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}.

 

side splitter theorem

 

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}.

side splitter theorem proof

\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.

Leave a Reply