8.5 Triangle Proportionality
Notes
A line parallel to one side of a triangle divides the other two sides proportionally.
- Triangle Proportionality Theorem: if $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$, then $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$. The converse is true too.
- Three or more parallel lines cut two transversals proportionally.
- Triangle Angle Bisector Theorem: an angle bisector splits the opposite side into pieces proportional to the other two sides: $\displaystyle \frac{BD}{DC} = \frac{AB}{AC}$.
- Example: $AD = 4$, $DB = 6$, $AE = 5$. Then $\displaystyle \frac{4}{6} = \frac{5}{EC}$, so $EC = 7.5$.
Practice
In $\triangle ABC$, $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$. $AD = 6$, $DB = 9$ and $AE = 8$. Find $EC$.
$\displaystyle \frac{6}{9} = \frac{8}{EC}$, so $6 \cdot EC = 72$.
In $\triangle ABC$, $\overline{AD}$ bisects $\angle A$ with $D$ on $\overline{BC}$. $AB = 10$, $AC = 15$ and $BC = 20$. Find $BD$.
$\displaystyle \frac{BD}{DC} = \frac{10}{15}$. Let $BD = 2x$, $DC = 3x$: $5x = 20$, so $BD = 8$.
In $\triangle PQR$, $S$ is on $\overline{PQ}$ and $T$ is on $\overline{PR}$. $PS = 3$, $SQ = 6$, $PT = 4$ and $TR = 8$. Is $\overline{ST} \parallel \overline{QR}$?
The sides are cut proportionally, so by the converse the segment is parallel.
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