Angle-Angle Similarity Theorem & Side-Side-Side Similarity Theorem Explained

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.

Regular pentagon ABCDE with all five diagonals drawn

Explore

Investigating Triangles With Two Pairs of Congruent Angles

In the applet, move points $Q$ and $R .$ The applet copies $∠ B$ and $∠ C$ on $\overline{Q R}$ 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.

Triangles ABC and DEF with congruent angles marked at A and D and B and E.

If $∠ A ≅ ∠ D$ and $∠ B ≅ ∠ E,$ then $△ A B C \sim △ D E F .$

Proof

Angle-Angle Similarity Theorem

Consider two triangles $△ A B C$ and $△ D E F,$ 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, $△ D E F$ can be dilated with the scale factor $k = \frac{A B}{D E}$ about $D,$ forming the new triangle $△ D E^{'} F^{'} .$

Since a dilation is a similarity transformation, it can be concluded that

△ D E^{'} F^{'}

and

△ D E F

are similar triangles. Next, it has to be proven that a rigid motion that maps

△ D E^{'} F^{'}

onto

△ A B C

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.

\frac{D E ^{'}}{D E} = k

In this case, the scale factor

k

\frac{A B}{D E} .

Applying the Transitive Property of Equality, an equation can be formed and simplified.

\frac{D E ^{'}}{D E} = \frac{A B}{D E} ⇕ D E^{'} = A B

It has been obtained that the two angles and the included side of

△ D E^{'} F^{'}

are congruent to the corresponding two angles and the included side of

△ A B C .

Therefore, by the Angle-Side-Angle (ASA) Congruence Theorem, the two triangles are congruent.

△ A B C ≅ △ D E^{'} F^{'}

Since congruent figures can be transformed into each other using rigid motions, and

△ A B C

and

△ D E^{'} F^{'}

are congruent triangles, there is a rigid motion placing

△ D E^{'} F^{'}

onto

△ A B C .

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps

△ D E F

onto

△ A B C .

Therefore, it can be concluded that $△ A B C$ and $△ D E F$ are similar triangles.

$△ A B C \sim △ D E F$

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.

\frac{x}{5} = \frac{2 4}{8}

The solution of this equation defines the value of

x

— the height of the street lamp.

\frac{x}{5} = \frac{2 4}{8}

▼

Solve for

x

CalcQuot

Calculate quotient

\frac{x}{5} = 3

MultEqn

$LHS \cdot 5 = RHS \cdot 5$

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 $\frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{F D},$ then $△ A B C \sim △ D E F .$

Proof

Consider two triangles $△ A B C$ and $△ D E F,$ whose corresponding sides are proportional.

Because dilation is a similarity transformation, it can be concluded that

△ D E^{'} F^{'}

and

△ D E F

are similar triangles. Now, it has to be proven that a rigid motion that maps

△ D E^{'} F^{'}

onto

△ A B C

exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.

\frac{D E ^{'}}{D E} = \frac{D F ^{'}}{D F} = \frac{E ^{'} F ^{'}}{E F} = k

In this case, the scale factor

k

\frac{A B}{D E} .

Since all of the sides of

△ A B C

and

△ D E F

are proportional, the scale factor can be expressed by any of the following ratios.

k = \frac{A B}{D E} = \frac{B C}{E F} = \frac{C A}{D F}

Applying the Transitive Property of Equality, three equations can be formed and simplified.

\frac{D E ^{'}}{D E} \frac{D F ^{'}}{D F} \frac{E ^{'} F ^{'}}{E F} = \frac{A B}{D E} = \frac{C A}{D F} = \frac{B C}{E F} \Rightarrow D E^{'} D F^{'} E^{'} F^{'} = A B = C A = B C

These relations imply that the three sides of

△ D E^{'} F^{'}

are congruent to the three sides of

△ A B C .

Therefore, by the Side-Side-Side (SSS) Congruence Theorem, the two triangles are congruent.

△ A B C ≅ △ D E^{'} F^{'}

Since congruent figures can be transformed into each other using rigid motions, and

△ A B C

and

△ D E^{'} F^{'}

are congruent triangles, there is a rigid motion placing

△ D E^{'} F^{'}

onto

△ A B C .

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△ D E F$ onto $△ A B C .$

Therefore, it can be concluded that $△ A B C$ and $△ D E F$ are similar triangles.

$△ A B C \sim △ D E F$

The proof is now complete.

Example

Solving Problems Using Similar Triangles

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

Triangle ACE and point B on AC and D on AE. Labelled distances: AB=459, BC=1275, AD=405, DB=360, DE=1125, BE=1275, EC=1360

Answer

$∠ D B A ≅ ∠ B C E ≅ ∠ B E C ≅ ∠ D B E$

Hint

Look for similar triangles and an isosceles triangle.

Solution

Step 1

First, notice that segments $\overline{B E}$ and $\overline{B C}$ are equal in length.

These are two sides of

△ B C E,

so by the Isosceles Triangle Theorem, the opposite angles are congruent.

∠ B C E ≅ ∠ B E C

Step 2

Two of the triangles, $△ A B D$ and $△ A C E$ look similar.

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

Ratio	Expression	Simplification
$\frac{A C}{A B}$	$\frac{4 5 9 + 1 2 7 5}{4 5 9} = \frac{1 7 3 4}{4 5 9}$	$\frac{3 4}{9}$
$\frac{A E}{A D}$	$\frac{4 0 5 + 1 1 2 5}{4 0 5} = \frac{1 5 3 0}{4 0 5}$	$\frac{3 4}{9}$
$\frac{E C}{D B}$	$\frac{1 3 6 0}{3 6 0}$	$\frac{3 4}{9}$

The last column of the table shows that the corresponding sides of

△ A C E

and

△ A B D

are proportional.

\frac{A C}{A B} = \frac{A E}{A D} = \frac{E C}{D B}

According to the Side-Side-Side (SSS) Similarity Theorem, the two triangles are similar.

△ A C E \sim △ A B D

Corresponding angles of similar triangles are congruent.

One pair of these angles is marked on the figure.

∠ B C E ≅ ∠ D B A

Step 3

In addition to the proportions in Step 2 showing that $△ A C E$ and $△ A B D$ 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 $\overline{D B}$ and $\overline{E C}$ are parallel.

Furthermore, since

\overline{E B}

is a transversal to two parallel lines, the Alternate Interior Angles Theorem guarantees that the angles at

E

and

B

are congruent.

∠ B E C ≅ ∠ D B E

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.

∠ D B A ≅ ∠ B C E ≅ ∠ B E C ≅ ∠ D B E

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.

Triangles ABC and DEF with congruent angles marked at A and D.

If $\frac{A B}{D E} = \frac{A C}{D F}$ and $∠ A ≅ ∠ D,$ then $△ A B C \sim △ D E F .$

Proof

Consider two triangles $△ A B C$ and $△ D E F,$ 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

△ D E^{'} F^{'}

and

△ D E F

are similar triangles. Now, it has to be proven that a rigid motion that maps

△ D E^{'} F^{'}

onto

△ A B C

exists. The ratios of the corresponding side lengths of similar polygons are the same and equal to the scale factor.

\frac{D E ^{'}}{D E} = \frac{D F ^{'}}{D F} = k

In this case, the scale factor

k

\frac{A B}{D E} .

Since

\overline{A B}

and

\overline{A C}

are proportional to

\overline{D E}

and

\overline{D F}

respectively, the scale factor can be expressed by any of the following ratios.

k = \frac{A B}{D E} = \frac{A C}{D F}

Applying the Transitive Property of Equality, three equations can be formed and simplified.

\frac{D E ^{'}}{D E} = \frac{A B}{D E} \frac{D F ^{'}}{D F} = \frac{A C}{D F} \Rightarrow \Rightarrow D E^{'} = A B D F^{'} = A C

These relations imply that the two sides of

△ D E^{'} F^{'}

are congruent to the corresponding two sides of

△ A B C .

Moreover, the included angles

∠ A

and

∠ D

are also congruent.

Therefore, by the Side-Angle-Side (SAS) Congruence Theorem, the two triangles are congruent.

△ A B C ≅ △ D E^{'} F^{'}

Since congruent figures can be transformed into each other using rigid motions, and

△ A B C

and

△ D E^{'} F^{'}

are congruent triangles, there is a rigid motion placing

△ D E^{'} F^{'}

onto

△ A B C .

The combination of this rigid motion and the dilation performed earlier forms a similarity transformation that maps $△ D E F$ onto $△ A B C .$

Therefore, it can be concluded that $△ A B C$ and $△ D E F$ are similar triangles.

$△ A B C \sim △ D E F$

The proof is now complete.

Example

Proving Similarity Between Triangles Given Sides

The diagram shows the distances between points on a figure.

Triangle ADE with point C on AE and B on AD. The distances given are AB=7, BD=3, AC=5, CE=9, BC=3.

Show that $△ A B C$ and $△ A E D$ are similar triangles. Then find $D E .$

Hint

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

Solution

Triangles $△ A B C$ and $△ A E D$ 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
$\frac{A B}{A E}$	$\frac{7}{5 + 9} = \frac{7}{1 4}$	$\frac{1}{2}$
$\frac{A C}{A D}$	$\frac{5}{7 + 3} = \frac{5}{1 0}$	$\frac{1}{2}$

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.

△ A B C \sim △ A E D

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.

\frac{C B}{D E} = \frac{1}{2}

Since the length of

\overline{C B}

is given in the diagram, this equation can be solved for

D E .

\frac{C B}{D E} = \frac{1}{2}

Substitute

$C B = 3$

\frac{3}{D E} = \frac{1}{2}

▼

Solve for

D E

MultEqn

$LHS \cdot D E = RHS \cdot D E$

3 = \frac{1}{2} D E

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

6 = D E

RearrangeEqn

Rearrange equation

D E = 6

The length of

\overline{D E}

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 = \frac{5 4 0}{5} = 108

Furthermore, since the sides of a regular pentagon are congruent,

△ A B E

is an isosceles triangle.

According to the Isosceles Triangle Theorem, the base angles of

△ A B E

are congruent.

∠ A B E ≅ ∠ A E B

Applying the information found so far to the Triangle Angle Sum Theorem, the measure of the base angles can be found.

m ∠ A B E + m ∠ A E B + m ∠ A = 180

SubstituteII

$m ∠ A E B = m ∠ A B E$ , $m ∠ A = 108$

m ∠ A B E + m ∠ A B E + 108 = 180

▼

Solve for

m ∠ A B E

SubEqn

$LHS - 108 = RHS - 108$

m ∠ A B E + m ∠ A B E = 72

SumToProdTwoTerms

$a + a = 2 a$

2 m ∠ A B E = 72

DivEqn

$LHS / 2 = RHS / 2$

m ∠ A B E = 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

∠ C A D .

m ∠ B A C + m ∠ C A D + m ∠ D A E = m ∠ A

SubstituteValues

Substitute values

36 + m ∠ C A D + 36 = 108

SubEqn

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

m ∠ C A D = 36

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

Next, focus on $△ A C E .$ In this triangle, $\overline{A C}$ and $\overline{E C}$ are diagonals of the pentagon, and $\overline{A E}$ is a side.

In the diagram, a smaller triangle labeled

△ A F E

is also present. These two triangles share a common angle at

A

and congruent angles at

C

and

E .

∠ C A E ∠ A C E ≅ ∠ E A F ≅ ∠ A E F

According to the Angle-Angle (AA) Similarity Theorem, that means the two triangles are similar. Therefore, the corresponding sides are proportional.

\frac{C A}{A E} = \frac{A E}{A F}

Continuing forward, notice that triangles

△ A F E

and

△ C E F

are isosceles. Therefore, their legs have equal lengths.

A E = E F = F C

Using

s

for the length of the sides,

A E = E F = F C = s

, as indicated on the figure. Also, using

d

for the length of the diagonal,

C E = d

and

F A = C A - F C = d - s .

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

\frac{C A}{A E} = \frac{A E}{A F}

SubstituteExpressions

Substitute expressions

\frac{d}{s} = \frac{s}{d - s}

The question asks for the ratio of

d

s .

A new variable can be introduced for to represent this unknown ratio.

x = \frac{d}{s}

The variable

d

in the expression can then be replaced with

x .

The result can then be simplified.

\frac{d}{s} = \frac{s}{d - s}

Substitute

$d = x s$

\frac{x s}{s} = \frac{s}{x s - s}

▼

Simplify

SimpQuot

Simplify quotient

x = \frac{s}{x s - s}

FactorOut

Factor out $s$

x = \frac{s}{( x - 1 ) s}

ReduceFrac

$\frac{a}{b} = \frac{a / s}{b / s}$

x = \frac{1}{x - 1}

The next step is to solve this equation.

x = \frac{1}{x - 1}

▼

Solve for

x

MultEqn

$LHS \cdot (x - 1) = RHS \cdot (x - 1)$

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 = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 1 )}{2 ( 1 )}

NegNeg

$- (- a) = a$

x = \frac{1 \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 1 )}{2 ( 1 )}

CalcPowProd

Calculate power and product

x = \frac{1 \pm 1 - ( - 4 )}{2}

SubNeg

$a - (- b) = a + b$

x = \frac{1 \pm 1 + 4}{2}

AddTerms

Add terms

x = \frac{1 \pm 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.

$\frac{Length of diagonal}{Length of side} = \frac{1 + 5}{2} \approx 1.618$

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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Catch-Up and Review

Proof

Hint

Solution

Proof

Answer

Hint

Solution

Step 1

Step 2

Step 3

Answering the Question

Proof

Hint

Solution

Hint

Solution

Extra