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 \begin{align*}XA = 16, XY = 18, XB = 32,\end{align*} find \begin{align*}XZ\end{align*}.
3. If \begin{align*}YZ = 27\end{align*}, find \begin{align*}AB\end{align*}.
4. Find \begin{align*}AY\end{align*} and \begin{align*}BZ\end{align*}.

Know What? To the right is a street map of part of Washington DC. \begin{align*}R\end{align*} Street, \begin{align*}Q\end{align*} Street, and \begin{align*}O\end{align*} Street are parallel and 7\begin{align*}^{th}\end{align*} Street is perpendicular to all three. All the measurements are given on the map. What are \begin{align*}x\end{align*} and \begin{align*}y\end{align*}?

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 \begin{align*}\triangle ABC\end{align*}. Label the vertices.

2. Draw \begin{align*}\overline {XY}\end{align*} so that \begin{align*}X\end{align*} is on \begin{align*}\overline {AB}\end{align*} ̅and \begin{align*}Y\end{align*} is on \begin{align*}\overline {BC}\end{align*}. \begin{align*}X\end{align*} and \begin{align*}Y\end{align*} can be anywhere on these sides.

3. Is \begin{align*}\triangle XBY \sim \triangle ABC\end{align*}? Why or why not? Measure \begin{align*}AX, XB, BY\end{align*}, and \begin{align*}YC\end{align*}. Then set up the ratios \begin{align*}\frac{AX}{XB}\end{align*} and \begin{align*}\frac{YC}{YB}\end{align*}. Are they equal?

4. Draw a second triangle, \begin{align*}\triangle DEF\end{align*}. Label the vertices.

5. Draw \begin{align*}\overline {XY}\end{align*} so that \begin{align*}X\end{align*} is on \begin{align*}\overline {DE}\end{align*} and \begin{align*}Y\end{align*} is on \begin{align*}\overline{EF}\end{align*} AND \begin{align*}\overline {XY} \| \overline {DF}\end{align*}.

6. Is \begin{align*}\triangle XEY \sim \triangle DEF\end{align*}? Why or why not? Measure \begin{align*}DX, XE, EY,\end{align*} and \begin{align*}YF\end{align*}. Then set up the ratios \begin{align*}\frac{DX}{XE}\end{align*} and \begin{align*}\frac{FY}{YE}\end{align*}. Are they equal?

From this investigation, we see that if \begin{align*}\overline {XY} \| \overline{DF}\end{align*}, then \begin{align*}\overline {XY}\end{align*} 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 \begin{align*}\overline{DE} \| \overline{AC}\end{align*}, then \begin{align*}\frac{BD}{DA} = \frac{BE}{EC}\end{align*}. (\begin{align*}\frac{DA}{BD} = \frac{EC}{BE}\end{align*} is also a true proportion.)

For the converse:

If \begin{align*}\frac{BD}{DA} = \frac{BE}{EC}\end{align*}, then \begin{align*}\overline {DE} \| \overline{AC}\end{align*}.

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: \begin{align*}\triangle ABC\end{align*} with \begin{align*}\overline {DE} \| \overline{AC}\end{align*}

Prove: \begin{align*}\frac{AD}{DB} = \frac{CE}{EB}\end{align*}

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

We will not prove the converse; it is basically this proof but in the reverse order.

Example 2: In the diagram below, \begin{align*}\overline {EB} \| \overline {BD}\end{align*}. Find \begin{align*}BC\end{align*}.

Solution: Set up a proportion.

\begin{align*}\frac{10}{15} = \frac{BC}{12} \longrightarrow \ 15(BC) &= 120\\ BC &= 8\end{align*}

Example 3: Is \begin{align*}\overline{DE} \| \overline{CB}\end{align*}?

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

\begin{align*}\frac{6}{18} = \frac{8}{24} = \frac{1}{3}\end{align*}

Because the ratios are equal, \begin{align*}\overline {DE} \| \overline{CB}\end{align*}.

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 \begin{align*}l \parallel m \parallel n\end{align*}, then \begin{align*}\frac{a}{b} = \frac{c}{d}\end{align*} or \begin{align*}\frac{a}{c} = \frac{b}{d}\end{align*}.

Example 4: Find \begin{align*}a\end{align*}.

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

\begin{align*}\frac{a}{20} &= \frac{9}{15}\\ 180 &= 15a\\ a &= 12\end{align*}

Example 5: Find \begin{align*}b\end{align*}.

Solution: Set up a proportion.

\begin{align*}\frac{12}{9.6} &= \frac{b}{24}\\ 288 &= 9.6b\\ b &= 30\end{align*}

Example 6: Algebra Connection Find the value of \begin{align*}x\end{align*} that makes the lines parallel.

Solution: Set up a proportion and solve for \begin{align*}x\end{align*}.

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

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 \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*}.

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

\begin{align*}\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\end{align*}

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 \begin{align*}\triangle BAC \cong \triangle CAD\end{align*}, then \begin{align*}\frac{BC}{CD} = \frac{AB}{AD}\end{align*}.

Example 8: Find \begin{align*}x\end{align*}.

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

\begin{align*}\frac{9}{x} &= \frac{21}{14}\\ 21x &= 126\\ x &= 6\end{align*}

Example 9: Algebra Connection Find the value of \begin{align*}x\end{align*} that would make the proportion true.

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

\begin{align*}\frac{5}{3} &= \frac{4x+1}{15}\\ 75 &= 3(4x+1)\\ 75 &= 12x+3\\ 72 &= 12x\\ 6 &= x\end{align*}

Know What? Revisited To find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, you need to set up a proportion using parallel the parallel lines.

\begin{align*}\frac{2640}{x} = \frac{1320}{2380} = \frac{1980}{y}\end{align*}

From this, \begin{align*}x = 4760 \ ft\end{align*} and \begin{align*}y = 3570 \ ft\end{align*}.

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. \begin{align*}\overline{DB} \| \overline{FE}\end{align*}.

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

Use the diagram to answer questions 6-12. \begin{align*}\overline{AB} \| \overline {DE}\end{align*}.

1. Find \begin{align*}BD\end{align*}.
2. Find \begin{align*}DC\end{align*}.
3. Find \begin{align*}DE\end{align*}.
4. Find \begin{align*}AC\end{align*}.
5. What is \begin{align*}BD:DC\end{align*}?
6. What is \begin{align*}DC:BC\end{align*}?
7. Why \begin{align*}BD:DC \neq DC:BC\end{align*}?

Use the given lengths to determine if \begin{align*}\overline{AB} \| \overline{DE}\end{align*}.

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

Find the value of each variable in the pictures below.

1. \begin{align*}\triangle AXB \sim \triangle YXZ\end{align*} by AA Similarity Postulate
2. \begin{align*}\frac{16}{18} = \frac{32}{XZ}, XZ = 36\end{align*}
3. \begin{align*}\frac{16}{18} = \frac{AB}{27}, AB = 24\end{align*}
4. \begin{align*}AY = 18-16 = 2, BZ = 36-32 = 4\end{align*}

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