5.4 Proportional Lengths
In a triangle, if a line is parallel to one side and intersects the other two sides in two points, then the line cuts off a triangle that is similar to the original triangle. There is another theorem that we will now prove that states that if a line is parallel to a side of a triangle it cuts the other two sides in proportional parts.
- Theorem 5.2:
The Side-Splitter Theorem
If a line is parallel to a side of a triangle and intersects the other two sides in two points, then it divides those two sides proportionally.
Given: In $\triangle
ABC$, $\overline{PQ}\parallel \overline{BC}$ Prove: $\dfrac{AP}{PB}=\dfrac{AQ}{QC}$ |
Proof:
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. \overline{PQ}\parallel \overline{BC} & 1. \text{ Given} \\
& 2. \triangle ABC\sim\triangle APQ & 2. \text{ If a line is parallel to a side} \\
&&\text{ of a triangle, it cuts off a triangle}\\
&&\text{ similar to the original triangle}\\
& 3. \frac{AB}{AP}=\frac{AC}{AQ} & 3. \text{ Def. of similar triangles}\\
& 4. \frac{AP+PB}{AP}=\frac{AQ+QC}{AQ} & 4. \text{ Substitution}\\
& 5. \frac{PB}{AP}=\frac{QC}{AQ} & 5. \text{ Properties of proportions & algebra}\\
&6. \frac{AP}{PB}=\frac{AQ}{QC} & 6. \text{ Property of proportion}
\end{array}
This theorem can be very useful in working with any proportions involving the sides of a triangle that have been cut by a line parallel to the third side of the triangle.
Example: In $\triangle ABC$, if $\overline{PQ}\parallel\overline{BC}$, then $\dfrac{a}{b}=\dfrac{c}{d}$.
If you want to use $p$ and $q$ as part of a proportion, then you will need to use Theorem 5.1 which says $\frac{p}{q}=\frac{a}{a+b}$ or $\frac{p}{q}=\frac{c}{c+d}$. It is important to remember that if you want to use an entire side of two triangles, then you need to use Theorem 5.1 (similar triangles), whereas if you are interested in cutting proportional parts of sides of a triangle, then you use Theorem 5.2.
Example: If
$\overline{AB}\parallel\overline{PQ}$, $a=4$, $b=8$, $c=3$, and $p=5$,
then find $d$ and $q$. Solution: By Theorem 5.4, $\frac{4}{8}=\frac{3}{d}$, so $d=6$. However, by Theorem 5.1, $\frac{5}{q}=\frac{4}{12}$, so $q=15$. |
In the drawing below, drag point $Y$ to see the Side Splitter Theorem in action. Now, drag point $X$. What stays the same and what changes?
This leads to the following theorem involving three or more parallel lines.
- Theorem 5.3:
If three parallel lines intersect two transversals, then those
transversals are cut proportionally.
Given:
$\overline{RX}\parallel\overline{SY}\parallel\overline{TZ}$ Prove: $\dfrac{RS}{ST}=\dfrac{XY}{YZ}$ |
Example: If the three lines
$\ell$, $m$, and $n$ are parallel, then find the value of $x$. Solution: We use Theorem 5.3 $$\frac{x}{20}=\frac{24}{16}\qquad\text{ so }\qquad x=30.$$ |
The final theorem of this section deals with the bisector of an angle of a triangle.
- Theorem 5.4:
The Triangle Angle-Bisector Theorem
If a ray bisects an angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the adjacent two sides of the triangle.
Given:
$\overline{AR}$ bisects $\angle BAC$. Prove: $\dfrac{BR}{AB}=\dfrac{RC}{AC}$ |
Proof: Through $B$ draw a line parallel to $\overline{AR}$ and intersecting $\overrightarrow{CA}$ at $K$. Since $\overline{AR}\parallel\overline{BK}$, by Theorem 5.2, we know that $\frac{AC}{AK}=\frac{RC}{RB}$. Using the properties of parallel lines, $\angle 1\cong \angle 3$ and $\angle 2 \cong \angle 4$. Since $\overline{AR}$ bisects $\angle BAC$, $\angle 1\cong \angle 2$, and therefore $\angle 3 \cong \angle 4$. Then by the Converse of the Isosceles Triangle Theorem, $AB=AK$. By substitution into $\frac{AC}{AK}=\frac{RC}{RB}$, we have $\frac{AC}{AB}=\frac{RC}{RB}$. It then follows from the properties of proportions that $\frac{BR}{AB}=\frac{RC}{AC}$.
Example: Given that
$\overline{AR}$ bisects $\angle BAC$, $BR=4$, $RC=8$, and $AB=6$,
determine $AC$. Solution: By Theorem 5.4, we can write the proportion $$\frac{4}{6}=\frac{8}{AC}.$$Therefore $AC=12$. |