When a line parallel to one side of a triangle cuts the other two sides, it creates a small triangle similar to the big one, as you saw in Proving AA Similarity and SSS and SAS Similarity. That similarity says something neat about the pieces of the sides.
💡 Why it works
In $\triangle ABC$ with $\overline{DE} \parallel \overline{BC}$, $\triangle ADE \sim \triangle ABC$ by AA, so $\frac{AB}{AD} = \frac{AC}{AE}$. Write $AB = AD + DB$ and $AC = AE + EC$: $\frac{AD + DB}{AD} = \frac{AE + EC}{AE}$, which is $1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}$. Subtract 1 and flip both sides: $\frac{AD}{DB} = \frac{AE}{EC}$. The converse is true too: if the sides are cut in the same ratio, the segment is parallel to the third side.
$\overline{DE} \parallel \overline{BC}$ cuts both sides in the same ratio: $\frac{AD}{DB} = \frac{AE}{EC}$ (red over gold on each side).
🔑 Key idea
Triangle Proportionality Theorem: a line parallel to one side of a triangle divides the other two sides proportionally, $\frac{AD}{DB} = \frac{AE}{EC}$. Conversely, a segment that divides two sides proportionally is parallel to the third side.
$\overline{DE} \parallel \overline{BC}$, $AD = x$, $DB = x + 3$, $AE = 10$ and $EC = 16$. Find $x$.
$\frac{x}{x + 3} = \frac{10}{16}$.
Cross-multiply: $16x = 10x + 30$, so $6x = 30$ and $x = 5$.
Check: $\frac{5}{8} = \frac{10}{16}$.
Answer: $x = 5$
✏️ Worked example: Is it parallel?
$D$ is on $\overline{AB}$ and $E$ is on $\overline{AC}$ with $AD = 5$, $DB = 7$, $AE = 8$ and $EC = 11$. Is $\overline{DE} \parallel \overline{BC}$?
Compare $\frac{5}{7}$ and $\frac{8}{11}$ by cross-multiplying: $5 \cdot 11 = 55$ and $7 \cdot 8 = 56$.
They are not equal, so the sides are not cut proportionally.
Answer: No (though it is close to parallel).
A useful corollary. If three or more parallel lines cut two transversals, they cut them proportionally. Join the transversals with a segment and apply the theorem in each of the two triangles it makes. So if the parallel lines cut one transversal into pieces $4$ and $6$, and the matching first piece on the other transversal is $10$, the second is $\frac{6}{4} \cdot 10 = 15$.
⚠️ Common mistake
Mixing a part on one side with a whole on the other, like $\frac{AD}{AB} = \frac{AE}{EC}$. Use parts with parts ($\frac{AD}{DB} = \frac{AE}{EC}$) or wholes with wholes ($\frac{AD}{AB} = \frac{AE}{AC}$). Check by labeling every segment in your proportion on the picture.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 3 XP.