The Triangle Proportionality (Side-Splitter) Theorem

What the theorem says

Suppose $\overline{DE}$ is parallel to side $\overline{BC}$ in $\triangle ABC$, with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$. The parallel segment splits both sides in the same ratio: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

The parallel mark is the whole reason this works. Because $\overline{DE} \parallel \overline{BC}$, the small triangle $ADE$ is similar to the big triangle $ABC$ (they share $\angle A$, and the parallel lines create congruent corresponding angles). Proportional pieces fall straight out of that similarity.

Setting up the proportion correctly

Match part to part. Top piece over bottom piece on the left side equals top piece over bottom piece on the right side: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

The classic mistake is mixing a part with a whole — writing $\dfrac{AD}{DB} = \dfrac{AE}{AC}$, where $AC$ is the entire side. Parts compare to parts. If a problem gives you a whole side, subtract to find the missing piece first, or use the part-to-whole version on both sides: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$. Just never mix the two forms in one proportion.

The converse works too

The theorem runs in both directions. If a segment divides two sides of a triangle proportionally, then that segment is parallel to the third side. So when a problem asks "is $\overline{DE}$ parallel to $\overline{BC}$?", compute both ratios — equal ratios mean parallel, unequal ratios mean not parallel.

Worked examples

Try one yourself

Common questions