# The Proof of the Side Splitter Theorem

Two polygons are similar if all their corresponding angles are congruent, or all their corresponding sides are proportional. For instance, we can say that triangle $ABC$ is similar to triangle $DEF$ if one of the following conditions hold:

Condition 1: $\angle A \cong \angle D$ and $\angle B \cong \angle E$ and $\angle C \cong \angle F$

Condition 2: $\frac{AB}{DE}$ and $\frac{AC}{DF}$ and $\frac{BC}{EF}$.

In this post, we are going to use the concept of similarity to prove the Side Splitter Theorem. The Side Splitter theorem states that if $PQR$ is any triangle, and $\overline{AB}$ is drawn parallel to $\overline{QR}$, then $\frac{PA}{AQ} = \frac{PB}{PR}$. The Side Splitting Theorem

If $PQR$ is any triangle, and $\overline{AB}$ is drawn parallel to $\overline{QR}$, then $\frac{PA}{AQ} = \frac{PB}{PR}$

Proof

Draw $\overline{AC}$ parallel to $\overline{AR}$.  $\angle PAB \cong \angle PQR$ and $PBA \cong \angle ACQ$ by the Parallel Line Postulate. Note that $\overline{AB}$ is parallel to $\overline{QR}$ and $\overline{AC}$ acts a transversal.

Now, by AA Similarity, triangle $PQR$ is similar to triangle $AQC$.

By definition of similarity, $\displaystyle \frac{PA}{AQ} = \frac{PB}{AC}$.

But notice that $ABRQ$ is a parallelogram (why?), so it follows that $\overline{AC} \cong \overline{BR}$.

Therefore, $\frac{PA}{AQ} = \frac{PB}{BR}$ which is what we want to prove.