Theorem 6.9: If triangles are similar then the ratios the lengths of their corresponding sides are all the same.
Proof: We show that if the angles are congruent that the ratios of the lengths of the corresponding sides are all the same.
In the illustration
Since D ≅ A, we can move D to be on top of A, by the definition of congruent. If E is not on AB, (or it's extension) then rotate ΔDEF about the angle bisector of / D, and we will be able to assume that it is. Then rotate the figure about A until BC is vertical.
Since AB and AC intersect BC , they are not vertical, and will, thus, have finite real slopes with which we can do arithmetic.
Since / AEF ≅ / B, EF || BC, by Theorem 6.2, so |EF| is also vertical.
Let us define coordinates for these points. Let
A = D = (x_{o}, y_{o})
B = (x_{1}, y_{1})
C = (x_{2}, y_{2})
E = (x_{3}, y_{3})
F = (x_{4}, y_{4})
Since BC is vertical, x_{1} = x_{2}, and, since EF is vertical, x_{3} = x_{4}. Let
m_{1} be the slope of AB
and
m_{2} be the slope of AC
Then by Theorem 1.5,
so
and
But, since x_{1} = x_{2}, and x_{3} = x_{4} This gives us
As required. Since A and D were an arbitrary pair of congruent angles, the same proof could be repeated for the other pairs of corresponding sides.
For the converse, assume that the corresponding sides in two triangles are proportional. In our example,
we are now assuming that
We can also assume, after, perhaps, relabeling, that
α < 1.
(If α = 1, then the triangles are congruent by SSS, and that angles are the same size because they would then be corresponding parts of congruent triangles.)
Let B′ be the point on AB such that
|AB′| = α|AB
Draw the line through B′ parallel to AB.
Let C′ be where this line intersects BC.
Since B′C′ || BC ΔABC ∼ ΔAB′C′ so by the proof of the first part of the theorem, the sides in ΔAB′C′ are proportional to the sides in ΔABC. As a result,
|AB′|= α|AB| = |DE|,
|AC′| = α|AC| = |DF|,
|B′C′| = α|BC| = |EF|.
As a result, ΔAB′C′ ≅ ΔDEF, by SSS, and so they have the same angles. Conclude that ΔDEF ∼ ΔABC.