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Chapter 2
EUCLIDEAN PARALLEL POSTULATE
2.1 INTRODUCTION. There is a well-developed theory for a geometry based solely on the five Common Notions and first four Postulates of Euclid. In other words, there is a geometry in which neither the Fifth Postulate nor any of its alternatives is taken as an axiom. This geometry is called Absolute Geometry , and an account of it can be found in several textbooks - in Coxeter’s book “Introduction to Geometry”, for instance, - or in many college textbooks where the focus is on developing geometry within an axiomatic system. Because nothing is assumed about the existence or multiplicity of parallel lines, however, Absolute Geometry is not very interesting or rich. A geometry becomes a lot more interesting when some Parallel Postulate is added as an axiom! In this chapter we shall add the Euclidean Parallel Postulate to the five Common Notions and first four Postulates of Euclid and so build on the geometry of the Euclidean plane taught in high school. It is more instructive to begin with an axiom different from the Fifth Postulate.
2.1.1 Playfair’s Axiom. Through a given point, not on a given line, exactly one line can be drawn parallel to the given line.
Playfair’s Axiom is equivalent to the Fifth Postulate in the sense that it can be deduced from Euclid’s five postulates and common notions, while, conversely, the Fifth Postulate can deduced from Playfair’s Axiom together with the common notions and first four postulates.
2.1.2 Theorem. Euclid’s five Postulates and common notions imply Playfair’s Axiom. Proof. First it has to be shown that if P is a given point not on a given line l , then there is at least one line through P that is parallel to l. By Euclid's Proposition I 12, it is possible to draw a line t through P perpendicular to l. In the figure below let D be the intersection of l with t.
t
n
m
l
E
A
F
B
P
D
By Euclid's Proposition I 11, we can construct a line m through P perpendicular to t. Thus by construction t is a transversal to l and m such that the interior angles on the same side at P and D are both right angles. Thus m is parallel to l because the sum of the interior angles is 180 °. (Note: Although we used the Fifth Postulate in the last statement of this proof, we could have used instead Euclid's Propositions I 27 and I 28. Since Euclid was able to prove the first 28 propositions without using his Fifth Postulate, it follows that the existence of at least one line through P that is parallel to l , can be deduced from the first four postulates. For a complete list of Euclid's propositions, see “College Geometry” by H. Eves, Appendix B.) To complete the proof of 2.1.2, we have to show that m is the only line through P that is parallel to l. So let n be a line through P with m ≠ n and let E ≠ P be a point on n. Since m ≠ n , ∠ EPD cannot be a right angle. If m∠ EPD < 90°, as shown in the drawing, then m∠ EPD + m ∠ PDA is less than 180°. Hence by Euclid’s fifth postulate, the line n must intersect l on the same side of transversal t as E, and so n is not parallel to l. If m∠ EPD > 90°, then a similar argument shows that n and l must intersect on the side of l opposite E. Thus, m is the one and only line through P that is parallel to l. QED
are piecewise linear figures as is the example of nested polygons below.
This example is a particularly interesting one because we can think of it as a figure containing a ‘hole’. But is it clear what is meant by the interior angles of such figures? For such a polygon as the following:
we obviously mean the angles indicated. But what about a piecewise linear figure containing holes? For the example above of nested polygons, we shall mean the angles indicated below
This makes sense because we are really thinking of the two polygons as enclosing a region so that interior angle then refers to the angle lying between two adjacent sides and inside the enclosed region. What this suggests is that for piecewise linear figures we will also need to specify what is meant by its interior. The likely formula for the sum of the interior angles of piecewise linear figures can be obtained from Theorem 2.2.1 in conjunction with Sketchpad. In the case of polygons this was probably done in high school. For instance, the sum of the angles of any quadrilateral, i. e ., any four-sided figure, is 360°. To see this draw any diagonal of the quadrilateral thereby dividing the quadrilateral into two triangles. The sum of the angles of the quadrilateral is the sum of the angles of each of the two triangles and thus totals 360°. If the polygon has n sides, then it can be divided into n -2 triangles and the sum of the angles of the polygon is equal to the sum of the angles of the n -2 triangles. This proves the following result.
2.2.2 Theorem. The sum of the interior angles of an n -sided polygon, n ≥ 3 , is (n − 2) ⋅ 180 °.
2.2.2a Demonstration. We can use a similar method to determine the sum of the angles of the more complicated piecewise linear figures. One such figure is a polygon having “holes”, that is, a polygon having other non-overlapping polygons (the holes) contained totally within its interior. Open a new sketch and draw a figure such as
2.2.3 Theorem. When an n -sided piecewise linear figure consists of a polygon with one polygonal hole inside it then the sum of its interior angles is ________________________. Note: Here, n equals the number of sides of the outer polygon plus the number of sides of the polygonal hole. End of Demonstration 2.2.2a.
Try to prove Theorem 2.2.3 algebraically using Theorem 2.2.2. The case of a polygon containing h polygonal holes is discussed in Exercise 2.5.1.
2.3 SIMILARITY AND THE PYTHAGOREAN THEOREM
Of the many important applications of similarity, there are two that we shall need on many occasions in the future. The first is perhaps the best known of all results in Euclidean plane geometry, namely Pythagoras’ theorem. This is frequently stated in purely algebraic terms as
a^2 + b^2 = c^2 , whereas in more geometrically descriptive terms it can be interpreted as saying that, in area, the square built upon the hypotenuse of a right-angled triangle is equal to the sum of the squares built upon the other two sides. There are many proofs of Pythagoras’ theorem, some synthetic, some algebraic, and some visual as well as many combinations of these. Here you will discover an algebraic/synthetic proof based on the notion of similarity. Applications of Pythagoras’ theorem and of its isosceles triangle version to decorative tilings of the plane will be made later in this chapter.
2.3.4 Theorem. (The Pythagorean Theorem) In any triangle containing a right angle, the square of the length of the side opposite to the right angle is equal to the sum of the squares of the lengths of the sides containing the right angle. In other words, if the length of the
hypotenuse is c and the lengths of the other two sides are a and b , then a^2 + b^2 = c^2.
Proof : Let ∆ ABC be a right-angled triangle with right angle at C , and let CD be the
perpendicular from C to the hypotenuse AB as shown in the diagram below.
c
b (^) a
A B
C
D
- Show∆ CAB is similar to ∆ DAC.
- Show∆ CAB is similar to ∆ DCB.
- Now let BD have length x , so that AD has length c − x. By similar triangles, x a = a c and c^ −^ x b
- Now eliminate x from the two equations to show a^2 + b^2 = c^2.
There is an important converse to the Pythagorean theorem that is often used.
2.3.5 Theorem. (Pythagorean Converse) Let ∆ ABC be a triangle such that a^2 + b^2 = c^2. Then ∆ ABC is right-angled with ∠ ACB a right angle.
2.3.5a Demonstration (Pythagorean Theorem with Areas) You may be familiar with the geometric interpretation of Pythagoras’ theorem. If we build squares on each side of ∆ ABC then Pythagoras’ theorem relates the area of the squares.
- Open a new sketch and draw a right-angled triangle ∆ ABC. Using the ‘Square By Edge’ tool construct an outward square on each edge of the triangle having the same edge length as the side of the triangle on which it is drawn.
- Measure the areas of these 3 squares: to do this select the vertices of a square and then construct its interior using “Construct Polygon Interior” tool. Now compute the area of each of these squares and then use the calculator to check that Pythagoras’ theorem is valid for the right-angled triangle you have drawn. End of Demonstration 2.3.5a.
This suggests a problem for further study because the squares on the three sides can be thought of as similar copies of the same piecewise linear figure with the lengths of the sides determining the edge length of each copy. So what does Pythagoras’ theorem become when the
2.3.7 Demonstration. Reformulate the result corresponding to Theorem 2.3.6 when the regular n -gons constructed on each side of a right-angled triangle are replaced by similar triangles.
This demonstration presents an opportunity to explain another feature of Custom Tools called Auto-Matching. We will be using this feature in Chapter 3 when we use Sketchpad to explore the Poincaré Disk model of the hyperbolic plane. In this problem we can construct the first isosceles triangle and then we would like to construct two other similar copies of the original one. Here we will construct a “similar triangle script” based on the AA criteria for similarity.
Tool Composition using Auto-Matching
- Open a new sketch and construct ∆ ABC with the vertices labeled.
- Next construct the line (not a segment) DE.
- Select the vertices B - A-C in order and choose "Mark Angle B - A -C" from the Transform Menu. Click the mouse to deselect those points and then select the point D. Choose “Mark Center D ” from the Transform Menu. Deselect the point and then select the line DE. Choose “Rotate…” from the Transform Menu and then rotate by Angle B - A - C.
- Select the vertices A - B - C in order and choose “Mark Angle A - B - C ” from the Transform Menu. Click the mouse to deselect those points and then select the point E. Choose “Mark Center E ” from the Transform Menu. Deselect the point and then select the line DE. Choose “Rotate…” from the Transform Menu and rotate by Angle A - B - C.
- Construct the point of intersection between the two rotated lines and label it F. ∆ DEF is similar to ∆ ABC. Hide the three lines connecting the points D , E , and F and replace them with line segments.
- Now from the Custom Tools menu, choose Create New Tool and in the dialogue box, name your tool and check Show Script View. In the Script View, double click on the Given “Point A ” and a dialogue box will appear. Check the box labeled Automatically Match Sketch Object. Repeat the process for points B and C.
In the future, to use your tool, you need to have three points labeled A , B , and C already constructed in your sketch where you want to construct the similar triangle. Then you only
need to click on or construct the points corresponding to D and E each time you want to use the script. Your script will automatically match the points labeled A , B , and C in your sketch with those that it needs to run the script. Notice in the Script View that the objects which are automatically matched are now listed under “Assuming” rather than under “Given Objects”. If there are no objects in the sketch with labels that match those in the Assuming section, then Sketchpad will require you to match those objects manually, as if they were “Given Objects.”
- Now open a new sketch and construct a triangle with vertices labeled A, B, and C.
- In the same sketch, construct a right triangle. Use the “similar triangle” tool to build triangles similar to ∆ ABC on each side of the right triangle. For each similar triangle, select the three vertices and then in the Construct menu, choose “construct polygon interior”. Measure the areas of the similar triangles and see how they are related. End of Demonstration 2.3.7.
2.4 INSCRIBED ANGLE THEOREM: One of the most useful properties of a circle is related to an angle that is inscribed in the circle and the corresponding subtended arc. In the figure below, ∠ ABC is inscribed in the circle and Arc ADC is the subtended arc. We will say that ∠ AOC is a central angle of the circle because the vertex is located at the center O. The measure of Arc ADC is defined to be the angle measure of the central angle, ∠ AOC.
D
O (^) A
C
B
2.4.0 Demonstration. Investigate the relationship between an angle inscribed in a circle and the arc it intercepts (subtends) on the circle.
- Open a new script in Sketchpad and draw a circle, labeling the center of the circle by O.
End of Demonstration 2.4.0.
The result you have discovered in Corollary 2.4.2 is a very useful one, especially in constructions, since it gives another way of constructing right-angled triangles. Exercises 2.5. and 2.5.5 below are good illustrations of this. The Inscribed Angle Theorem can also be used to prove the following theorem, which is useful for proving more advanced theorems.
2.4.4 Theorem. A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. (A quadrilateral that is inscribed in a circle is called a cyclic quadrilateral. )
2.5 Exercises
Exercise 2.5.1. Consider a piecewise linear figure consisting of a polygon containing h holes (non-overlapping polygons in the interior of the outer polygon) has a total of n edges, where n includes both the interior and the exterior edges. Express the sum of the interior angles as a function of n and h. Prove your result is true.
Exercise 2.5.2. Prove that if a quadrilateral is cyclic, then the opposite angles of the quadrilateral are supplementary, i. e ., the sum of opposite angles is 180°. [ This will provide half of the proof of Theorem 2.4.4. ]
Exercise 2.5.3. Give a synthetic proof of the Inscribed Angle Theorem 2.4.1 using the properties of isosceles triangles in Theorem 1.4.6. Hint: there are three cases to consider: here ψ is the angle subtended by the arc and is the angle subtended at the center of the circle. The problem is to relate ψ to.
Case 1: The center of the circle lies on the subtended angle.
Case 2: The center of the circle lies within the interior of the inscribed circle.
Case 3: The center of the circle lies in the exterior of the inscribed angle.
End of Exercise 2.5.3.
For Exercises 2.5.4, 2.5.5, and 2.5.6, recall that any line tangent to a circle at a particular point must be perpendicular to the line connecting the center and that same point. For all three of these exercises, the Inscribed Angle Theorem is useful.
Exercise 2.5.4. Use the Inscribed Angle Theorem to devise a Sketchpad construction that will construct the tangents to a given circle from a given point P outside the circle. Carry out your construction. (Hint: Remember Corollary 2.4.2).
Exercise 2.5.5. In the following figure
Exercise 2.5.7. Prove that the vertices of a regular polygon always lie on a circumscribing circle. (Be careful! Don’t assume that your polygon has a center; you must prove that there is a point equidistant from all the vertices of the regular polygon.)
Exercise 2.5.8. Now suppose that the edge length of a regular n -gon is l and let R be the radius of the circumscribing circle for the n -gon. The Apothem of the n -gon is the perpendicular distance from the center of the circumscribing circle to a side of the n -gon.
A
l
R
The Apothem
(a) With this notation and terminology and using some trigonometry complete the following R = l _____________ , l = R ______________, Apothem = R __________. Use this to deduce
(b) area of n -gon =^1 2 n R^2 sin 2 n
,^ (c) perimeter of^ n -gon = 2 n R sin^ n
(d) Then use the well-known fact from calculus that
limsin 1 0
→
to derive the formulas for the area of a circle of radius R as well as the circumference of such a circle.
Exercise 2.5.9. Use Exercise 2.5.8 together with the usual version of Pythagoras’ theorem to give an algebraic proof of the generalized Pythagorean Theorem (Theorem 2.3.6).
Exercise 2.5.10 Prove the converse to the Pythagorean Theorem stated in Theorem 2.3.5.
2.6 RESULTS REVISITED. In this section we will see how the Inscribed Angle Theorem can be used to prove results involving the Simson Line, the Miquel Point, and the Euler Line. Recall that we discovered the Simson Line in Section 1.8 while exploring Pedal Triangles.
2.6.1 Theorem (Simson Line). If P lies on the circumcircle of ∆ ABC , then the perpendiculars from P to the three sides of the triangle intersect the sides in three collinear points. Proof. Use the notation in the figure below.
- Why do P , D , A , and E all lie on the same circle? Why do P , A , C , and B all lie on another circle? Why do P , D , B and F all lie on a third circle? Verify all three of these statements using Sketchpad.
A
B
C
P E
F
D
- In circle PDAE, m ∠ PDE ≅ m ∠ PAE = m ∠ PAC. Why?
- In circle PACB, m ∠ PAC ≅ m ∠ PBC = m ∠ PBF. Why?
- In circle PDBF, m ∠ PBF ≅ m ∠ PDF. Why?
Since m ∠ PDE ≅ m ∠ PDF , points D , E , and F must be collinear. QED
In Exercise 1.9.4, the Miquel Points of a triangle were constructed.
2.6.2 Theorem (Miquel Point) If three points are chosen, one on each side of a triangle, then the three circles determined by a vertex and the two points on the adjacent sides meet at a point called the Miquel Point. Proof. Refer to the notation in the figure below.
circumcenter. The line containing the centroid, orthocenter, and circumcenter of a triangle is called the Euler Line. Proof. In the diagram below, A ′is the midpoint of the side opposite to A and O, G, and H are the circumcenter, centroid, and orthocenter, respectively. Since A, G, and A ′are collinear, we can show that O, G, and H are also collinear, by showing that ∠ AGH ≅∠ A GO ′. To do this, it
suffices to show that ∆ AHG^^ ~^ ∆ A ′ OG. If we also show that the ratio of similiarity is 2:1, then
we will also prove that G trisects OH.
H
O G
A'
C
A
B
The proof that ∆ AHG ~ ∆ A OG ′ with ratio 2:1 proceeds as follows: Let I be the point where the
ray CO intersects the circumcircle of ∆ ABC. Then IB ⊥ CB (why?). It follows that ∆ BCI ~ ∆ A CO ′ with ratio 2:1 (why?) It is also true that AIBH is a parallelogram (why?) and
I
H G O
A'
C
A
B
hence AH = IB = 2( O A ′ ). Since G is the median, we know that AG = 2( G A ′). Thus we have two corresponding sides proportional. The included angles are congruent because they are
alternate interior angles formed by the parallel lines AH and O A ′ and the transversal A A ′.
(Why are AH and O A ′ parallel?) Thus, ∆ AHG^^ ~^ ∆ A ′ OG with ratio 2:1 by SAS. Of course, as we noted in Chapter 1, we must be careful not to rely too much on a picture when proving a theorem. Use Sketchpad to find examples of triangles for which our proof breaks down, i.e. triangles in which we can’t form the triangles ∆ AHG and ∆ A OG ′. What sorts of triangles arise? You should find two special cases. Finish the proof of Theorem 2.6. by proving the result for each of these cases (see Exercise 2.8.2 ).
2.7 THE NINE POINT CIRCLE. Another surprising triangle property is the so-called Nine- Point Circle, sometimes credited to K.W. Feuerbach (1822). Sketchpad is particularly well adapted to its study. The following Demonstration will lead you to its discovery.
2.7.0 Demonstration: Investigate the nine points on the Nine Point Circle.
The First set of Three points:
- In a new sketch construct ∆ ABC. Construct the midpoints of each of its sides; label these midpoints D , E , and F.
- Construct the circle that passes through D , E , and F. (You know how to do this!)
- This circle is called the Nine-Point Circle. Complete the statement: The nine-point circle passes through _________________________________.
The Second set of Three points: In general the nine-point circle will intersect ∆ ABC in three more points. If yours does not, drag one of the vertices around until the circle does intersect ∆ ABC in three other points. Label these points J , K , and L.
- Construct the line segment joining J and the vertex opposite J. Change the color of this segment to red. What is the relationship between the red segment and the side of the triangle containing J? What is an appropriate name for the red segment?
- Construct the corresponding segment joining K and the vertex opposite K and the segment joining L to the vertex opposite L. Color each segment red. What can you say about the three red segments?
- Place a point where the red segments meet; label this point M and complete the following statement: The nine-point circle also passes through ____________________________. .
