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Triangle Proportionality

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What if you were given a triangle with a line segment drawn through it from one side to the other? How could you use information about the triangle's side lengths to determine if that line segment is parallel to the third side? After completing this Concept, you'll be able to answer questions like this one.

Watch This

CK-12 Foundation: Chapter7TriangleProportionalityA

James Sousa: Triangle Proportionality Theorem

James Sousa: Using the Triangle Proportionality Theorem to Solve for Unknown Values

Guidance

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.

Investigation: 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

Example A

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

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.

Example B

In the diagram below, \overline{EB}  \ || \  \overline{CD} . Find BC .

Use the Triangle Proportionality Theorem.

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

Example C

Is \overline{DE}  \ || \  \overline{CB} ?

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} .

Watch this video for help with the Examples above.

CK-12 Foundation: Chapter7TriangleProportionalityB

Vocabulary

A line segment that connects two midpoints of the sides of a triangle is called a midsegment . A midpoint is a point that divides a segment into two equal pieces. Pairs of numbers are proportional if they are in the same ratio.

Guided Practice

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}

Answers:

1. \triangle DBC \sim \triangle FEC

2. DF

3. DC

4. FE

5. DF; DB

Explore More

Use the diagram to answer questions 1-7. \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. 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} .

Find the unknown length.

  1. What is the ratio that the midsegment divides the sides into?

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