Section5.3The 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.
Figure 5.3.2 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.
Recall that the scale factor is a ratio. It compares the lengths of corresponding sides of two similar shapes. Do the given ratios above compare corresponding sides?
Remember that the scale factor is a ratio that compares corresponding sides of two similar shapes. Again, determine if the given ratios above compare corresponding sides.
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.
The corresponding angles of similar triangles are congruent, meaning they have the same measure. Use the order of the letters in the similarity statement \(\Delta FGH\sim\Delta IJK\) to identify the angle in the second triangle that corresponds to \(m\angle G\text{.}\)
To find the angle that corresponds to \(\angle H\text{,}\) once again use the similarity statement. See which angle lines up with \(\angle H\) and is congruent.
The scale factor is the ratio of corresponding sides. Use the similarity statement, \(\Delta FGH\sim\Delta IJK\text{,}\) to identify which sides match up, then find their lengths. The ratio of their lengths represents the scale factor.
To find the length of side \(IJ\text{,}\) once again use the similarity statement. Find its corresponding side in \(\Delta FGH\) and apply the scale factor.
To find the measure of each of these angles, use the interactive GeoGebra tool (Figure 5.3.5). On the options bar at the top, click on the sixth button from the left, indicated by a small red angle symbol. Then proceed by clicking on the three vertices of the angle. Remember that the second point clicked should be the vertex of the angle. If the measurement given is the larger, outside angle, simply subtract it from 360 to find the interior angle.
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.
Think of this line segment from point \(C\) to \(\overline{AB}\) as the altitude. How does it divide the larger triangle into two smaller right triangles?
To prove that these two triangles are similar, show two of their angles are congruent using the Angle-Angle (AA) rule. Recall the definition for constant of proportionality.
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)?
Look at the line segment that includes both \(AD\) and \(DB\text{.}\) What is the name of that entire line segment? By which lowercase letter is it represented?
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 \(c^2?\) Show all work.
Recall the expressions for \(a^2\) and \(b^2\) found previously. Substitute those expressions into the equation, then add them up. Observe and take note of the result.
The right triangle in Exploration 5.3.4 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.
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.
The Desmos Scientific Calculator at desmos.com 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.
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.
Bisecting \(\angle CAB\) splits \(\Delta ABC\) into two triangles. Use GeoGebra’s Polygon tool to highlight the subtriangle that is similar to \(\Delta ABC\text{.}\)
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.
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 has not already been done and record at least four digits after the decimal point.
Now use the angle bisector tool to bisect angle \(\angle ABD\) in Figure 5.3.7. Let \(E\) be the point where this angle bisector intersects segment \(\overline{AD}\text{.}\) Draw segment \(\overline{BE}\) and hide ray \(\overrightarrow{BE}\text{.}\)
The new bisector at angle \(\Delta ABD\) creates another isosceles triangle \(\Delta ABE\text{.}\) Which other triangle in the sketch is similar to \(\Delta ABE?\)
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?
Again, review the definition of constant of proportionality. Experiment with moving the vertices \(A\) and \(B\text{.}\) Use the ratio of corresponding sides to see if changing side lengths has any impact on the constant of proportionality.
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?
The area tool is under the angle menu. What is the ratio of the areas of similar triangles \(\Delta ABC\) and \(\Delta BDA?\) How is this number related to the ratio of the corresponding sides?
The isosceles triangle \(\Delta ABC\) explored in Exploration 5.3.6 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 2
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.
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 \(\angle EAB\text{.}\)
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 \(\overleftrightarrow{CD}\text{.}\)
Notice how extending the sides creates new points of intersection outside the pentagon. Consider what shape is created when these points are connected in order.
In Figure 5.3.10, \(D\text{,}\)\(E\text{,}\) 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, \(m\angle ACB=30^{\circ}\) and \(m\angle DEF=108^{\circ}\text{.}\) Answer the following:
Based on the Triangle Angle Sum Theorem, \(m\angle BAC=108\) degrees. Use triangle similarity to find \(m\angle EFC\) and \(m\angle BDE\text{.}\) Hint...Both \(\angle EFC\) and \(\angle BDE\) correspond to \(m\angle BAC\text{.}\)
Use angle similarity and match corresponding angles. Which angle in the larger triangle \(\Delta ABC\) corresponds with \(\angle FDE?\) What about \(\angle AFD?\) Use these relationships to find missing measures.
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 \(m\angle EFD\text{.}\)
Imagine \(\Delta ABC\) as the big triangle. If lines are drawn inside it that are parallel to its sides, smaller triangles are made. These smaller triangles will have the same angles as the big one, \(\Delta ABC\text{.}\)
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.
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.
