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

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.

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

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Challenge

Using Similarity of Triangles to Solve Problems

Discussion

Angle-Angle Similarity Theorem

Example

Solving Problems Using Angle-Angle-Similarity Theorem

Hint

Solution

Pop Quiz

Practice Solving Problems Using Similar Triangles

Discussion

Side-Side-Side Similarity Theorem

Discussion

Side-Angle-Side Similarity Theorem

Closure

Applying Triangle Similarity Theorems to Solve Problems

Hint

Solution

Extra

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