The Pythagorean Theorem and the Golden Triangle

Section 5.3 The Pythagorean Theorem and the Golden Triangle

In this section, we will see how similar right triangles can be used to derive yet another proof of the Pythagorean Theorem. We also will learn about self-similarity and the Golden Triangle. First, we review the key concepts of dilations and similarity already explored in this chapter.

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Subsection 5.3.1 A Review of Similarity

Exploration 5.3.1.

(a)

Figure 5.3.1 gives a triangle \(\Delta ABC\) and its image \(\Delta A'B'C'\text{.}\) Use either the GeoGebra applet or the coordinate system to complete the following.

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Figure 5.3.1. GeoGebra applet for Task 5.3.1.a

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(i)

Determine the center \(O\) of the dilation that takes \(\Delta ABC\) to \(\Delta A'B'C'\text{.}\)

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Hint.

What lines must \(O\) lie on?

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(ii)

Determine \(\frac{OA'}{OA}\text{,}\) \(\frac{OB'}{OB}\text{,}\) and \(\frac{OC'}{OC}\text{.}\) What do these ratios tell you about the dilation?

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Hint.

Besides the center, what else defines a dilation?

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(iii)

Is it true that \(\frac{OA'}{AA'}=\frac{OB'}{BB'}=\frac{OC'}{CC'}\) ? Are these equal to the scale factor?

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(iv)

Is it true that \(\frac{A'B'}{AB}=\frac{B'C'}{BC}=\frac{A'C'}{AC}\) ? Are these equal to the scale factor?

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(v)

Which pairs of angles must be congruent?

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Hint.

There are three pairs in the original sketch of the two triangles. If you want an added challenge you can find more pairs using your center as a side point.

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(b)

In Figure 5.3.2, \(\Delta FGH\sim\Delta IJK\)

The triangles are similar

. Also, \(FG=4.4, FH=13, IK=7.8, JK=9.06, m\angle GFH=110.34\deg\) and \(m\angle IJK=53.8\deg\)Determine the following:

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Triangles FGH and IJK are similar. Angle GFH has measure 110.34, angle IJK has measure 53.8. FH has length 13, IK has length 7.8, and, JK has length 9.06. — Figure 5.3.2. Illustration for Task 5.3.1.b
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(i)

\(m\angle FGH\)

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(ii)

\(m\angle FHG\)

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(iii)

\(m\angle JIK\)

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(iv)

\(m\angle KIJ\)

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(v)

The scale factor of the dilation that takes \(\Delta FGH\) to \(\Delta IJK\)

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(vi)

\(GH\)

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(vii)

\(IJ\)

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Subsection 5.3.2 Applications of Similarity

We now apply our knowledge of similarity to two topics: another proof of the Pythagorean Theorem and a triangle known as the Golden Triangle.

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Exploration 5.3.2. A Proof of the Pythagorean Theorem.

Use Figure 5.3.3 for this exploration.

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Figure 5.3.3. GeoGebra applet of a Right Triangle

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(a) Right Triangles within a Right Triangle.

Complete the following:

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(i)

Measure angles \(\angle ABC\text{,}\) \(\angle BCA\text{,}\) and \(\angle CAB\text{.}\)

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(ii)

Draw a line segment from point \(C\) meeting side \(\overline{AB}\) perpendicularly at a point \(D\text{.}\) How many triangles are now in the sketch? Name them.

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(iii)

To what other angle(s) is \(\angle CDA\) congruent?

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(iv)

Explain how you know triangle \(\Delta ACD\) is similar to \(\Delta ABC\text{.}\) What is the constant of proportionality?

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(v)

What other triangle is similar to triangle \(\Delta ABC\text{?}\) Be sure to write the letters in the order of corresponding vertices.

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(b) A Similar Triangle Proof of the Pythagorean Theorem.

We continue to use Figure 5.3.3 and the constructions from above in this task.

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(i)

Explain why \(\frac{AC}{AB}=\frac{AD}{AC}\) and \(\frac{BC}{AB}=\frac{BD}{BC}\text{.}\)

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(ii)

Since \(\frac{AC}{AB}=\frac{AD}{AC}\text{,}\) we see \(AC\cdot AC=AB\cdot AD\text{.}\) Why?

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(iii)

Use the equation\(\frac{BC}{AB}=\frac{BD}{BC}\text{,}\) to complete the equation \(BC\cdot BC=\text{.}\)

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(iv)

So that the Pythagorean Theorem will be more obvious, we will assign each length a single letter; namely, \(a=BC\text{,}\) \(b=AC\text{,}\) and \(c=AB\text{.}\) What is \(AD+DB\) using these lower-case letter(s)?

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(v)

Equating \(a^2+b^2\) with \(BC\cdot BC+AC\cdot AC\text{,}\) substitute the righthand expressions from the equations in the first two parts of this task. Then add the expressions. Do you get \(c^2\text{?}\) Show all work.

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Subsection 5.3.3 Self-Similar Triangles

The right triangle in Exploration 5.3.2 has the property that the altitude drawn from the right angle to the hypotenuse splits the triangle into two smaller triangles. Both of these triangles are similar to the original triangle. This property is true for all right triangles since the measure of the two acute angles did not matter. In particular, each new triangle has a right angle and one of the original angles. Since the sum of the angles in any triangle is 180 degrees, the newly created angle has the same measure as the remaining angle of the original triangle.

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Will this happen with other triangles? We explore this question in the next activity.

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Exploration 5.3.3. A Special Triangle.

In Figure 5.3.4, an isosceles triangle with two 72-degree angles is provided. Use the GeoGebra applet to answer the following:

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Figure 5.3.4. A Golden Triangle

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(a)

What is the measure of the third angle \(\angle ACB\text{?}\)

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(b)

Compute \(\frac{AC}{AB}\text{,}\) the ratio of the long side to the short side. As you record this number, write down at least four digits after the decimal point. Adjust your calculator settings to show more digits if fewer than five digits are visible.

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Hint.

The Desmos Scientific Calculator at desmos.com

desmos.com/scientific

may be used to perform these calculations. If you enter your computation, the application will give you a new calculation line so you can compare results.

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(c)

Use the angle bisector tool to bisect angle \(\angle CAB\text{.}\) Let \(D\) be the point where this bisector intersects side \(BC\) and create line segment \(\overline{AD}\text{.}\) Use the show/hide tool to hide bisector \(\overrightarrow{AD}\text{.}\) Segment \(\overline{AD}\) should still be visible.

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(i)

Bisecting \(\angle CAB\) splits \(\Delta ABC\) into two triangles. Use GeoGebra’s Polygon tool to highlight the subtriangle that is similar to \(\Delta ABC\text{.}\)

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(ii)

List the three pairs of corresponding sides.

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(iii)

What is the constant of proportionality, \(k\) for this similarity? As you record this number, write down at least four digits after the decimal point.

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(iv)

Note that the constant of proportionality depends on whether you chose to divide the shorter length by the longer or to divide the longer length by the shorter. Perform whichever computation have not already done and record at least four digits after the decimal point.

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(v)

What do you notice about these two numbers?

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Hint.

There is some slight unavoidable error in the numbers due to GeoGebra measurements and estimation. Five significant digits seems to work here.

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(d)

Now use the angle bisector tool to bisect angle \(\angle ABD\) in Figure 5.3.4. Let \(E\) be the point where this angle bisector intersects segment \(\overline{AD}\text{.}\) Draw segment \(\overline{BE}\) and hide ray \(\overrightarrow{BE}\text{.}\)

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(i)

Identify another triangle in the sketch which is similar to \(\Delta ABE\text{.}\)

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(ii)

Identify two triangles in the sketch that are similar to \(\Delta DEB\text{.}\)

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(e) Reflective Questions and Extensions.

Based on your work in this exploration, discuss the following:

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(i)

If you use the arrow tool to move vertex \(A\) or \(B\text{,}\) the lengths of the sides of triangles will change, but the angle measures will not. Do the constants of proportionality change?

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(ii)

Suppose you continued this process, bisecting \(\angle BDE\) to create triangle \(\Delta DEF\text{,}\) then bisecting \(\angle DEF\) to create \(\Delta EFG\text{,}\) and so forth. What type of design would you get? What properties would it have?

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(iii)

The area tool is under the angle menu. What is the ratio of the areas of similar triangles \(\Delta ABC\) and \(\Delta BDA\text{?}\) How is this number related to the ratio of the corresponding sides?

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Hint.

Use the ‘Polygon’ tool to create \(\Delta BDA\) before using the GeoGebra ‘Area’ tool.

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Also, review Section 3.5 for one relationship.

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(iv)

Use a calculator to estimate the number \(\frac{1+\sqrt{5}}{2}\text{.}\) What do you notice?

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The isosceles triangle \(\Delta ABC\) explored in Exploration 5.3.3 is known as a Golden Triangle. When we create a smaller, but similar, triangle by bisecting a base angle, the constant of proportionality equals the ratio of the long side to the short side (base) of the original triangle. We can repeat this process of bisecting the base angle of our new triangle to form even smaller similar triangles indefinitely. At every stage of the process, the ratio of the long side to the base will remain \(\frac{1+\sqrt{5}}{2}\approx 1.618033989\) and equal to the constant of proportionality for the pair of similar triangles. This number is known as the Golden Ratio and is an irrational number

An irrational number cannot be written as the ratio of integers. The decimal expansion of an irrational number is infinite and does not become an infinitely repeating block of digits.

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Subsection 5.3.4 A Connection to a Regular Pentagon

We explored regular polygons in Section 2.4.

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Exploration 5.3.4.

The polygon \(ABCDE\) in Figure 5.3.5 is a regular pentagon.

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Regular pentagon ABCDE — Figure 5.3.5. A Triangle in a Regular Pentagon
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(a)

In Section 2.4, we learned that each vertex angle of a regular \(n\)-gon measures \(\frac{180(n-2)}{n}\) degrees. Assume that \(ABCDE\) is a regular pentagon and determine the measure of \(\angle EAB\text{.}\)

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(b)

Draw segment \(\overline{AD}\text{.}\) What type of triangle is \(\Delta EAD\text{?}\)

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(c)

Compute \(m\angle{EAD}\text{.}\) How did you determine this?

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(d)

Determine \(m\angle{DAB}\) showing all work.

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(e)

Construct \(\overline{BD}\text{.}\) What is the measure of \(\angle CDE\text{?}\) How do you know?

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(f)

Describe triangle \(\Delta DAB\) as specifically as possible. Why does this exploration belong in this section?

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(g)

Now extend lines \(\overleftrightarrow{ED}\) and \(\overleftrightarrow{BC}\) so that they meet at a point \(F\text{.}\) Similarly, let \(G\) be the point where \(\overleftrightarrow{DC}\) meets \(\overleftrightarrow{AB}\text{,}\) let \(H\) be the point where \(\overleftrightarrow{CB}\) and \(\overleftrightarrow{EA}\) intersect, let \(I\) be the intersection of \(\overleftrightarrow{BA}\) and \(\overleftrightarrow{DE}\text{,}\) and let \(J\) be the intersection of \(\overleftrightarrow{AE}\) and \(\overleftrightarrow{CD}\text{.}\)

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(i)

Describe the shape of the decagon \(DFCGBHAIEJ\text{.}\)

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(ii)

What type of triangle is \(\Delta DFC\text{?}\) How is it related to \(\Delta ADB\text{?}\)

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Exercises 5.3.5 Exercises

Skills and Recall

1.

In Figure 5.3.6, \(D\text{,}\) \(E\text{,}\) and \(F\) are the midpoints of \(\overline{AB}\text{,}\) \(\overline{BC}\text{,}\) and \(\overline{AC}\text{,}\) respectively. Also, \(AB=8\text{,}\) \(BC=14\text{,}\) and \(AC=10\text{.}\) Angles \(\angle ABC\text{,}\) \(\angle ADF\text{,}\) and \(\angle FEC\) have measure 42 degrees; whereas, \(m\angle ACB=30\deg\) and \(m\angle DEF=108\deg\text{.}\) Answer the following:

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Triangle ABC with midpoints DEF as described — Figure 5.3.6.
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(a)

Determine \(m\angle DEB\) and \(m\angle BAC\text{.}\)

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Hint.

Look at the angles meeting at \(E\text{.}\)

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Answer.

30 degrees

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Solution.

\(180-(108+42)=180-150=30\)

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(b)

Determine \(m\angle EFC\text{,}\) \(m\angle BDE\text{,}\) and \(m\angle BAC\text{.}\)

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Hint.

In the previous problem, it was determined that \(m\angle BAC\) = 108 degrees. Use triangle similarity to find \(m\angle EFC\) and \(m\angle BDE\text{.}\) Hint - Both \(\angle EFC\) and \(\angle BDE\) correspond to \(m\angle BAC\text{.}\)

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Answer.

\(m\angle EFC\) = 108 degrees \(m\angle BDE\) = 108 degrees \(m\angle BAC\) = 108 degrees

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Solution.

Triangles \(\Delta EFC\) and \(\Delta BDE\) are similar to triangle \(\Delta BAC\text{.}\) This means their corresponding angles are equal. Since \(m\angle BAC\) = 108 degrees, both \(m\angle EFC\) and \(m\angle BDE\) are also 108 degrees.

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(c)

Determine \(m\angle FDE\) and \(m\angle AFD\text{.}\)

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Hint.

Use angle similarity and match corresponding angles. Which angle in the larger triangle \(\Delta ABC\) corresponds with \(\angle FDE\text{?}\) What about \(\angle AFD\text{?}\) Use these relationships to find missing measures.

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Answer.

\(m\angle FDE\) = 30 degrees \(m\angle AFD\) = 30 degrees

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Solution.

When D, E, and F are midpoints on the sides of \(\Delta ABC\text{,}\) the smaller triangles like \(\Delta DEF\) and \(\Delta ADF\) are similar to \(\Delta ABC\text{.}\) This means their corresponding angles are equal. So \(\angle FDE\) and \(\angle AFD\) both correspond to \(m\angle C\text{.}\) Since \(m\angle C\) is 30 degrees, both \(m\angle FDE\) and \(m\angle AFD\) are also 30 degrees.

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(d)

Determine \(m\angle EFD\text{.}\)

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Hint.

The interior angles of any triangle always add up to 180 degrees (Triangle Angle Sum Theorem). For \(\Delta FDE\text{,}\) \(m\angle FDE\) and \(m\angle DEF\) are already known. Use these two angles measures to find \(m\angle EFD\text{.}\)

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Answer.

\(m\angle EFD\) = 42 degrees

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Solution.

It is given that \(m\angle FDE\) = 30 degrees and \(m\angle DEF\) = 108 degrees. Using the Triangle Angle Sum Theorem, \(m\angle EFD\) is calculated as: 180 - (108 + 30) = 42 degrees.

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(e)

Identify four triangles that are similar to \(\Delta ABC\text{.}\) Be sure to list the vertices in order.

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Hint.

Use the Midpoint Theorem to identify four triangles formed by connecting midpoints. These triangles are all similar to the original triangle, \(\Delta ABC\text{.}\)

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Answer.

Four triangles similar to \(\Delta ABC\) include: \(\Delta EFD\text{,}\) \(\Delta ADF\text{,}\) \(\Delta DBE\text{,}\) and \(\Delta FEC\text{.}\)

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Solution.

When D, E, and F are the midpoints of the sides of triangle \(\Delta ABC\text{,}\) connecting them forms four smaller triangles. All four of these triangles (\(\Delta EFD\text{,}\) \(\Delta ADF\text{,}\) \(\Delta DBE\text{,}\) and \(\Delta FEC\)) are similar to the original triangle \(\Delta ABC\text{.}\) This means that each smaller triangle is a miniature copy of the larger triangle \(\Delta ABC\text{.}\) Therefore, all angles within these four smaller triangles perfectly match the corresponding angles in \(\Delta ABC\text{.}\)

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(f)

Determine the length of \(\overline{AF}\text{.}\) Support your answer.

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Hint.

Use the definition of a midpoint. What does a midpoint do to a line segment?

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Answer.

The length of \(\overline{AF}\) is 5.

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Solution.

Point F is the midpoint of side \(\overline{AC}\text{,}\) meaning it divides the side exactly in half. Since the total length of \(\overline{AC}\) is 10, the length of \(\overline{AF}\) is half of that. Therefore, the length of \(\overline{AC}\) is 5.

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(g)

Determine the length of \(\overline{DF}\text{.}\) Support your answer.

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Hint.

Line segment \(\overline{DF}\) connects points D and F. Since both are midpoints, the Midsegment Theorem applies. This theorem helps find the length of a line segment that connects two midpoints of a triangle.

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Answer.

The length of \(\overline{DF}\) is 7.

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Solution.

When D and F are the midpoints of sides \(\overline{AB}\) and \(\overline{AC}\text{,}\) they create a midsegment called \(\overline{DF}\text{.}\) The Midsegment Theorem states that this segment is always half the length of the triangle’s third side, \(\overline{BC}\text{.}\) Since \(\overline{BC}\) measures 14, \(\overline{DF}\) is exactly half of that, making it 7.

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(h)

There is a dilation taking \(\Delta ABC\) to \(\Delta FEC\text{.}\) What is the center and scale factor of this dilation?

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Hint.

For the center of this dilation, find the vertex that is shared by both \(\Delta ABC\) and \(\Delta FEC\text{.}\) For scale factor, consider how the sides of \(\Delta FEC\) are smaller because F and E are midpoints.

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Answer.

Center of Dilation: Point C Scale Factor: \(\frac{1}{2}\)

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Solution.

Point C is the center of dilation because it is the shared corner between \(\Delta ABC\) and \(\Delta FEC\text{.}\) The scale factor is \(\frac{1}{2}\text{.}\) This is because the new triangle \(\Delta FEC\) has sides such as \(\overline{FC}\) that are exactly are half the length of the matching side \(\overline{AC}\) in the original triangle.

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(i)

There is an isometry taking \(\Delta DEB\) to \(\Delta FCE\text{.}\) What type of isometry? Identify its defining object(s).

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Hint.

See Table 4.4.12.

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Answer.

Type of Isometry: Translation Defining Object(s): Translation Vector

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Solution.

This movement is a translation. It means \(\Delta DEB\) simply slides over to become \(\Delta FCE\text{,}\) without rotating or flipping. Every point in \(\Delta DEB\) moves the same distance and in the same direction to align exactly with \(\Delta FCE\text{.}\)

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(j)

There is an isometry taking \(\Delta ADF\) to \(\Delta EFD\text{.}\) What type of isometry? Identify its defining object(s).

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Hint.

See Table 4.4.12. You may want to check orientation as well.

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Answer.

Type of Isometry: Rotation Defining Object(s): Center and Rotational Angle

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Solution.

This a rotation with exactly one fixed point, the midpoint of \(\overline{DF}\text{.}\) Rotating \(\Delta ADF\) 180 degrees about this point aligns it exactly with \(\Delta EFD\text{.}\)

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Extending the Concepts

2. Calculating the Golden Ratio.

The ratio \(\frac{1+\sqrt{5}}/2\) that was the constant of proportionality in Exploration 5.3.3 is known as the Golden Ratio.

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(a)

Add an algebraic exploration into the number \(\frac{1+\sqrt{5}}{2}\text{,}\) its reciprocal, and its square.

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3. Surprising Facts about the Golden Ratio.

The ratio \(\frac{1+\sqrt{5}}{2}\) that was the constant of proportionality in Exploration 5.3.3 is known as the Golden Ratio.

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Writing Prompts

4.

Write a letter to a friend in which you use nested similar right triangles, such as in Exploration 5.3.2 to prove the Pythagorean Theorem.

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5.

Learn about the Golden Rectangle and the Golden Mean. How are these related to the Golden Triangle?

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