB′′∈[A,B andC′′∈[A,Cso that|A,B′′|=|A′,B′|,
|A,C′′|=|A′,C′|.Thenas|∠B′′AC′′|◦=|∠BAC|◦=|∠B′A′C′|◦, by the SAS principle
we see that the triangles[A,B′′,C′′],[A′,B′,C′]are congruent in the correspondence
(A,B′′,C′′)→(A′,B′,C′). In particular|∠AB′′C′′|◦=|∠A′B′C′|◦and so|∠AB′′C′′|◦=
|∠ABC|◦. These are corresponding angles in the sense of 5.1.1, soB′′C′′‖BCand then
by 5.3.1
|A,B′′|
|A,B|
=
|A,C′′|
|A,C|
,
so
|A′,B′|
|A,B|
=
|A′,C′|
|A,C|
.
By a similar argument on taking a triangle[B,E,F]which is congruent to[B′,C′,A′],
we have
|B′,C′|
|B,C|
=
|B′,A′|
|B,A|
.
COMMENT. Triangles like these, which have the degree-measures of correspond-
ing angles equal and so the ratios of the lengths of corresponding sides are equal, are
saidtobesimilarin the correspondence(A,B,C)→(A′,B′,C′).
Let A,B,C and A′,B′,C′be two sets of non-collinear points such that|A′,B′|
|A,B|=
|A′,C′|
|A,C|
,|∠B′A′C′|◦=|∠BAC|◦.
Then the triangles are similar.
Proof. ChooseB′′∈[A,B,C′′∈[A,Cso that|A,B′′|=|A′,B′|,
|A,C′′|=|A′,C′|.Then as|∠B′A′C′|◦=|∠BAC|◦=|∠B′′AC′′|◦, by the SAS principle
we see that the triangles[A′,B′,C′],[A,B′′,C′′]are congruent. We note that
|A,B′′|
|A,B|