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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.

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

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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.

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=DEAB $ 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.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.

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?

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

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

For the given diagram, find the missing length.

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 $DEAB =EFBC =FDCA ,$ then △ABC∼△DEF.

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

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=DEAB $ about D, forming the new triangle $△DE_{′}F_{′}.$

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.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.

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

∠DBA≅∠BCE≅∠BEC≅∠DBE

Look for similar triangles and an isosceles triangle.

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.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 |
---|---|---|

$ABAC $ | $459459+1275 =4591734 $ | $934 $ |

$ADAE $ | $405405+1125 =4051530 $ | $934 $ |

$DBEC $ | $3601360 $ | $934 $ |

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.

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 $DEAB =DFAC $ and ∠A≅∠D, then △ABC∼△DEF.

Consider two triangles △ABC and △DEF, whose two pairs of corresponding sides are proportional and the included 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=DEAB $ about D, forming the new triangle $△DE_{′}F_{′}.$

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.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.

The diagram shows the distances between points on a figure.

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

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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 |
---|---|---|

$AEAB $ | $5+97 =147 $ | $21 $ |

$ADAC $ | $7+35 =105 $ | $21 $ |

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.

Begin by determining the angle measures of the figure.

The Polygon Angle Sum Theorem identifies the sum of the interior angle measures of a pentagon.
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.
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.
Applying the information found so far to the Triangle Angle Sum Theorem, the measure of the base angles can be found.
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.
This information and the measures of the angles in symmetrical position can now be labeled in the diagram.
According to the Angle-Angle (AA) Similarity Theorem, that means the two triangles are similar. Therefore, the corresponding sides are proportional.
Continuing forward, notice that triangles △AFE and △CEF are isosceles. Therefore, their legs have equal lengths.
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.
The question asks for the ratio of d to s. A new variable can be introduced for to represent this unknown ratio.
The variable d in the expression can then be replaced with x. The result can then be simplified.
The next step is to solve this equation.
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.

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

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.$x=x−11 $

Solve for x

MultEqn

LHS⋅(x−1)=RHS⋅(x−1)

x(x−1)=1

MultPar

Multiply parentheses

x2−x=1

SubEqn

LHS−1=RHS−1

x2−x−1=0

UseQuadForm

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

$x=2(1)-(-1)±(-1)_{2}−4(1)(-1) $

NegNeg

-(-a)=a

$x=2(1)1±(-1)_{2}−4(1)(-1) $

CalcPowProd

Calculate power and product

$x=21±1−(-4) $

SubNeg

a−(-b)=a+b

$x=21±1+4 $

AddTerms

Add terms

$x=21±5 $

$Length of sideLength of diagonal =21+5 ≈1.618$

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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