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7.5: Proportionality Relationships

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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

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Grades:

8 , 9 , 10

Date Created:

Feb 22, 2012

Last Modified:

Aug 21, 2014
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CK.MAT.ENG.SE.1.Geometry-Basic.7.5

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