Chapter 2

The Conditions for Triangle Similarity

Student Learning Objectives:
Identify and verify similar triangles using angles and side lengths

Why

Understanding the conditions for triangle similarity is pivotal in geometry. The lesson delves into the Angle-Angle Similarity Theorem, which states that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. Additionally, the Side-Side-Side Similarity Theorem is explored, emphasizing that if the corresponding sides of two triangles are proportional, the triangles are deemed similar. These conditions provide a foundation for determining the similarity of triangles, aiding in various geometric applications and problem-solving scenarios.

Lesson Settings & Tools

	11 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Image Credits

The Conditions for Triangle Similarity

Slide of 11

In this lesson some conditions will be developed that guarantee the similarity of triangles.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

The concept of similarity.
Conditions for similarity of polygons.
The concept of congruence.
Conditions that guarantee the congruence of triangles.

Challenge

Using Similarity of Triangles to Solve Problems

Find the ratio of the length of a diagonal and a side of a regular pentagon.

Explore

Investigating Triangles With Two Pairs of Congruent Angles

In the applet, move points Q and R. The applet copies ∠ B and ∠ C on QR and labels the intersection of the corresponding rays by P.

What do you notice about the angles and side lengths of the two triangles?
Are the two triangles similar?

Discussion

Angle-Angle Similarity Theorem

Two polygons are similar if corresponding angles are congruent and corresponding sides are proportional. For triangles, the congruence of two angles already implies similarity.

If two angles of a triangle are congruent to two angles of another triangle, then the triangles are similar.

If ∠ A ≅ ∠ D and ∠ B ≅ ∠ E, then △ ABC ~ △ DEF.

Proof

Angle-Angle Similarity Theorem

Consider two triangles △ ABC and △ DEF, whose two corresponding angles are congruent.

These triangles can be proven to be similar by identifying a similarity transformation that maps one triangle onto the other. First, △ DEF can be dilated with the scale factor k= ABDE about D, forming the new triangle △ DE'F'.

Since a dilation is a similarity transformation, it can be concluded that △ DE'F' and △ DEF are similar triangles. Next, it has to be proven that a rigid motion that maps △ DE'F' onto △ ABC exists. The corresponding angles of similar figures are congruent, so ∠ E' and ∠ E are congruent angles. ∠ E'≅ ∠ E Additionally, since ∠ E is congruent to ∠ B, by the Transitive Property of Congruence, ∠ E' is congruent to ∠ B. ∠ E'≅ ∠ B The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. DE'/DE=k In this case, the scale factor k is ABDE. Applying the Transitive Property of Equality, an equation can be formed and simplified. DE'/DE=AB/DE ⇕ DE'=AB It has been obtained that the two angles and the included side of △ DE'F' are congruent to the corresponding two angles and the included side of △ ABC.

Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, the two triangles are congruent. △ ABC≅△ DE'F' Since congruent figures can be transformed into each other using rigid motions, and △ ABC and △ DE'F' are congruent triangles, there is a rigid motion placing △ DE'F' onto △ ABC.

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps △ DEF onto △ ABC.

Therefore, it can be concluded that △ ABC and △ DEF are similar triangles.

△ ABC~ △ DEF

The proof is now complete.

Example

Solving Problems Using Angle-Angle-Similarity Theorem

The Grim Reaper, who is 5 feet tall, stands 16 feet away from a street lamp at night. The Grim Reaper's shadow cast by the streetlamp light is 8 feet long. How tall is the street lamp?

Hint

Both the lamp post and the Grim Reaper stand vertically on horizontal ground.

Solution

A sketch of the situation is helpful for finding the solution. Under the assumption that the lamp post and the Grim Reaper make right angles in relation to the ground, two right triangles can be drawn. The unknown height of the lamp post is labeled as x.

As these triangles both have a right angle and share the angle on the right-hand side, they are similar by the Angle-Angle (AA) Similarity Theorem. Notice that the base of the larger triangle measures to be 24 feet.

Since the triangles are similar, the ratios between corresponding side lengths are the same. \begin{gathered} \dfrac x 5 = \dfrac{24}8 \end{gathered} The solution of this equation defines the value of x — the height of the street lamp.

\dfrac x 5 = \dfrac{24}8

▼

Solve for x

CalcQuot

Calculate quotient

\dfrac x 5 = 3

MultEqn

LHS * 5=RHS* 5

x = 15

The street lamp at 15 feet high towers over The Grimp Reaper.

Pop Quiz

Practice Solving Problems Using Similar Triangles

For the given diagram, find the missing length.

Discussion

Side-Side-Side Similarity Theorem

A second theorem allows for determining triangle similarity when only the lengths of corresponding sides are known.

If corresponding sides of two triangles are proportional, then the triangles are similar.

If AB/DE=BC/EF=CA/FD, then △ ABC ~ △ DEF.

Proof

Consider two triangles △ ABC and △ DEF, whose corresponding sides are proportional.

Because dilation is a similarity transformation, it can be concluded that △ DE'F' and △ DEF are similar triangles. Now, it has to be proven that a rigid motion that maps △ DE'F' onto △ ABC exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. DE'/DE= DF'/DF= E'F'/EF=k In this case, the scale factor k is ABDE. Since all of the sides of △ ABC and △ DEF are proportional, the scale factor can be expressed by any of the following ratios. k= AB/DE= BC/EF= CA/DF Applying the Transitive Property of Equality, three equations can be formed and simplified. DE'/DE&=AB/DE [0.3cm] DF'/DF&=CA/DF [0.3cm] E'F'/EF&=BC/EF ⇒ DE' &= AB DF' &= CA E'F' &= BC These relations imply that the three sides of △ DE'F' are congruent to the three sides of △ ABC. Therefore, by the Side-Side-Side (SSS) Congruence Theorem, the two triangles are congruent. △ ABC≅△ DE'F' Since congruent figures can be transformed into each other using rigid motions, and △ ABC and △ DE'F' are congruent triangles, there is a rigid motion placing △ DE'F' onto △ ABC.

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps △ DEF onto △ ABC.

Therefore, it can be concluded that △ ABC and △ DEF are similar triangles.

△ ABC~ △ DEF

The proof is now complete.

Example

Solving Problems Using Similar Triangles

There are four congruent angles in the figure. Try to identify them.

Answer

∠ DBA ≅ ∠ BCE ≅ ∠ BEC ≅ ∠ DBE

Hint

Look for similar triangles and an isosceles triangle.

Solution

Step 1

First, notice that segments BE and BC are equal in length.

These are two sides of △ BCE, so by the Isosceles Triangle Theorem, the opposite angles are congruent. ∠ BCE≅∠ BEC

Step 2

Two of the triangles, △ ABD and △ ACE look similar.

Because the lengths of the sides are given, the ratio of corresponding sides can be calculated.

Ratio	Expression	Simplification
AC/AB	459+1275/459=1734/459	34/9
AE/AD	405+1125/405=1530/405	34/9
EC/DB	1360/360	34/9

The last column of the table shows that the corresponding sides of △ ACE and △ ABD are proportional. AC/AB=AE/AD=EC/DB According to the Side-Side-Side (SSS) Similarity Theorem, the two triangles are similar. △ ACE~△ ABD Corresponding angles of similar triangles are congruent.

One pair of these angles is marked on the figure. ∠ BCE≅∠ DBA

Step 3

In addition to the proportions in Step 2 showing that △ ACE and △ ABD are similar, they also show the two triangles are dilations of each other from the common vertex A. Since dilations map a segment to a parallel segment, segments DB and EC are parallel.

Furthermore, since EB is a transversal to two parallel lines, the Alternate Interior Angles Theorem guarantees that the angles at E and B are congruent. ∠ BEC≅∠ DBE These angles are marked on the figure.

Answering the Question

The previous three steps showed three pairs of congruent angles. The transitive property of congruence shows that all four angles mentioned in these pairs are congruent to each other. ∠ DBA ≅ ∠ BCE ≅ ∠ BEC ≅ ∠ DBE The congruent angles are marked on the figure.

Discussion

Side-Angle-Side Similarity Theorem

Two theorems have been covered, now a third theorem that can be used to prove triangle similarity will be investigated. This third theorem allows for determining triangle similarity when the lengths of two corresponding sides and the measure of the included angles are known.

If two sides of a triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar.

If AB/DE=AC/DF and ∠ A ≅ ∠ D, then △ ABC ~ △ DEF.

Proof

Consider two triangles △ ABC and △ DEF, whose two pairs of corresponding sides are proportional and the included angles are congruent.

Because dilation is a similarity transformation, it can be concluded that △ DE'F' and △ DEF are similar triangles. Now, it has to be proven that a rigid motion that maps △ DE'F' onto △ ABC exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor. DE'/DE= DF'/DF=k In this case, the scale factor k is ABDE. Since AB and AC are proportional to DE and DF respectively, the scale factor can be expressed by any of the following ratios. k= AB/DE= AC/DF Applying the Transitive Property of Equality, three equations can be formed and simplified. l c r DE'/DE=AB/DE & ⇒ & DE'=AB [0.3cm] DF'/DF=AC/DF & ⇒ & DF'=AC These relations imply that the two sides of △ DE'F' are congruent to the corresponding two sides of △ ABC. Moreover, the included angles ∠ A and ∠ D are also congruent.

Therefore, by the Side-Angle-Side (SAS) Congruence Theorem, the two triangles are congruent. △ ABC≅△ DE'F' Since congruent figures can be transformed into each other using rigid motions, and △ ABC and △ DE'F' are congruent triangles, there is a rigid motion placing △ DE'F' onto △ ABC.

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps △ DEF onto △ ABC.

Therefore, it can be concluded that △ ABC and △ DEF are similar triangles.

△ ABC~ △ DEF

The proof is now complete.

Example

Proving Similarity Between Triangles Given Sides

The diagram shows the distances between points on a figure.

Show that △ ABC and △ AED are similar triangles. Then find DE.

Hint

Use the Side-Angle-Side (SAS) Similarity Theorem.

Solution

Triangles △ ABC and △ AED have a common angle at A.

The table below contains the ratios of two pairs of corresponding sides of the two triangles.

Ratio	Expression	Simplified Form
AB/AE	7/5+9=7/14	1/2
AC/AD	5/7+3=5/10	1/2

The simplified forms of each ratio are the same. That means these two pairs of sides are proportional. Since the included angle is common in both triangles, the Side-Angle-Side (SAS) Similarity Theorem implies that the triangles are similar. △ ABC~△ AED In similar triangles, all pairs of corresponding sides are proportional. Therefore, the ratio of the two pairs of corresponding sides, found in the table, also applies to the third pair of corresponding sides. CB/DE=1/2 Since the length of CB is given in the diagram, this equation can be solved for DE.

CB/DE=1/2

Substitute

CB= 3

3/DE=1/2

▼

Solve for DE

MultEqn

LHS * DE=RHS* DE

3=1/2DE

MultEqn

LHS * 2=RHS* 2

6=DE

RearrangeEqn

Rearrange equation

DE=6

The length of DE is 6 units.

Closure

Applying Triangle Similarity Theorems to Solve Problems

Through applying the theorems of similar triangles, the ratio of the lengths of a diagonal and the sides of a regular pentagon can be found.

Hint

Begin by determining the angle measures of the figure.

Solution

The Polygon Angle Sum Theorem identifies the sum of the interior angle measures of a pentagon. (5-2)180=540 In a regular pentagon, all five angles are equal measures. Therefore, one of the measures of the angles is a fifth of the sum of the five angle measures. m∠ A = 540/5=108 Furthermore, since the sides of a regular pentagon are congruent, △ ABE is an isosceles triangle.

According to the Isosceles Triangle Theorem, the base angles of △ ABE are congruent. ∠ ABE≅∠ AEB Applying the information found so far to the Triangle Angle Sum Theorem, the measure of the base angles can be found.

m∠ ABE + m∠ AEB + m∠ A = 180

SubstituteII

m∠ AEB= m∠ ABE, m∠ A= 108

m∠ ABE + m∠ ABE + 108 = 180

▼

Solve for m∠ ABE

SubEqn

LHS-108=RHS-108

m∠ ABE + m∠ ABE = 72

SumToProdTwoTerms

a+a=2a

2m∠ ABE = 72

DivEqn

.LHS /2.=.RHS /2.

m∠ ABE = 36

Due to the symmetry of the regular pentagon, there are ten angles with the same measure in the diagram.

Since the measure of ∠ A is known, this gives the measure of ∠ CAD.

m∠ BAC + m∠ CAD + m∠ DAE = m∠ A

SubstituteValues

Substitute values

36 + m∠ CAD + 36 = 108

SubEqn

LHS-(36+36)=RHS-(36+36)

m∠ CAD = 36

This information and the measures of the angles in symmetrical position can now be labeled in the diagram.

Next, focus on △ ACE. In this triangle, AC and EC are diagonals of the pentagon, and AE is a side.

In the diagram, a smaller triangle labeled △ AFE is also present. These two triangles share a common angle at A and congruent angles at C and E. ∠ CAE&≅∠ EAF ∠ ACE&≅∠ AEF According to the Angle-Angle (AA) Similarity Theorem, that means the two triangles are similar. Therefore, the corresponding sides are proportional. CA/AE=AE/AF Continuing forward, notice that triangles △ AFE and △ CEF are isosceles. Therefore, their legs have equal lengths. AE=EF=FC Using s for the length of the sides, AE=EF=FC=s, as indicated on the figure. Also, using d for the length of the diagonal, CE=d and FA=CA-FC=d-s.

These expressions can be substituted in the proportionality relationship previously obtained.

CA/AE=AE/AF

SubstituteExpressions

Substitute expressions

d/s=s/d-s

The question asks for the ratio of d to s. A new variable can be introduced for to represent this unknown ratio. x=d/s The variable d in the expression can then be replaced with x. The result can then be simplified.

d/s=s/d-s

Substitute

d= xs

xs/s=s/xs-s

▼

Simplify

SimpQuot

Simplify quotient

x=s/xs-s

FactorOut

Factor out s

x=s/(x-1)s

ReduceFrac

a/b=.a /s./.b /s.

x=1/x-1

The next step is to solve this equation.

x=1/x-1

▼

Solve for x

MultEqn

LHS * (x-1)=RHS* (x-1)

x(x-1)=1

MultPar

Multiply parentheses

x^2-x=1

SubEqn

LHS-1=RHS-1

x^2-x-1=0

UseQuadForm

Use the Quadratic Formula: a = 1, b= - 1, c= - 1

x=-( -1)±sqrt(( -1)^2-4( 1)( -1))/2( 1)

NegNeg

- (- a)=a

x=1±sqrt((-1)^2-4(1)(-1))/2(1)

CalcPowProd

Calculate power and product

x=1±sqrt(1-(-4))/2

SubNeg

a-(- b)=a+b

x=1±sqrt(1+4)/2

AddTerms

Add terms

x=1±sqrt(5)/2

The ratio of two lengths is positive, so the positive solution gives the ratio of the length of the diagonal and the side of a regular pentagon.

Length of diagonal/Length of side=1+sqrt(5)/2≈ 1.618

Extra

Construction of a regular pentagon

The ratio of the diagonal to the side of a regular pentagon can be used to prove that the following construction creates a regular pentagon. This is a construction created by Yosifusa Hirano in the 19th century.

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Level 3

The Conditions for Triangle Similarity

Exercise 3.1

You have a white rectangular shaped sheet of paper with a gray backside. The paper is folded such that the lower right corner meets the upper horizontal side and the crease goes through the lower left corner.

a sheet of paper which is about to be folded

Calculate the area of the backside that becomes visible after the paper has been folded. The unfolded paper is 15 by 12 centimeters. Round the answer to the nearest square centimeter.

Let's first draw a diagram showing how the paper will look like when folded. At the same time, we will introduce some labels to make it easier to refer to the diagram.

In order to calculate the area of △ DEF, we need to know two side lengths. That means we need the length of EF. From the diagram, we see that x makes the leg of a right triangle where the hypotenuse and second leg are known. Therefore, we can determine x by using the Pythagorean Theorem.

When we know the length of the side labeled x, we can determine the length of EB by subtracting this from 15. 15-9 = 6 Examining the diagram, we can identify two right triangles, △ AED and △ EBF. We need to prove that these triangles are similar. Let's label the acute angles of these triangles.

At point E we can see that the sum of u, v and the right angle makes a straight angle. Therefore, we can write the following equation. m∠ u +m∠ v+90^(∘)= 180^(∘) ⇓ m∠ u = 90^(∘)- m∠ v Next, we will write an equation for the sum of the angles in △ AED. Then we can use the relationship just found to show that ∠ v and ∠ w are congruent.

As we have shown, ∠ w and ∠ v are congruent. Furthermore, since △ ADE and △ BFE have two pairs of congruent angles, we can claim similarity by the Angle-Angle Similarity Theorem.

Using this similarity, we can write an equation preparing to solve for EF.. EF/15=6/12 Let's go for it!

When we know the length of EF, we can calculate the gray triangle's area.

The visible gray area is about 56 square centimeters.

The city council is planning to build a school on a plot of land. Below we see a diagram of the school as seen from above. The school's two shorter buildings each have an entrance facing the courtyard. These are labeled E_1 and E_2.

The city planner wants to place a third entrance, E_3, somewhere along the longer building's side facing the courtyard such that the distance between E_1, E_2, and E_3 is minimized. What is the minimum distance between the three entrances so that can be achieved? The council has asked for the city planner to give their answer in exact form.

Let's start by drawing an outline of the school and adding some lengths to the buildings. We will also label the buildings as A, B, and C.

To determine the minimal distance, we remember that the shortest distance between two points is a straight line. Therefore, we need to place E_3 somewhere along the side of the longest building such that the distance from E_1 to E_3 to E_2 is as short as if it was a straight line. To find where we must place E_3, we can reflect E_2 across the longer side of B.

The distance from E'_2 to E_1 is minimized by a straight segment between them. Where this segment intersects the longer side of B, is where we should place the third entrance to minimize the distance. If we draw this segment we can identify two similar triangles by the Angle-Angle Similarity Theorem.

Using the similarity between the triangles, we can write an equation. x/250-x=100/150 Let's solve this equation for x.

The third door should be placed 100 feet to the left of where B and C meet on the courtyard.

Now we can determine the minimum distance by using the Pythagorean Theorem. 150^2+150^2=c^2 &⇔ c =150sqrt(2) 100^2+100^2=d^2 &⇔ d=100sqrt(2) Finally, we will add the distance of c and d to obtain the minimal distance between the three entrances. c+d=150sqrt(2)+100sqrt(2) ⇓ c+d=250sqrt(2)feet

The Conditions for Triangle Similarity

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