A HISTORICAL INTRODUCTION TO ELEMENTARY GEOMETRY

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1 i MATH 119 A HISTORICAL INTRODUCTION TO ELEMENTARY GEOMETRY Geometry is an word derived from ancient Greek meaning earth measure ( ge = earth or land ) + ( metria = measure ). Euclid wrote the Elements of geometry between 330 and 320 B.C. It was a compilation of the major theorems on plane and solid geometry presented in an axiomatic style. Near the beginning of the first of the thirteen books of the Elements, Euclid enumerated five fundamental assumptions called postulates or axioms which he used to prove many related propositions or theorems on the geometry of two and three dimensions. POSTULATE 1. Any two points can be joined by a straight line. POSTULATE 2. Any straight line segment can be extended indefinitely in a straight line. POSTULATE 3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center. POSTULATE 4. All right angles are congruent. POSTULATE 5. (Parallel postulate) If two lines intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. The circle described in postulate 3 is tacitly unique. Postulates 3 and 5 hold only for plane geometry; in three dimensions, postulate 3 defines a sphere. Postulate 5 leads to the same geometry as the following statement, known as Playfair's axiom, which also holds only in the plane: Through a point not on a given straight line, one and only one line can be drawn that never meets the given line.

2 ii Abraham Lincoln, the 16 th president of the United States, once told a friend: In fact, Lincoln tried legal cases in the courtroom and structured his most famous speeches as president based on the same six elements of a proposition that Euclid used when composing solutions to geometrical problems. After the time of Euclid, Archimedes and the other great geometers had passed in ancient Greece, the Hindu-Arabic decimal based number system along with the subjects of arithmetic, algebra and modern trigonometry were developed and promoted throughout India, China and Western Asia during the first millennia A.D.

3 iii It was only quite recently, in 1637, that René Descartes published his most innovative text on geometry, Discourse on Method: The Geometry, which introduced the subject of Analytic geometry to mathematics. Analytic or Cartesian geometry utilizes both the algebra of the real numbers and the theory of Euclidean geometry to describe and investigate the properties of geometrical figures such as lines, polygons, curves, surfaces and polyhedrons. Although René Descartes is credited with inventing Analytic Geometry with his 1637 publication, it can be argued that the subject actually originated around 240B.C. when Apollonius of Perga wrote the Conics, Books 1-5.

4 iv In order to understand how Apollonius was essentially able to describe a parabolic conic section in terms of the relative horizontal and vertical displacements of its points away from the vertex point, we must first review some fundamental theorems of Euclidean geometry. Theorem 1. The sum of the interior angles in any triangle is equal to two right angles, i.e., a straight angle or 180 o. Proof: Let ABC be any triangle and thru the vertex A draw line DAE parallel to side BC and extend side BA to segment BAB. Then since BCsDE it follows that: (1) :B AE = :ABC If this were not so, then one of the two angles must be greater than the other. Suppose that :B AE is greater than :ABC, then :ABC + :EAB < :B AE + :EAB = two right angles. So the lines DE and BC extended indefinitely would meet due to Euclid s 5 th Postulate. This is a contradiction. So :B AE is not greater than :ABC. In a similar way, we may establish that :ABC is not greater than :B AE. Therefore, :B AE=:ABC. and because :B AE + :DAB = a straight angle = :BAD + :DAB it follows that (2) :B AE = :BAD. Hence, from (1) and (2) we see that (3) :ABC =:BAD. Arguing in the same way, it follows that the pair of alternate interior angles :EAC and :ACB are also equal: (4) :EAC =:ACB. Now the angles :BAD, :BAC and :EAC combine to form a straight angle: (5) :BAD + :BAC +:EAC = a straight angle (180 o in measure). and by substituting from (3) and (4) into (5) we find that (6) :ABC + :BAC +:ACB = a straight angle.

5 v Therefore, the sum of the three interior angles of a triangle is always equal to a straight angle or 180 o. The above proof of theorem 1 is essentially the one that Eudemus had attributed to the Pythagoreans. Theorem 2. In any isosceles triangle the interior angles opposite the equal sides are equal. Proof: In ABC let AB = AC. It suffices to verify that the angles :ABC and :ACB are corresponding parts of congruent triangles. To do this, we may use a straight edge and compass to mark point P : the midpoint of side BC. Now join A to P. Since AB = AC, BP = PC and AP = AP it follows that ABP y ACP by SSS*. r :ABC = :ACB. *The Elements: Proposition 1.8 (SSS Criterion for Congruent Triangles) If two triangles have two sides equal to two sides respectively, and also have the base equal to the base, then they will also have equal the angles encompassed by the equal straight lines. (Euclid proved Proposition 1.8 using superposition, i.e. the technique of placing one triangle on top of the other to see if they are congruent.) Theorem 3. Every triangle inscribed in a semicircle is a right triangle. Proof: Let ABC be inscribed in a semicircle having diameter AC and center O as in the figure below.

6 vi Observe that AOB and COB are both isosceles since OA = OB = OC = r h radius of semi-circle. Consequently, and :OAB = :ABO h α :OBC = :BCO h β being angles opposite equal sides in isosceles triangles by Theorem 2. Moreover, :CAB + :ABC + :BCA = 2 right angles as the sum of the angles in ABC is equal to 2 right angles due to Theorem 1. That is, :CAB + (:ABO + :OBC ) + :BCA = 2 right angles. So we have that :ABO + (:ABO + :OBC ) + :BCO = 2 right angles by substitution of equal angles for equal angles. Equivalently, :ABO + (:ABO + :OBC ) + :OBC = 2 right angles. Regrouping the angles in the sum on the left side, we have (:ABO + :OBC ) + (:ABO + :OBC ) = 2 right angles. 0 :ABC + :ABC = 2 right angles. Therefore, :ABC is a right angle.

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8 AB In this figure, assume that point A is directly above the midpoint of segment BC. Thus, we have generated a right circular cone. We also require that this cutting plane intersects the base of the cone in a line that is perpendicular to diameter BC.

9 3 First, we establish that the parabola has a line of symmetry. Due to the orientation of the cutting plane, observe that PR t LL. Moreover, OL = OL, both being radii of the circle with center O. By Theorem 2 on isosceles triangles, we have that (a) :OLM = :OL M Now in OLM and OL M we see that (b) :LOM + :OLM = right angle = :L OM +:OL M From (a) and (b) it follows that :LOM = :L OM by subtracting equals from equals. Therefore, LOM y L OM by SAS*. Thus, ML = ML being corresponding sides of congruent triangles. It now follows immediately that line EM is the line of symmetry for the parabola. *The Elements: Proposition 1.4 (SAS Criterion for Congruent Triangles) If two triangles have: two sides equal to two sides respectively; the angles contained by the equal straight lines equal they will also have: their third sides equal; the remaining two angles equal to their respective remaining angles, namely, those which the equal sides subtend.

10 Now we shall prove that the segment length EM is always proportional to the square of the segment length ML. 4

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