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# 7.5: Proportionality Relationships

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Identify proportional segments within triangles.
• 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 $BZ$.

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. All the measurements are given on the map. What are $x$ and $y$?

## 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 splits the sides evenly. The ratio would be 8:8 or 10:10, which both 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, we see that if $\overline {XY} \| \overline{DF}$, 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.

If $\overline{DE} \| \overline{AC}$, then $\frac{BD}{DA} = \frac{BE}{EC}$. ($\frac{DA}{BD} = \frac{EC}{BE}$ is also a true proportion.)

For the converse:

If $\frac{BD}{DA} = \frac{BE}{EC}$, then $\overline {DE} \| \overline{AC}$.

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 basically this proof but in the reverse order.

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

Solution: Set up a proportion.

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

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

Solution: If the ratios are equal, then the lines are parallel.

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

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.

If $l \parallel m \parallel n$, then $\frac{a}{b} = \frac{c}{d}$ or $\frac{a}{c} = \frac{b}{d}$.

Example 4: Find $a$.

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

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

Example 5: Find $b$.

Solution: Set up a proportion.

$\frac{12}{9.6} &= \frac{b}{24}\\288 &= 9.6b\\b &= 30$

Example 6: Algebra Connection Find the value of $x$ that makes the lines parallel.

Solution: Set up a proportion and solve for $x$.

$\frac{5}{8} = \frac{3.75}{2x-4} \longrightarrow \ 5(2x-4) &= 8(3.75)\\10x-20 &= 30\\10x &= 50\\x &= 5$

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 7: Find $a, b,$ and $c$.

Solution: Line up the segments that are opposite each other.

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

## Proportions with Angle Bisectors

The last proportional relationship we will explore is how an angle bisector intersects the opposite side of a triangle.

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.

If $\triangle BAC \cong \triangle CAD$, then $\frac{BC}{CD} = \frac{AB}{AD}$.

Example 8: Find $x$.

Solution: The ray is the angle bisector and it splits the opposite side in the same ratio as the sides. The proportion is:

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

Example 9: Algebra Connection Find the value of $x$ that would make the proportion true.

Solution: You can set up this proportion 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 $y = 3570 \ ft$.

## Review Questions

• Questions 1-12 are similar to Examples 1 and 2 and review.
• Questions 13-18 are similar to Example 3.
• Questions 19-24 are similar to Examples 8 and 9.
• Questions 25-30 are similar to Examples 4-7.

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 6-12. $\overline{AB} \| \overline {DE}$.

1. Find $BD$.
2. Find $DC$.
3. Find $DE$.
4. Find $AC$.
5. What is $BD:DC$?
6. What is $DC:BC$?
7. Why $BD:DC \neq DC:BC$?

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.

## Review Queue Answers

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. $AY = 18-16 = 2, BZ = 36-32 = 4$

8 , 9 , 10

## Date Created:

Feb 22, 2012

Dec 11, 2014
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