Proportionality Theorems
Apply the Triangle Proportionality Theorem and its converse: a line parallel to one side of a triangle divides the other two sides proportionally. Use the Side-Splitter theorem and Angle Bisector theorem in problem-solving.
A lesson of Geometry, in Triangle Relationships, about 10 minutes.
By the end of this lesson, you will be able to apply the Triangle Proportionality Theorem and its converse, the Angle Bisector Proportionality Theorem, and proportionality in similar triangles to solve problems.
The Big Idea
When a line parallel to one side of a triangle cuts the other two sides, it doesn't just create parallel segments — it divides the sides proportionally. This is the Triangle Proportionality Theorem, and it's the algebraic version of similarity in action.
Triangle Proportionality Theorem
If a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.
If $\overline{DE} \parallel \overline{BC}$ in $\triangle ABC$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Equivalently: $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ (segment to whole side).
Converse
If a line divides two sides of a triangle proportionally, it is parallel to the third side.
$$\frac{AD}{DB} = \frac{AE}{EC} \implies \overline{DE} \parallel \overline{BC}$$
Example 1 — Finding a Missing Segment
$\overline{DE} \parallel \overline{BC}$, $AD = 6$, $DB = 4$, $AE = 9$. Find $EC$.
$$\frac{6}{4} = \frac{9}{EC} \implies EC = \frac{4 \times 9}{6} = 6$$
Example 2 — Verifying Parallel Lines
$AD = 3$, $DB = 5$, $AE = 6$, $EC = 10$. Is $\overline{DE} \parallel \overline{BC}$?
$$\frac{3}{5} = 0.6 \quad \text{and} \quad \frac{6}{10} = 0.6$$
Equal ratios → yes, $\overline{DE} \parallel \overline{BC}$ by the converse.
Corollary — Three Parallel Lines
If three parallel lines intersect two transversals, they divide the transversals proportionally.
$$\frac{AB}{BC} = \frac{DE}{EF}$$
This is used in problems involving equally spaced parallel lines.
Angle Bisector Theorem (Proportionality Version)
When an angle bisector from vertex $A$ meets side $\overline{BC}$ at point $D$:
$$\frac{BD}{DC} = \frac{AB}{AC}$$
The angle bisector divides the opposite side in the ratio of the adjacent sides.
Example 3 — Angle Bisector Theorem
In $\triangle ABC$, $\overrightarrow{AD}$ bisects $\angle A$. $AB = 8$, $AC = 12$, $BC = 15$. Find $BD$ and $DC$.
$$\frac{BD}{DC} = \frac{8}{12} = \frac{2}{3}$$
$BD + DC = 15$ and $BD = \dfrac{2}{5}(15) = 6$, $DC = 9$
Example 4 — Algebra with the Theorem
$\overline{DE} \parallel \overline{BC}$, $AD = x + 2$, $DB = x - 1$, $AE = x + 3$, $EC = x$. Find $x$.