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7.5: Proportionality Relationships

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Identify proportional segments when two sides of a triangle are cut by a segment parallel to the third side.
• Extend triangle proportionality to parallel lines.

Review Queue

1. Write a similarity statement for the two triangles in the diagram. Why are they similar?
2. If $XA = 16, XY = 18, XB = 32,$ find $XZ$.
3. If $YZ = 27$, find $AB$.
4. Find $AY$ and $BZ$.
5. Is $\frac{AY}{AX}=\frac{BZ}{BX}$?

Know What? To the right is a street map of part of Washington DC. $R$ Street, $Q$ Street, and $O$ Street are parallel and $7^{th}$ Street is perpendicular to all three. $R$ and $Q$ are one “city block” (usually $\frac{1}{4}$ mile or 1320 feet) apart. The other given measurements are on the map. What are $x$ and $y$?

What is the distance from:

• $R$ and $7^{th}$ to $R$ and Florida?
• $Q$ and $7^{th}$ to $Q$ and Florida?
• $O$ and $7^{th}$ to $O$ and Florida?

Triangle Proportionality

Think about a midsegment of a triangle. A midsegment is parallel to one side of a triangle and divides the other two sides into congruent halves. The midsegment divides those two sides proportionally.

Example 1: A triangle with its midsegment is drawn below. What is the ratio that the midsegment divides the sides into?

Solution: The midsegment’s endpoints are the midpoints of the two sides it connects. The midpoints split the sides evenly. Therefore, the ratio would be $a:a$ or $b:b$. Both of these reduce to 1:1.

The midsegment divides the two sides of the triangle proportionally, but what about other segments?

Investigation 7-4: Triangle Proportionality

Tools Needed: pencil, paper, ruler

1. Draw $\triangle ABC$. Label the vertices.
2. Draw $\overline{XY}$ so that $X$ is on $\overline{AB}$ and $Y$ is on $\overline{BC}$. $X$ and $Y$ can be anywhere on these sides.
3. Is $\triangle XBY \sim \triangle ABC$? Why or why not? Measure $AX, XB, BY,$ and $YC$. Then set up the ratios $\frac{AX}{XB}$ and $\frac{YC}{YB}$. Are they equal?
4. Draw a second triangle, $\triangle DEF$. Label the vertices.
5. Draw $\overline{XY}$ so that $X$ is on $\overline{DE}$ and $Y$ is on $\overline{EF}$ AND $\overline{XY} \ || \ \overline{DF}$.
6. Is $\triangle XEY \sim \triangle DEF$? Why or why not? Measure $DX, XE, EY,$ and $YF$. Then set up the ratios $\frac{DX}{XE}$ and $\frac{FY}{YE}$. Are they equal?

From this investigation, it is clear that if the line segments are parallel, then $\overline{XY}$ divides the sides proportionally.

Triangle Proportionality Theorem: If a line parallel to one side of a triangle intersects the other two sides, then it divides those sides proportionally.

Triangle Proportionality Theorem Converse: If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

Proof of the Triangle Proportionality Theorem

Given: $\triangle ABC$ with $\overline{DE} \ || \ \overline{AC}$

Prove: $\frac{AD}{DB}=\frac{CE}{EB}$

Statement Reason
1. $\overline{DE} \ || \ \overline{AC}$ Given
2. $\angle 1 \cong \angle 2, \angle 3 \cong \angle 4$ Corresponding Angles Postulate
3. $\triangle ABC \sim \triangle DBE$ AA Similarity Postulate
4. $AD + DB = AB\!\\EC + EB = BC$ Segment Addition Postulate
5. $\frac{AB}{BD}=\frac{BC}{BE}$ Corresponding sides in similar triangles are proportional
6. $\frac{AD+DB}{BD}=\frac{EC+EB}{BE}$ Substitution PoE
7. $\frac{AD}{BD}+\frac{DB}{DB}=\frac{EC}{BE}+\frac{BE}{BE}$ Separate the fractions
8. $\frac{AD}{BD}+1=\frac{EC}{BE}+1$ Substitution PoE (something over itself always equals 1)
9. $\frac{AD}{BD}=\frac{EC}{BE}$ Subtraction PoE

We will not prove the converse, it is essentially this proof but in the reverse order. Using the corollaries from earlier in this chapter, $\frac{BD}{DA}=\frac{BE}{EC}$ is also a true proportion.

Example 2: In the diagram below, $\overline{EB} \ || \ \overline{BD}$. Find $BC$.

Solution: Use the Triangle Proportionality Theorem.

$\frac{10}{15} = \frac{BC}{12} \longrightarrow 15(BC) &= 120\\BC &= 8$

Example 3: Is $\overline{DE} \ || \ \overline{CB}$?

Solution: Use the Triangle Proportionality Converse. If the ratios are equal, then the lines are parallel.

$\frac{6}{18}=\frac{1}{3}$ and $\frac{8}{24}=\frac{1}{3}$

Because the ratios are equal, $\overline{DE} \ || \ \overline{CB}$.

Parallel Lines and Transversals

We can extend the Triangle Proportionality Theorem to multiple parallel lines.

Theorem 7-7: If three parallel lines are cut by two transversals, then they divide the transversals proportionally.

Example 4: Find $a$.

Solution: The three lines are marked parallel, so you can set up a proportion.

$\frac{a}{20} &= \frac{9}{15}\\180 &= 15a\\a &= 12$

Theorem 7-7 can be expanded to any number of parallel lines with any number of transversals. When this happens all corresponding segments of the transversals are proportional.

Example 5: Find $a, b,$ and $c$.

Solution: Look at the corresponding segments. Only the segment marked “2” is opposite a number, all the other segments are opposite variables. That means we will be using this ratio, 2:3 in all of our proportions.

$\frac{a}{2} &= \frac{9}{3} && \ \ \frac{2}{4}=\frac{3}{b} && \ \frac{2}{3}=\frac{3}{c}\\3a &= 18 && \ 2b=12 && 2c=9\\a &= 6 && \ \ \ b=6&& \ \ c=4.5$

There are several ratios you can use to solve this example. To solve for $b$, you could have used the proportion $\frac{6}{4}=\frac{9}{b}$, which will still give you the same answer.

Proportions with Angle Bisectors

The last proportional relationship we will explore is how an angle bisector intersects the opposite side of a triangle. By definition, $\overrightarrow{AC}$ divides $\angle BAD$ equally, so $\angle BAC \cong \angle CAD$. The proportional relationship is $\frac{BC}{CD}=\frac{AB}{AD}$. The proof is in the review exercises.

Theorem 7-8: If a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the lengths of the other two sides.

Example 6: Find $x$.

Solution: Because the ray is the angle bisector it splits the opposite side in the same ratio as the sides. So, the proportion is:

$\frac{9}{x} &= \frac{21}{14}\\ 21x &= 126\\x &= 6$

Example 7: Algebra Connection Determine the value of $x$ that would make the proportion true.

Solution: You can set up this proportion just like the previous example.

$\frac{5}{3} &= \frac{4x+1}{15}\\75 &= 3(4x+1)\\75 &= 12x+3\\72 &= 12x\\6 &= x$

Know What? Revisited To find $x$ and $y$, you need to set up a proportion using parallel the parallel lines.

$\frac{2640}{x}=\frac{1320}{2380}=\frac{1980}{y}$

From this, $x = 4760 \ ft$ and $y = 3570 \ ft$.

To find $a, b,$ and $c$, use the Pythagorean Theorem.

$2640^2+a^2 &= 4760^2\\3960^2+b^2 &= 7140^2\\5940^2+c^2 &= 10710^2$

$a = 3960.81, \ b = 5941.21, \ c = 8911.82$

Review Questions

Use the diagram to answers questions 1-5. $\overline{DB} \ || \ \overline{FE}$.

1. Name the similar triangles. Write the similarity statement.
2. $\frac{BE}{EC}=\frac{?}{FC}$
3. $\frac{EC}{CB}=\frac{CF}{?}$
4. $\frac{DB}{?}=\frac{BC}{EC}$
5. $\frac{FC+?}{FC}=\frac{?}{FE}$

Use the diagram to answer questions 6-10. $\overline{AB} \ || \ \overline{DE}$.

1. Find $BD$.
2. Find $DC$.
3. Find $DE$.
4. Find $AC$.
5. We know that $\frac{BD}{DC}=\frac{AE}{EC}$ and $\frac{BA}{DE}=\frac{BC}{DC}$. Why is $\frac{BA}{DE} \neq \frac{BD}{DC}$?

Use the given lengths to determine if $\overline{AB} \ || \ \overline{DE}$.

Algebra Connection Find the value of the missing variable(s).

Find the value of each variable in the pictures below.

Find the unknown lengths.

1. Error Analysis Casey attempts to solve for a in the diagram using the proportion $\frac{5}{a}=\frac{6}{5}$ What did Casey do wrong? Write the correct proportion and solve for $a$.
2. Michael has a triangular shaped garden with sides of length 3, 5 and 6 meters. He wishes to make a path along the perpendicular bisector of the angle between the sides of length 3 m and 5 m. Where will the path intersect the third side?
3. ${\;}$ This is a map of lake front properties. Find $a$ and $b$, the length of the edge of Lot 1 and Lot 2 that is adjacent to the lake.
4. Fill in the blanks of the proof of Theorem 7-8. Given: $\triangle BAD$ with $\overrightarrow{AC}$ is the angle bisector of $\angle BAD$ Auxiliary lines $\overrightarrow{AX}$ and $\overleftrightarrow{XD}$, such that $X, A, B$ are collinear and $\overrightarrow{AC} \ || \ \overleftrightarrow{XD}$. Prove: $\frac{BC}{CD}=\frac{BA}{AD}$
Statement Reason
1. $\overrightarrow{AC}$ is the angle bisector of $\angle BAD$ $X, A, B$ are collinear and $\overrightarrow{AC} \ || \ \overleftrightarrow{XD}$
2. $\angle BAC \cong \angle CAD$
3. Corresponding Angles Postulate
4. $\angle CAD \cong \angle ADX$
5. $\angle X \cong \angle ADX$
6. $\triangle XAD$ is isosceles
7. Definition of an Isosceles Triangle
8. Congruent segments are also equal
9. Theorem 7-7
10.

1. $\triangle AXB \sim \triangle YXZ$ by AA Similarity Postulate
2. $\frac{16}{18} = \frac{32}{XZ}, XZ = 36$
3. $\frac{16}{18} = \frac{AB}{27}, AB = 24$
4. $AY = 18-16 = 2, BZ = 36-32 = 4$
5. $\frac{2}{16} = \frac{4}{32}$. Yes, this is a true proportion.

Feb 22, 2012

Aug 22, 2014