Mar 12, 2014 - Varignon's theorem on the other hand is similarly upheld when two of a ... theoretical definitions of line (as without width) and point (as without radius), ... parallel postulate in Euclidean geometry led to the discovery of ... triangles formed as a result of intersecting sides 1 and 3 which is ... m RS = 4.79 cm.
DEFINITIONAL CONFLICTS BETWEEN EUCLIDEAN GEOMETRY AND DYNAMIC GEOMETRY ENVIRONMENTS: VARIGNON THEOREM AS AN EXAMPLE Mdutshekelwa Ndlovu Stellenbosch University Centre for Pedagogy, Stellenbosch University (SOUTH AFRICA)
Abstract The purpose of this paper is to illustrate some of the obstacles or theoretical-computational conflicts that the integration of technology in the geometry classrooms can potentially bring. The question is whether we have to accept Euclidean definitions as immutable or we have to revisit them. The traditional school geometry quadrilateral, for example, is understood to be convex and many school syllabi present examples of these without counter-examples of alternative interpretations of the definitions. Monaghan stresses that there is a connection between the perception and description of a given geometrical figure. By allowing the user to manipulate, experiment and conjecture with shapes through the dragging and animation capabilities Dynamic Geometry Environments or DGEs have the potential to influence the perception of geometrical figures in new ways that challenge the legitimacy of some of the traditionally accepted geometrical definitions. For example, a quadrilateral with a reflex angle (concave quad) perceptually appears to be a triangle to many learners. However learners exposed to DGEs can readily perceive the shape as a quadrilateral they have previously constructed. Similarly, a crossed quadrilateral may be perceived not to be a quadrilateral at all by many learners until its vertices are dragged experimentally and discovered to fulfill all the properties of its convex counterpart. Varignon’s theorem on the other hand is similarly upheld when two of a quadrilateral’s sides are dragged to be co-linear surprisingly permitting us to re-define a triangle as a quad with two collinear vertices in direct contravention of this forbidden Euclidean condition. Keywords: dynamic geometry environments, technology, geometry, quadrilateral, varignon’s theorem.
1
INTRODUCTION
The introduction of information and communication technologies (ICTs) is encouraged by the mathematics education community as it has the potential to enhance conceptual understanding of mathematical concepts. Dynamic geometry environments or DGEs are one such ICT or digital technology tool that can be manipulated interactively by learners to illustrate and even experiment with mathematical ideas in real time in the classroom. The dynamic nature of mathematics software makes it possible for learners to construct and dynamically explore geometric figures satisfying particular properties in keeping with the high school curriculum. However, in the process of such manipulations by means of dragging or even animating surprises often occur and engender cognitive conflict in learners and educators alike resulting in breaches of some longstanding definitions. The Euclidian or theoretical definitions of line (as without width) and point (as without radius), for example, are frequently violated, as even in paper and pencil situations, for the sake of visualisation. The purpose of this paper is to illustrate some of the cognitive obstacles or theoretical-computational conflicts (e.g. Giraldo et al [1]) brought on by the introduction of technology in the context of geometry. The question is whether we have to accept Euclidean definitions as immutable from a formalist/absolutist accounts of mathematical knowledge as certain, superhuman and incorrigible (e.g. Ernest [2]). Or can we from fallibilist, humanist and social constructivist accounts of mathematical knowledge as tentative and humanly created (e.g. [2]), rather revisit and modify some of them, extend and relax some of their longstanding constraints/conditions for validity. That the negation of the parallel postulate in Euclidean geometry led to the discovery of equally valid non-Euclidian geometries on alternative surfaces (e.g. spherical and hyperbolic) strengthens the case for re-examination and consequently reaffirms the tentative nature of scientific and mathematical knowledge emphasized by the fallibilists. The advent of digital technologies also brings with it an experimental science flavour to the nature of mathematical and geometrical knowledge.
Proceedings of INTED2014 Conference 10th-12th March 2014, Valencia, Spain
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ISBN: 978-84-616-8412-0
2 2.1
THE CASE OF THE QUADRILATERAL The convex and concave quadrilaterals
In traditional school mathematics/geometry the quadrilateral, defined as any plane shape or polygon with four sides, or vertices, is often presented as a convex shape and many school syllabi and texts present convex examples to the exclusion of counter-examples illustrating alternative meanings and interpretations of the definitions. Monaghan [3] attests that there is a connection between the perception and description of a given geometrical figure. By allowing the user to manipulate, experiment and conjecture with shapes in real-time DGEs (such as Sketchpad, Cabri, and Geogebra) have, through their dragging and animation affordances, the potential to influence the perception of geometrical figures in a non-traditional paper and pencil way that challenges the legitimacy of some of the traditionally accepted geometrical definitions of a quadrilateral. For example, to many learners a quadrilateral with a reflex angle (concave quad) perceptually appears to be a triangle as learners consider only the convex or acute angled vertices to be relevant.
Fig. 1: a) A concave quadrilateral
Fig. 1: b) A convex quadrilateral However, learners exposed to dynamic geometry environments in which they interactively manipulate the virtual shapes can readily perceive the shape as a quadrilateral when they drag the obtuse angled vertex to obtain a convex quadrilateral (Fig. 1b) from the same shape. Furthermore, the number of sides and angles can be counted and angle measures introduced and manipulated to confirm that the convex quadrilateral satisfies the critical properties of a normal
°
∠
°
360 -m ADC = 265.01 it is quadrilateral. With due care taken to measure the reflex angle as easily verified that the sum of the interior angles is 360º, hence proving the quadrilateral to be a less
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familiar but legitimate version that exists in real life in many forms including as an arrow in many instances.
B Side 2
m∠ BAD = 26.40° m∠ ABC = 39.90°
C
Side 3
m∠ BCD = 28.69°
Side 1
m∠ ADC = 94.99°
D
360° -m∠ ADC = 265.01°
Side 4
m∠ BAD+m∠ ABC+m∠ BCD+( 360° -m∠ ADC) = 360.00°
A
Fig. 2: a) Angle measures for concave quadrilateral
B Side 2
m∠ BAD = 95.74° m∠ ABC = 39.90°
C m∠ BCD = 130.87°
Side 1
m∠ ADC = 93.48°
Side 3 D m∠ BAD+m∠ ABC+m∠ BCD+m∠ ADC = 360.00°
Side 4
A Fig. 2: b) Angle measures for a convex quadrilateral The uniqueness of each learner’s construction adds to the coherence of the logic behind the constructions, a move away from the answer driven approach. Learners are able to explain why the answer is rather than only tell what it is. That is, the learners can explain why a convex quadrilateral is a valid quadrilateral.
2.2
The crossed quadrilateral
Similarly, a crossed quadrilateral may be perceived not to be a proper quadrilateral at all by many learners until its vertices are dragged experimentally and discovered to fulfil all the properties of its convex counterpart. By dragging the point D in Fig. 2a across Side 1 (AB) it is easily seen that the sum of the interior angles of the quadrilateral is equal to the sum of the interior angles of the two triangles formed as a result of intersecting sides 1 and 3 which is equal to 180º + 180º = 360º. Fig. 3 shows this scenario; hence a crossed quadrilateral is a legitimate quadrilateral.
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m∠ ABC = 42.28°
B Side 2
m∠ BCD = 15.03° m∠ BOC = 122.69°
C
O
m∠ ABC+m∠ BCD+m∠ BOC = 180.00°
Side 3 D
Side 1
m∠ BAD = 21.36° m∠ ADC = 35.94°
Side 4
m∠ AOD = 122.69° m∠ BAD+m∠ ADC+m∠ AOD = 180.00° ( m∠ ABC+m∠ BCD+m∠ BOC) +( m∠ BAD+m∠ ADC+m∠ AOD) = 360.00°
A
Fig. 3: The sum of the interior angles of a crossed quadrilateral is 360º. The uniqueness of each learner’s construction again adds to the cogency of the logic behind the constructions and learners are ultimately able to explain why the why a crossed quadrilateral is a valid quadrilateral.
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THE CASE OF VARIGNON’S THEOREM1
Varignon’s theorem states that the midpoints of the sides of any quadrilateral join to form a parallelogram. A parallelogram is defined as a quadrilateral with its opposite sides parallel or equivalently, as a quadrilateral with opposite sides equal. Varignon’s theorem is valid for a convex quadrilateral as Fig. 4 shows.
B
C
Q
m PQ = 4.79 cm m RS = 4.79 cm
R
P
m QR = 5.39 cm
D
m SP = 5.39 cm
S A
Fig. 4. Varignon’s Theorem in a convex quadrilateral. The theorem is also true when the quadrilateral is concave as Fig. 5 shows.
1
See Yu [4] or the appendix for a proof of this theorem.
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C
Q
m PQ = 3.99 cm m RS = 3.99 cm
R
B
m QR = 2.10 cm
A
m SP = 2.10 cm
P
S
D Fig. 5. Varignon’s Theorem in a concave quadrilateral. The theorem is also true when the quadrilateral is a crossed version as in Fig. 6.
C
Q B m PQ = 3.99 cm
P
m RS = 3.99 cm m QR = 3.40 cm m SP = 3.40 cm
R
A
S
D Fig. 6: Varignon’s theorem for a crossed quadrilateral
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Even more surprisingly, the theorem even upheld when two of a quadrilateral’s sides are dragged to be co-linear as in Fig. 7.
C
Q B m PQ = 3.99 cm m RS = 3.99 cm
P
R
m QR = 4.05 cm m SP = 4.05 cm
A
S
D Fig. 7: Varignon’s theorem for three coo-linear points in a quadrilateral. Usiskin [5] has also made this observation in his regular lecturer to the ICME2012 conference. However this paper suggests that DGEs permit us to define a triangle as a quadrilateral with three of its vertices co-linear a condition that is forbidden in the Euclidian definition of a polygon. The condition that non of the sides should be co-linear thus impoverishes the concept of quadrilateral whereas allowing it, enriches the concept. The interior angles continue to add to 360º given that the fourth angle is 180º. By the same logic, it appears plausibly logical that a quadrilateral can be redefined as a pentagon with three of its sides co-linear and carrying on like this the concepts of polygons can be enriched as Giraldo et al [1] posit for the concept of derivative.
4
CONCLUSION
The above discussion illustrates how the affordances of DGEs can help learners and educators to explore properties of geometrical shapes and even experience definitional conflicts that challenge the traditional geometry resulting in more sophisticated understandings and richer concept images. It therefore makes sense to suggest that student’s movement from a spatio-graphic geometry (of physical models, diagrams, and computer image...) to pre-axiomatic geometry (of axioms and definitions) referred to by Parzysz [6] can be aided by the multiple cases of the same figure that DGEs afford them. In this case we see even multiple cases that give rise to traditional (Euclidian) definitional conditions being abrogated. DGEs such as Sketchpad which are now ubiquitously present in many classrooms allow learners to create this so-called traditionally forbidden geometry and thus challenge mathematics educators and curriculum writers to reframe the Euclidean surface in dynamic terms to belie some of the seemingly immutable definitions in the Euclidean stable.
REFERENCES [1]
Giraldo, V., Tall, D. & Carvalho, L. M. 2003. Using theoretical computational conflicts to enrich the concept image of the derivative. Research in Mathematics Education. 5(1):63-78.
[2]
Ernest, P. 2013. What is ‘first philosophy’ in mathematics education? Philosophy of Mathematics Education Journal, 27: April Issue
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[3]
Monaghan, F. 2000. What difference does it make? Children’s views of the differences between some quadrilaterals. Educational Studies in Mathematics, 42:179-196.
[4]
Yu, S. 2013. Varignon’s Thoerem Proof. Avalable at: http://www.youtube.com/watch?v=pTuQ_gd9-GA
[5]
Usiskin, Z. 2012. The shapes of geometry and their implications for the school geometry th curriculum. 12 International Congress on Mathematical Education, Topic Study Group 10. pp.2336-2345, 8-15 July 2012, COEX, Seoul, Korea.
[6]
Parzysz, B. 2003. Pre-service elementary teachers and the fundamental ambiguity of diagrams in geometry problem-solving. European Research in Mathematics Education III, Thematic Group 1.
APPENDIX. PROOF OF VARIGNON’S THEOREM2 Prior knowledge From Euclid’s fifth postulate parallel lines form corresponding angles that are equal. Two triangles are similar (///) if and only if their corresponding angles are equal. The corresponding sides of similar triangles are proportional. Example: If, in ∆ ABC and ∆ A’B’C’,
A=
A’,
B=
B’ and
C=
C’,
A' A b C then,
! !!
=
! !!
b'
c a
B
C'
a'
c' B'
!
= . !!
Proof: (of the midpoint theorem) Take ∆ ABD, Construct a midpoint, P on side AD, and draw a line from P parallel to DB to meet AB at Q.. Required to prove (RTP): That Q is the midpoint of AB.
2
Although in South Africa Varignon’s Theorem is not prescribed in the high school syllabus, the latter applies the midpoint theorem which is prescribed in Grade 11. The study of the theorem will therefore enhance rather than exceed students’ understanding.
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A
A
Q
P D
B
As PQ//DB, then ∆ AQP
///
D=
APQ,
B
D B =
AQP. Also
A is common,
∆ ABD.
This means that corresponding sides must be in the same proportion too.
A
A'
Q
P B
D
Q
P
B'
D'
That is, AP:AD = PQ: DB = AQ:AB =1;2, which means Q is the midpoint of AB, and therefore AQ = QB. (QED) Conversely, PQ // DB if Q is a midpoint. Proof of Varignon’s Theorem: (As iterations of the proof of the midpoint theorem) Constructions: Point C to create quadrilateral ABCD,, Midpoints R and S of sides DC and BC, and Segments (PR, RS and QS) to join all midpoints. RTP: That the quadrilateral PQRS is a parallelogram.
A Q
P B
D S
R C 6165
As we just proved that Q must be a midpoint in the upper triangle we can similarly prove that S is a midpoint of BC if R is a midpoint of DC in the lower triangle and RS//DB. That is, we can turn the quadrilateral ABCD clockwise by 90⁰ and repeat the proof with PR and QS parallel to AC and midpoints P and Q to obtain midpoints R and S. Thus we get PQ//RS and PR//QS making quadrilateral PQRS a parallelogram. (QED)
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