Side-Angle-Side Similarity Theorem

Consider two triangles $△ABC$ and $△DEF,$ whose two pairs of corresponding sides are proportional and the included angles are congruent.

These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, $△DEF$ can be dilated with the scale factor $k=\frac{AB}{DE}$ about $D,$ forming the new triangle $△D{E}^{\prime }{F}^{\prime }.$

Because dilation is a similarity transformation, it can be concluded that $△D{E}^{\prime }{F}^{\prime }$ and $△DEF$ are similar triangles. Now, it has to be proven that a rigid motion that maps $△D{E}^{\prime }{F}^{\prime }$ onto $△ABC$ exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. In this case, the scale factor $k$ is $\frac{AB}{DE}.$ Since $\overline{AB}$ and $\overline{AC}$ are proportional to $\overline{DE}$ and $\overline{DF}$ respectively, the scale factor can be expressed by any of the following ratios. Applying the Transitive Property of Equality, three equations can be formed and simplified. These relations imply that the two sides of $△D{E}^{\prime }{F}^{\prime }$ are congruent to the corresponding two sides of $△ABC.$ Moreover, the included angles $\angle A$ and $\angle D$ are also congruent. Therefore, by the Side-Angle-Side (SAS) Congruence Theorem, the two triangles are congruent. Since congruent figures can be transformed into each other using rigid motions, and $△ABC$ and $△D{E}^{\prime }{F}^{\prime }$ are congruent triangles, there is a rigid motion placing $△D{E}^{\prime }{F}^{\prime }$ 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.