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.