Consider two triangles △ABC and △DEF, whose corresponding sides are .
These triangles can be proven to be similar by identifying a that maps one triangle onto the other. First, △DEF can be with the k=DEAB about D, forming the new triangle △DE′F′.
Because dilation is a similarity transformation, it can be concluded that
△DE′F′ and
△DEF are . Now, it has to be proven that a that maps
△DE′F′ onto
△ABC exists. The ratios of the corresponding side lengths of are the same and equal to the scale factor.
DEDE′=DFDF′=EFE′F′=k
In this case, the scale factor
k is
DEAB. Since all of the sides of
△ABC and
△DEF are proportional, the scale factor can be expressed by any of the following ratios.
k=DEAB=EFBC=DFCA
Applying the , three equations can be formed and simplified.
DEDE′DFDF′EFE′F′=DEAB=DFCA=EFBC⇒DE′DF′E′F′=AB=CA=BC
These relations imply that the three sides of
△DE′F′ are to the three sides of
△ABC. Therefore, by the , the two triangles are congruent.
△ABC≅△DE′F′
Since congruent figures can be transformed into each other using rigid motions, and
△ABC and
△DE′F′ are congruent triangles, there is a rigid motion placing
△DE′F′ onto
△ABC.
The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps △DEF onto △ABC.
Therefore, it can be concluded that △ABC and △DEF are similar triangles.
The proof is now complete.