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7. Similarity
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Geometry

7.04 Applications of similarity

Lesson
Practice

Proportionality theorems

Exploration

Consider a triangle ABC, where \overline{DE} is drawn parallel to one of the sides, in this case \overline{AC}. Drag the points of the triangle and move the parallel side within the triangle.

Loading interactive...
  1. What do you notice about \angle CAB and \angle EDB? Why do you think this is true?
  2. What do you notice about the relationships between segments \overline{BD},\,\overline{DA},\,\overline{BE} and \overline{EC}?
  3. Make a conjecture about the relationships between these line segments and formulate a plan for proving them.
Side-splitter theorem

If a line intersects two sides of a triangle and is parallel to the third side of the triangle, then it divides those two sides proportionally

A triangle. Triangle A B C with point D on A B, and point E on A C. A segment parallel to the B C is drawn from D to E.
Converse of the side-splitter theorem

If a line divides two sides of a triangle proportionally, then the line is parallel to the third side of the triangle

Triangle A B C with point D on A B, and point E on A C. A segment parallel to the B C is drawn from D to E.
Triangle midsegment theorem

The midsegment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this midsegment is half the length of the third side

Triangle A B C with midpoint D on A B, and midpoint E on A C. A segment parallel to the B C is drawn from D to E. B C has a length of x, and D E has a length of x divided by 2.

Examples

Example 1

Formulate a proof for the side-splitter theorem.

Worked Solution
Create a strategy

First, we draw a figure with the characteristics from the theorem, and state what we know and what we're trying to prove. We know \overline{DE} \parallel \overline{BC}. We need to prove the two sides are divided proportionally, or \dfrac{BD}{AD} = \dfrac{CE}{AE}.

Triangle A B C with point D on A B and point E on A C. Segment D E is drawn.

We know similar triangles have congruent angles and proportional side lengths, so the triangle similarity theorems (AA, SAS, SSS) may be helpful. We also know some things about congruent angles when parallel lines are cut by transversals.

Extend \overline{DE} to be a line with auxilary points: a point F such that D lies between F and E, and a point G such that E lies between D and G. \overline{AB} and \overline{AC} are transversals of parallel lines \overline{DE} and \overline{BC}.

Triangle A B C with point D on A B and point E on A C. A point F is outside the triangle and on the left of D. A point E is outside the triangle and on the right of E. A line F G is drawn passing through points D and E. Angles B D F, E D A, and C B D are congruent. Angles C E G, B C E, and D E A are congruent.
Apply the idea
To prove: \dfrac{BD}{AD}=\dfrac{CE}{AE}
StatementsReasons
1.\overline{DE} \parallel \overline{BC}Given
2.\begin{aligned} m \angle{BDF}= m \angle{CBD} \\ m \angle{CEG} = m \angle{BCE} \end{aligned}Alternate interior angles are congruent
3.\begin{aligned} m \angle{BDF}=m \angle{EDA} \\ m \angle{CEG} = m \angle{DEA} \end{aligned}Vertical angles are congruent
4.\begin{aligned} m \angle BDF =m \angle CBD = m \angle EDA \\ m \angle CEG = m \angle BCE = m \angle DEA \end{aligned}Transitive property of equality
5.\triangle{ADE} \sim \triangle{ABC}AA similarity theorem
6.\dfrac{AB}{AD}=\dfrac{AC}{AE}Definition of similar triangles
7.\begin{aligned} AB=AD+BD \\ AC=AE+CE \end{aligned}Definition of collinear line segments
8.\dfrac{AD+BD}{AD}=\dfrac{AE+CE}{AE}Substitution property of equality
9.\dfrac{AD}{AD}+\dfrac{BD}{AD}=\dfrac{AE}{AE} + \dfrac{CE}{AE}Common denominators
10.1 + \dfrac{BD}{AD}= 1 + \dfrac{CE}{AE}Definition of one
11.\dfrac{BD}{AD}= \dfrac{CE}{AE}Subtraction property of equality

Example 2

Use theorems to solve the problems that follow.

a

Determine whether \overline{KM} \parallel \overline{JN}. Justify your answer.

Triangle J L N with point K on J L, and point M on L N. A segment is drawn from K to M. Segment J K has a length of 1, K L has a length of 1.6, L M has a length of 8, and M N has a length of 5.
Worked Solution
Create a strategy

We can use the converse of the side-splitter theorem to determine whether \overline{KM} \parallel \overline{JN}.If \dfrac{LK}{KJ}=\dfrac{LM}{MN}, then the sides \overline{KM} and \overline{JN} are parallel.

Apply the idea

We have that \dfrac{LK}{KJ}=\dfrac{1.6}{1}=1.6 and \dfrac{LM}{MN}=\dfrac{8}{5}=1.6.

Therefore, \overline{KM} \parallel \overline{JN} based on the converse of the side-splitter theorem.

b

The Eastern Garbage Patch in the Pacific Ocean is a collection of marine debris that is difficult to measure directly. Using the diagram, how can we indirectly determine the length of the Eastern Garbage Patch?

The image shows an Eastern Garbage Patch. Triangle A E D with point B on A D, and point C on A E. A segment parallel to the E D is drawn from C to B.
Worked Solution
Create a strategy

We can use the side-splitter theorem to indirectly measure the length of the Eastern Garbage Patch.

Apply the idea

Using the side-splitter theorem, we see that \dfrac{AB}{BD}=\dfrac{AC}{CE}. We can calculate the distances of BD, AC, and CE and use those to find the length of the Eastern Garbage Patch.

c

Xiker is training for a triathalon. He wants to use a lake nearby to train for the swimming portion. To determine how far the length of the lake is, he paces out a triangle, counting his paces, as shown in the diagram below:

A figure of a lake. A triangle with its midsegment of length x is imposed over the figure. The midsegment represents the length of the lake. The midsegment divides the left leg of the triangle into two segments both of length 40, the right leg into two segment both of length 75. The base of the triangle is of length 125.

If Xiker's strides are 2.75 \text{ feet}, determine the distance he must swim across the lake for his training.

Worked Solution
Create a strategy

Use the triangle midsegment theorem to determine the length of the lake in paces, then convert the paces to feet.

Apply the idea

The triangle midsegment theorem states that the line connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle which is the midsegment, and the length of this midsegment is half the length of the third side.

The third side of the triangle is 125 paces, so the length of the lake is 62.5 paces. If each of Xiker's paces is 2.75 feet, then the length of the lake is approximately 172 feet.

Example 3

Consider the diagram shown where \overline{DE} \parallel \overline{BC}:

Triangle A B C with point D on A B and point E on A C. Segment D E is drawn and is parallel to B C. B D has a length of x plus 3, A D has a length of 5, A E has a length of 4, and C E has a length of 2 x.

Find x. What else can we say about \overline{DE}?

Worked Solution
Create a strategy

The side-splitter theorem states that if a line intersects two sides of a triangle and is parallel to the third side of the triangle, then it divides those two sides proportionally. Solve a proportion with the side lengths.

Apply the idea

We have:

\displaystyle \dfrac{BD}{AD}\displaystyle =\displaystyle \dfrac{CE}{AE}Since \overline{DE} \parallel \overline{BC}, the sides are split proportionally (side-splitter theorem)
\displaystyle \dfrac{x+3}{5}\displaystyle =\displaystyle \dfrac{2x}{4}Substitute the lengths of each segment
\displaystyle \dfrac{4x+12}{5}\displaystyle =\displaystyle 2xMultiply both sides by 4
\displaystyle 4x+12\displaystyle =\displaystyle 10xMultiply both sides by 5
\displaystyle 12\displaystyle =\displaystyle 6xSubtract 4x from both sides
\displaystyle 2\displaystyle =\displaystyle xDivide both sides by 6

By substituting 2 for x in the triangle, we know that BD=5 and CE=4. This means that \overline{DE} is also the midsegment of the large triangle.

Reflect and check

We couldn't assume that \overline{DE} was the midsegment with the information given initially and assume that we could solve the equations BD=AD \to x+3=5 and CE=AE \to 2x=4.

Idea summary

We can use these theorems to help us solve problems with triangles:

  • Side-splitter theorem: If a line intersects two sides of a triangle and is parallel to the third side of the triangle, then it divides those two sides proportionally
  • Triangle midsegment theorem: The midsegment connecting the midpoints of two sides of a triangle is parallel to the third side of the triangle, and the length of this midsegment is half the length of the third side

Right triangle similarity

Exploration

These four triangles are similar:

Right angled triangle with 45 degree between the sides height and hypotenuse.
Triangle A
Right angled triangle with lengths of 5 for base and height.
Triangle B
Right angled triangle with length of 5 for height and square root of 50 for hypotenuse.
Triangle C
Right angled triangle with 45 degree between the sides base and hypotenuse.
Triangle D
  1. How could we show that Triangle A and Triangle B are similar?
  2. How could we show that Triangle A and Triangle D are similar?
  3. How could we show that Triangle B and Triangle C are similar?
  4. If we know two triangles are right triangles, what additional information do we need to prove they are similar?

When we are dealing with right triangles, we only need to know that the hypotenuse and leg are proportional to determine similarity. This is because the Side-Side-Side similarity theorem is indirectly satisfied by the Pythagorean theorem.

Consider a right triangle ABC with leg a and hypotenuse c, and a second right triangle XYZ with a proportional side ka and proportional hypotenuse kc. The scale factor between the known sides is k.

\displaystyle b\displaystyle =\displaystyle \sqrt{c^2-a^2}Length of unknown leg in \triangle ABC
\displaystyle x\displaystyle =\displaystyle \sqrt{\left(kc^2\right)-\left(ka\right)^2}Length of unknown leg in \triangle XYZ
\displaystyle =\displaystyle \sqrt{k^2\left(c^2-a^2\right)}Factor out k^2
\displaystyle =\displaystyle k\sqrt{c^2-a^2}Product of radicals property

This shows the corresponding unknown sides are also proportional with a scale factor of k.

Hypotenuse-Leg similarity (HL \sim) theorem

If the ratio of the hypotenuse and leg of one right triangle is equal to the ratio of the hypotenuse and leg of another right triangle, then the two triangles are similar

Two right triangles. The hypotenuse and horizontal sides of the two triangles are congruent.

An altitude of a triangle is a perpendicular line segment drawn from one vertex to the opposite side of the triangle.

Right triangle similarity theorem

The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and to each other.

A right triangle with the altitude to the hypotenuse drawn, forming two similar triangles.

The altitude in a right triangle creates three similar right triangles:

The image shows triangles. At the top, there is a triangle labeled ABC with points A, B, and C. Point C is at the top vertex, and a dashed line extends downward to point D on segment AB. At the bottom, there are two separate triangles. The first triangle is labeled ACD, with C at the top and D at the right vertex on the base. The second triangle, is labeled CBD, with C at the top and D at the left vertex on the base. Ask your teacher for more information.

Examples

Example 4

Use theorems to solve the problems that follow.

a

A flagpole that stands 4.9 meters high casts a shadow of 4.5 meters. At the same time, the shadow of a nearby building falls at the same point S. The shadow cast by the building measures 13.5 meters. Find h, the height of the building.

A 4.9 meter flagpole casting a shadow of length 4.5 meters. The tip of the flagpole's shadow on the ground is labeled by point S. Next to the flagpole is a building. The building casts a shadow of length 13.5 meters. The tip of building's shadow at also at S. A diagonal segment is drawn from the top of the building to S.
Worked Solution
Create a strategy

The building and flagpole are both perpendicular to the ground, creating a set of congruent angles. Both triangles also share \angle S. So the AA similarity theorem would be used to show the two triangles are similar. Equate the ratios of the corresponding sides of the similar triangles.

Apply the idea

The corresponding sides are h and 4.9, and 13.5 and 4.5.

\displaystyle \dfrac{h}{4.9}\displaystyle =\displaystyle \dfrac{13.5}{4.5}Corresponding side lengths are proportional in similar triangles
\displaystyle h\displaystyle =\displaystyle 14.7Multiply both sides by 4.9

The height of the building is 14.7meters.

Reflect and check

We can check the similarity ratio by confirming that the triangles are proportional.

\dfrac{h}{4.9}=\dfrac{13.5}{4.5} \to \dfrac{14.7}{4.9}=\dfrac{13.5}{4.5}

Since the corresponding side lengths of the triangles are proportional, we can confirm that the height of the building is 14.7 meters.

b

Determine whether \triangle{ABC} \sim \triangle{ADE}. Justify your reasoning.

Triangle A B C with right angle B C A. A point D is on B A and point E on C A. Segment D E is drawn and is perpendicular to C A. D E has a length of 14 feet, B C has a length of 28 feet, B D has a length of 19 feet, and D A has a length of 19 feet.
Worked Solution
Create a strategy

The hypotenuse-leg similarity theorem states that if the ratio of the hypotenuse and leg of one right triangle is proportional to the ratio of another hypotenuse and leg of another triangle, the triangles are similar.

Apply the idea

If the ratio of the hypotenuse and leg of \triangle ABC is proportional to the ratio of the hypotenuse and leg of \triangle ADE, then we can state that the triangles are similar.\dfrac{AB}{BC}=\dfrac{AD}{DE} \to \dfrac{38}{28}=\dfrac{19}{14} \to \dfrac{19}{14}=\dfrac{19}{14}

Since the ratios of the hypotenuse and leg of each triangle are proportional, the triangles are similar by the HL similarity theorem.

Example 5

Solve for the length of \overline{CD}.

Right triangle A B C with right angle C and a point D on A B. A segment is drawn from C to D perpendicular to A B. Segment A D has a length of 8, and D B has a length of 4.
Worked Solution
Create a strategy

We know the triangles are similar by the right triangle similarity theorem. We can separate the similar triangles to help us visualize which ones to use to create the similarity ratios.

A larger triangle labeled with vertices A, B, and C, where C is at the peak. There is a perpendicular line from vertex C to the base AB, meeting at point D. The base AB is divided into segments, with AD labeled as 8 and DB labeled as 4. Ask your teacher for more information.

  • \triangle ABC \sim \triangle CBD

  • \triangle ABC \sim \triangle ACD

  • \triangle ACD \sim \triangle CBD

Notice that \triangle ABC does not have \overline{CD} has a side, so we want to use triangles ACD and CBD to set up the ratios.

Apply the idea

Since \triangle ACD \sim \triangle CBD, we know that \overline{AD} and \overline{CD} are corresponding sides and \overline{CD} and \overline{BD} are corresponding sides, so their proportions will be equal.

\displaystyle \frac{AD}{CD}\displaystyle =\displaystyle \frac{CD}{BD}Ratios of corresponding sides are equal
\displaystyle \frac{8}{CD}\displaystyle =\displaystyle \frac{CD}{4}Substitute AD=8 and BD=4
\displaystyle \left(CD\right)^2\displaystyle =\displaystyle 32Cross multiply
\displaystyle CD\displaystyle =\displaystyle \sqrt{32}Square root both sides
\displaystyle CD\displaystyle =\displaystyle 4\sqrt{2}Simplify the radical
Reflect and check

Recall there are two methods we can use to simplify radicals.

Prime factorization method:

\displaystyle \sqrt{32}\displaystyle =\displaystyle \sqrt{2\cdot2\cdot2\cdot2\cdot2}Find prime factorization of radicand
\displaystyle =\displaystyle \sqrt{2^2\cdot2^2\cdot2}Group factors according to the index
\displaystyle =\displaystyle \sqrt{2^2}\cdot\sqrt{2^2}\cdot\sqrt{2}Multiplication property of radicals
\displaystyle =\displaystyle 2\cdot2\cdot\sqrt{2}Simplify perfect squares
\displaystyle =\displaystyle 4\sqrt{2}Multiply the coefficients

Perfect square method:

\displaystyle \sqrt{32}\displaystyle =\displaystyle \sqrt{16\cdot2}Find largest perfect square factor of radicand
\displaystyle {}\displaystyle =\displaystyle \sqrt{16}\cdot\sqrt{2}Multiplication property of radicals
\displaystyle {}\displaystyle =\displaystyle 4\sqrt{2}Simplify perfect square
Idea summary

We can use these theorems to help us solve problems with triangles:

  • Hypotenuse-Leg similarity theorem: If the ratio of the hypotenuse and leg of one right triangle is equal to the ratio of the hypotenuse and leg of another right triangle, then the two triangles are similar
  • Right triangle similarity theorem: The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the orginal triangle and to each other.

Outcomes

G.TR.3

The student will, given information in the form of a figure or statement, prove and justify two triangles are similar using direct and indirect proofs, and solve problems, including those in context, involving measured attributes of similar triangles.

G.TR.3a

Use definitions, postulates, and theorems (including Side-Angle-Side (SAS); Side-Side-Side (SSS); and Angle-Angle (AA)) to prove and justify that triangles are similar.

G.TR.3b

Use algebraic methods to prove that triangles are similar.

G.TR.3e

Solve problems, including those in context involving attributes of similar triangles

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