High School Geometry Archives - Page 20 of 23

The Exterior Angle Theorem

December 26, 2009 GB High School Calculus, High School Geometry, High School Mathematics

In the angle sum of a triangle post, we have discussed that the angle sum of a triangle is $180$ degrees. In the angle sum of a polygon post, we also have discussed that and that the angle sum of a polygon with $n$ sides is $180(n-2)$ . For example, a pentagon has $5$ sides, so the sum of its interior angle is $180(5-2) = 180(3) = 540$ degrees.

Figure 1 – The interior and exterior angles a triangle and a quadrilateral.

The angle sums that we have discussed in both blogs refer to the sum of the interior angles. What about the exterior angles?

The exterior angle is formed when we extend a side of a polygon. In the triangle above, $\alpha$ is an exterior angle. The sum of the interior angle and the exterior angle adjacent to it is always $180$ degrees (Why?). Angles whose sum is $180$ degrees are called supplementary angles. If two angles are supplementary, we call them a linear pair. For example, angles $\alpha$ and $a_1$ are supplementary angles and at the same time a linear pair, so $\alpha + a_1 = 180$ degrees. Now this means, that $\alpha = 180 - a_1$ .Therefore, if we want to compute the measure of an exterior angle adjacent to an interior angle, we can always subtract the measure of the interior angle from $180$ as shown in Figure 1.

Observe the computation in the two diagrams. If we let $S_t$ be the angle sum of the exterior angles of a triangle, then $S_t = (180 - a_1) + (180 - a_2) + (180 - a_3) = 540$ . Rearranging the terms, we have $S_t = 540 - (a_1 + a_2 + a_3)$ . But $a_1 + a_2 + a_3$ is the sum of the interior angles of a triangle which is $180$ degrees, so $540 - (a_1 + a_2 + a_3) = 540 - 180 = 360$ degrees.

Now, try calculating for the sum of the exterior angles of the quadrilateral above. What is your answer?

To verify our hunch, we will try to compute for the sum of the exterior angles of a pentagon.

Let $S_p$ be the sum of the exterior angles of the pentagon in Figure 2. Then

$S_p =(180 - c_1)+ (180 - c_2) + (180 - c_3) +(180 - c_4) +(180 - c_5)$ . Simplifying, we have $S_p = 900 - (c_1 + c_2 + c_3 + c_4 + c_5)$ . But according to the angle sum theorem for polygons, $c_1 + c_2 + c_3 + c_4 + c_5 = 540$ . Therefore, $900 - (c_1 + c_2 + c_3 + c_4 + c_5) = 900 - 540 = 360$ degrees.

We have three polygons – triangle, quadrilateral, pentagon – whose angle sums of exterior angles are always $360$ degrees. Now, is this true for all polygons? Try to compute polygons up to $10$ sides and see if the sum is $360$ degrees.

Delving Deeper

We know that in a polygon, the number of exterior angles is equal to the number of interior angles. Furthermore, we know that the angle sum of an interior angle and the exterior angle adjacent to each is always latex 180 degrees. If we have a polygon with 5 sides, then

interior angle sum + exterior angle sum = 180(5)

In general, this means that in a polygon with n sides

interior angle sum* + exterior angle sum = 180n

But the interior angle sum = 180(n – 2). So, substituting in the preceding equation, we have

180(n – 2) + exterior angle sum = 180n

which means that the exterior angle sum = 180n – 180(n – 2) = 360 degrees. More formal proofs using these arguments are shown below.

Theorem: The sum of the measure of the exterior angles of a polygon with n sides is 360 degrees.

Proof 1:

Let $a_1, a_2, \cdots a_n$ be measures of the interior angles of a polygon with n sides. Let $b_1, b_2, \cdots b_n$ be measures of the exterior angles of the same polygon where all angle names with the same subscripts are adjacent angles from $a_1$ and $b_1$ all the way up through $a_n$ and $b_n$ . We know that adjacent interior and exterior angles are supplementary angles, so this implies that their measures add up to 180 degrees. Hence,

(a₁ + b₁) + (a₂ + b₂) + … + (a_n + b_n) = 180 + 180 + … +180 (n of them) = 180n

Regrouping the terms of the preceding equation, we have

(a₁ + a₂ + … + a_n) + (b₁ + b₂ + … + b_n) = 180n

But the sum of the interior angles is a₁ + a₂ + … + a_n = 180(n – 2)

So,

180(n – 2) + (b₁ + b₂ + … + b_n) = 180n

b₁ + b₂ + … + b_n = 180n – 180(n – 2) = 360

Therefore, the sum of the exterior angles of any polygon is equal to 360 degrees.

Proof 2:

Let a₁, a₂, …, a_n be measures of the interior angles of the polygon with n sides. Since each adjacent interior and exterior angle is a linear pair, it follows that the measure of the exterior angles adjacent to them respectively are 180 – a₁, 180 – a₂, …, 180 – a_n.

If we let S, be the sum of the measure of the exterior angles, we have

S = (180 – a₁) + (180 – a₂) + (180 – a₃) + … + (180 – a_n)

= (180 + 180 + 180 + … +180 (n of them)) – a– a₂ – a₃– … – a_n

S = 180n – (a₁+ a₂ + a₃+ … + a_n)

But a₁+ a₂ + a₃+ … + a_nis the sum of the measures of the interior angles of a polygon with n sides which equals

180(n – 2), so, S = 180n – 180(n – 2) = 360, which is want we want to show.

Therefore, the sum of the exterior angles of any polygon is equal to 360 degrees.

Leave a comment angle sum, angle sum of polygons, exterior angle math, exterior angle theorem, exterior angles convex polygon, exterior angles definition, interior angles, linear pair, polygon exterior angle sum theorem, sum of exterior angles of a polygon, sum of interior angles, triangle angle sum

Parallel Lines and Transversals

December 17, 2009 GB High School Geometry, Questions and Quandaries

Given a triangle and a line segment, let us place the triangle such that one of its side and the line segment coincide. Next, we replicate the triangle and slide to the right hand side such that two of their vertices meet as shown in Figure 1.

Figure 1 – Translated triangles have parallel corresponing sides.

The angles with the same color are basically the same angle, so we will call them corresponding angles. The sides which are parallel are also the same side, so we will call them corresponding sides.

What can you say about their corresponding angles? Their corresponding sides?

Figure 2 – The sum of the angles in point T is 180 degrees.

Next, we replicate one of the triangles, rotate it and place it between the two triangles as shown in Figure 2. Click here to rotate the triangles using GeoGebra. In the figure, we can observe and deduce the following:

The measure of the three adjacent angles in T is half the circle and is therefore 180 degrees. This implies that the sum of the interior angles of triangle PQT is also 180 degrees, since they are the sum of the three adjacent angles in T.
PQRT and QRST are parallelograms. (Explain why the opposite sides are parallel).
The opposite angles of parallelogram are congruent, and their opposite sides are also congruent. (Explain why).
We also observe that the sum of the consecutive angles of a parallelogram is 180 degrees. For example, the sum of angle PQR and angle QRT is 180 degrees.
The interior angles of a parallelogram consists of 2 green angles, 2 yellow angles and 2 red angles, which means that the angle sum of the interior angles of a parallelogram is twice 180 or 360 degrees.
PQRS is a trapezoid and the sum of its interior angles is also 360 degrees. (Explain why QR and PS are parallel).

If we extend QR and QT, then we will two parallel line segments and a transversal. In Figure 3, OR and PS are parallel lines a NU is a transversal. From Figure 1 and Figure 2, the yellow angles are corresponding angles, so RQT and PTQ are congruent. We can also see that the blue angle and yellow angle are supplementary (the sum of their measure is 180 degrees).

Figure 3 – NU is a transversal to the parallel lines OR and PS.

With the knowledge that the blue angle and the yellow angle are supplementary, we can deduce the following:

Interior angles on the same side of a transversal of parallel lines (RQT and STQ) are supplementary.
Angles OQN and OQT are supplementary so NQT must have the same measure as that of the yellow angle.
Angles RQT and RQN are supplementary so RQN must have have the same measure as that of the blue angle.
From 2 and 3, it follows that PTU is a blue angle and STU is a yellow angle.

With all the observations and deductions above, we came up with the following diagram. Note that angles symbols with the same color are congruent angles.

Figure 4 – Angle signs with the same color are congruent.

Pairs of these angles have special names for distinction purposes. Most of the names are descriptive, so it is easy to determine other pairs. Some pairs of angles and their names are listed below.

1.) TQR and QTS are same side interior angles.

2.) NQR and STU are same side exterior angles.

3.) RQT and PTQ are alternate interior angles.

4.) STU and NQO are alternate exterior angles.

5.) NQO and QTP are corresponding angles.

Parallel lines and its relationship with its interval was mentioned in Euclid’s book The Elements. It is called the Fifth postulate or the Parallel Postulate. The parallel postulate states that

If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

All the relationships we have deduced above can be derived from this single statement.

I have also written why the angle sum of the interior angles of a triangle is 180 degrees. C

Leave a comment angle sum, math parallel lines, parallel lines congruent, parallel lines postulate, parallel postulate geometry, triangle angle sum

Geometer’s Sketchpad Tutorial 1: Constructing an Equilateral Triangle

December 14, 2009 GB Geometer's Sketchpad, High School Geometry, Software Tutorials

In this tutorial, we will use the Geometer’s Sketchpad to mimic the compass and straightedge construction tool we use in elementary geometry. Geometer’s Sketchpad is a proprietary interactive geometry software developed by Key Curriculum Press.

The idea in constructing an equilateral triangle is to use the centers of two circles and one of their intersections as vertices of the triangle (see Figure 6). First, we construct a circle with center A passing through B, then construct a circle with center B and passing through point A, and then determine their intersections. Next, we hide all the unnecessary objects and connect the three points with segments. The final output of the tutorial is shown in Figure 1.

Figure 1 – The appearance of the expected of this tutorial.

In constructing our triangle, we will learn the following:

select, deselect and move objects
construct objects such as circles and segments
display the label of objects
hide objects
display the measure of segments
construct points of intersection of two objects

Construction Steps

1.) Click the circle tool, determine the center by clicking a location on the drawing pad, move the mouse to determine the radius, and click another location to draw the circle. If you made a mistake, just click the Edit menu and choose Undo from the list.

Figure 2 – The toolbar contains the different tools of Sketchpad used for drawing.

2.) Click the arrow selection tool and try to drag the two points. What do you observe?

3.) We now show the label of the two points. Click the two points, then click the Display menu from the menu bar, then click Show labels from the drop-down menu. Notice that Geometer’s Sketchpad names the objects in alphabetical order, the center of the circle is named A and the point on the circle is named B.

Figure 3 – The Display drop-down menu.

4.) To construct a circle with center point B and passing through A, click the compass tool, click point B and click point A. Your drawing should look like the one shown in Figure 4.

Figure 4 – Circe with center A and passing through B and circle with center B passing through A.

5.) Next, we construct the intersections of two circles. Click the selection arrow tool, click then select the two circles. (Be sure that only the two circles are selected. To deselect an object, click anywhere on the drawing pad).

Figure 5 – The two circles are selected.

6.) To construct the intersection, click Construct from the menu bar, then Intersections from the drop-down menu bar.

7.) Display the name of the two intersections by clicking both of them and clicking Show labels from the Display menu.

8.) Next, we hide the two circles and point D leaving only points A, B and C on the drawing area. To do this click the selection arrow tool, click the two circles and point D.

Figure 6 – The two circles and point D are selected.

9.) To hide the selected objects, click the Display menu and choose Hide Objects. Notice that the only left objects on the drawing area are points A, B and C.

10.) Select segment tool (be sure that you choose the segment tool, not the ray tool or line tool) from the straightedge tool, then connect the three points to form triangle ABC. Using the selection arrow tool, move the points on the vertices of the triangle.

What do you observe?
What kind of triangle is triangle ABC? Why do you say so?

11.) Let us try to see if your conjecture about triangle ABC is true. We will display the length of the segments. To do this, select one of the sides of the triangle, click Measure from the menu bar and choose Length from the drop-down menu.

Measure the other two sides. What do you observe?
Move the vertices of the triangle. Is your observation still true?
Explain why triangle ABC is always an equilateral triangle.

12.) If you want to save your file, click the File menu from the menu bar and click Save from the drop-down menu.

13.) Type the file name of your first sketch, choose a location to save, then click Save.

I also created a similar tutorial on creating an equilateral triangle but using GeoGebra. If you are interested, click here.

Congratulations, you just finished your first construction. Click here to go to Geometer’s Sketchpad Tutorial 2.

3 comments geometer's sketchpad constructions, geometer's sketchpad demo, geometer's sketchpad help, geometer's sketchpad learning guide, geometer's sketchpad lesson plans, geometer's sketchpad tutorial

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