Theorem 6.9: Two triangles are similar if and only if the ratios the lengths of their corresponding sides are all the same.
Proof: This proof is in two parts. The first part is to show that if the angles are the same size that the ratios of the lengths of the sides are all the same. The second part is to show that if the ratios of the lengths of the sides are all the same then all of the corresponding angles are the same size.
We first show that if the angles are the same size that the ratios of the lengths of the sides are all the same.
In the illustration
Let B" be the point on the ray from A through B which is the same distance from A as B' is from A', and let C" be the point on the ray from A through C which is the same distance from A as C' is from A'.
Triangle A"B"C" is then congruent to triangle A'B'C' by SAS, Theorem 6.8, so since / A"B"C" is congruent to / B', they have the same size by Theorem 6.6. Similarly, / A"C"B" has the same size as / C'.
As a result, m / A"B"C" = m/ B, and m/ A"C"B" = m/ C, so by Theorem 6.4, B"C" is parallel to BC.
But by Theorem 2.9, B"C" || BC if and only if there exists a real number t such that
and
Then by Theorem 2.2,
and
so
Since isometries preserve distances we have that
and
so
and the sides in the triangles are proportional.
To show that
we could simply reach the conclusion by relabeling, but we can derive it directly. Let
and
Then
and
so
But t is a ratio of two distances, so it is positive, and we get
For the converse, suppose that the ratios of the lengths of the sides are all the same. Let that common ratio be t. That is
Then
and
If we let
and
then we know the following things. First,
and
by Theorem 2.2. Moreover B"C" will be parallel to BC, by Theorem 2.9, so
and
by Theorem 6.4, and of course
so they have the same size.
Since triangle A'B'C' will be congruent to triangle A"B"C" by Theorem 6.7, triangle A'B'C' will have the same size angles as triangle ABC.