Download A HISTORICAL INTRODUCTION TO ELEMENTARY GEOMETRY

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MATH 119 – Fall 2012:
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.
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Abraham Lincoln, the 16th 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 arguments in support of his
geometrical assertions.
After the time of Euclid, Archimedes, and the other great geometers had
passed in ancient Europe, the Hindu-Arabic decimal based number system
along with the subjects of arithmetic, algebra and modern trigonometry
were developed and promoted throughout Asia during the first millennia
A.D.
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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.
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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.
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Theorem 1. The sum of the interior angles in any triangle is equal to
two right angles , i.e., a straight angle or 180o.
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Proof: Let UABC 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 5th 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 (180o in measure).
and by substituting from (3) and (4) into (5) we find that
(6)
:ABC + :BAC +:ACB = a straight angle .
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Therefore, the sum of the three interior angles of a triangle is always equal
to a straight angle or 180o .
The above proof of theorem 1 is essentially the one that Eudemus had attributed
to the Pythagoreans.
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Theorem 2. In any isosceles triangle the interior angles opposite the
equal sides are equal.
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Proof: In UABC 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
UABP y UACP by SSS*.
r
:ABC = :ACB .
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*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.)
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Theorem 3. Every triangle inscribed in a semicircle is a right triangle.
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Proof: Let UABC be inscribed in a semicircle having diameter AC and
center O as in the figure below.
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Observe that UAOB and UCOB are both isosceles since
OA = OB = OC = r h radius of semi-circle.
Consequently,
:OAB = :ABO h α
and
: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 UABC 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 ) + :OBC = 2 right angles
by substitution of equal angles for equal angles.
Regrouping the angles in the sum on the left side, we have
(:ABO + :OBC ) + (:ABO + :OBC ) = 2 right angles.
0 :ABC
+
:ABC
Therefore, :ABC is a right angle .
= 2 right angles.
AB
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