Using dynamic geometry software to investigate midpoint quadrilateral.

Introduction

The NCTM standards  argue that students should be provided with the opportunity to investigate mathematical problems as an aid to understanding mathematical ideas. Ponte  points out "a mathematical investigation stresses mathematical processes such as searching regularities, formulating, testing, justifying and proving conjectures, reflecting, and generalizing (p. 54)". To engage students in mathematical investigations, educators should provide students with tools that enhance the process. Dynamic geometry software such as the Geometer Sketchpad (GSP)  is a tool to reach that goal. As research studies (e.g., , , ) indicate, dynamic geometry software can be a cognitive tool to help students develop problem-solving abilities if used effectively. They identified different purposes for which students used dragging, the main feature of the dynamic geometry software, and different purposes for which students used measures. These purposes appeared to be influenced by students' mathematical understandings that were reflected in how they reasoned about the physical representations, the types of abstractions they made, and the reactive or proactive strategies employed. These issues are key to the investigation reported herein.

This article reports on the following class activity in the form of an investigation that was implemented by us in our college geometry class for high school presevice teachers using the Geometer Sketchpad:
Given any quadrilateral, construct a midpoint on each side. Connect
each consecutive midpoint with a segment. What are the properties
of the shape formed by joining the midpoints? Does the resulting
shape depend on the type of quadrilateral (e.g. convex & concave)?
How does the area of the midpoint quadrilateral compare to the area
of the original quadrilateral? This activity was adapted from the
Intermath website .

Part I. The investigation on the shape of the midpoint quadrilateral

We started the activity by asking students to construct any quadrilateral and its midpoint quadrilateral using GSP. They worked in a group of three with a computer. This was a rather straightforward task for most of them. One of the students did the construction by first constructing four points A, B, C and D, using the GSP selection tool to select these points in clockwise order, and then constructed segments [bar.AB], [bar.BC], [bar.CD] and [bar.DA], which formed a quadrilateral (Figure la). Then the student selected the four segments and constructed midpoints F, G, H, and E on the segments respectively. The student then constructed segments [bar.EF], [bar.FG], [bar.GH] and [bar.HE] to form the midpoint quadrilateral as shown in Figure lb. Other students constructed their midpoint quadrilaterals in a similar way.

[FIGURE 1 OMITTED]

[FIGURE 2 OMITTED]

[FIGURE 3 OMITTED]

We now asked the students to try to prove their conjecture. Some of the students thought they had already done the proof by stating that their reasoning was not based on one case but on all possible cases with the GSP dynamic movement feature. We challenged them to examine their reasoning by asking, "You indicated that the opposite sides of quadrilateral EFGH always have the same length and its opposite angles are always congruent. Did these properties come from your measurement using GSP, or from your logical reasoning?" Through serious discussion, the students realized that the measurement is very important in the investigation process, but it is still different from a mathematical proof that begins with a truth and proceeds by logical steps to a conclusion, which then must be true. They then worked hard on constructing a proof as a group in class and were given ample time as they shared ideas among themselves.

The following is the proof that one student came up with for the convex quadrilateral case (with minor help from us):

[FIGURE 4 OMITTED]
Start with figure 4a and construct segment [bar.DB] (diagonal of
quadrilateral [A.bar]BCD). In [DELTA]DAB, [bar.EF] is a
mid-segment? By the Mid-segment Theorem (4), [bar.EF] II [bar.DB]
and m ( [bar.EF]) = (l/2)*m([bar.DB]). In [DELTA]DCB, [bar.HG] is a
mid-segment, and again by the Mid-segment Theorem, [bar.HG] II
[bar.DB] and m([bar.HG]) = (l/2)*m([bar.DB]). By transitivity,
[bar.EF] // [bar.HG] and m([bar.EF]) = m([bar.HG]). Therefore the
midpoint quadrilateral is a parallelogram (as it has a pair of
opposite sides being both parallel and congruent).

While two groups of the students used the same method, the rest of the class worked out their proofs using slightly different ways. Instead of proving one pair of opposite sides parallel and congruent, they either proved two pairs of opposite sides being parallel, or proved two pairs of opposite sides being congruent. For either of the two ways, two diagonals (see Figure 4a and Figure 4b) rather than just one needed to be constructed. By sharing the proofs, the students not only developed multiple ways of proof, but also deepened their understanding of the properties and conditions of a parallelogram.

The discussion on the proof for the midpoint quadrilateral of the convex quadrilateral made sense to students in terms of visualizing mid-segments as the diagonal(s) are drawn. But it was a different story for the situations with the concave quadrilateral case and crossed quadrilateral case. It was difficult for some students to see where the diagonals of the quadrilateral are and how to use them to prove that the midpoint quadrilateral is a parallelogram.

One student suggested to start with a convex quadrilateral and its diagonals first, and then use the dynamic features of GSP to move one of the vertices inward to form the concave quadrilateral case (Figure 5), or even further to form the crossed quadrilateral case (Figure 6). This was a good idea to help the rest of the class to visualize the diagonals (DB and AC), and to make the proof process almost identical to that of the convex quadrilateral case. Thus, students came up with their proofs without difficulty in their written work, which was assessed in class by the instructors through discussing and evaluating the proofs with individual students.

[FIGURE 5 OMITTED]

[FIGURE 6 OMITTED]

Part II. The investigation on the area of the midpoint quadrilateral

[FIGURE 7 OMITTED]

The proof of this conjecture was not easy for most of the students. Except in the special cases (e.g., quadrilateral ABCD is either a rectangle or a square), the students have difficulties in coming up with the proof. To help students get started, we discussed the following idea:

Lemma 1: In an arbitrary triangle ABC, if D is the midpoint of AB, and E is the midpoint of AC, then Area([DELTA]ADE) = 1/4 Area([DELTA]ABC).

In doing this, we used a GSP sketch (Figure 8) to demonstrate the lemma and students noticed that the ratio of the area of [DELTA]ADE to the area of [DELTA]ABC was always 1:4 no matter how the figure changed in shape. The dynamic feature of GSP enabled students to click and drag a vertex and observe that the ratio of the two areas was always 1:4.

[FIGURE 8 OMITTED]

We asked the students how one can prove the lemma and they gave suggestions such as finding the heights of the two triangles, and considering trigonometric ratios. With minimal help, one student volunteered to present his proof to the rest of the class as follows:

In Figure 8, Area([DELTA]ADE) = 1/2 AD x AE x sin(<A),

Area([DELTA]ABC) = 1/2 AB x AC x sin(<A)

Since D is the midpoint of [bar.AB], and E is the midpoint of [bar.AC], AD = 1/2 AB, a nd

AE = 1/2 AC

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Using Lemma 1, the students discussed ways of proving the three cases (convex, concave and crossed quadrilaterals) in Figure 7. With some help (e.g., the suggestion of using backward reasoning) from us, the students developed the following proofs:

Case 1: Convex Quadrilateral ABCD

Proof: Construct diagonal AC. Since F is the midpoint of [bar.AB] and G is the midpoint of [bar.BC] then

Area(A[DELTA]FG) = 1/4 Area([DELTA]BAC) by Lemma 1 ... (1)

Use the same reasoning, we can prove the following:

Area([DELTA]DEH) = 1/4 Area([DELTA]DAC) ... (2),

Area([DELTA]AEF) = 1/4 Area([DELTA]ADB) ... (3), and

Area([DELTA]CHG) = 1/4 Area([DELTA]CDB) ... (4)

Proofs of 3 and 4 may require constructing diagonal [bar.BD] (Figure 9).

[FIGURE 9 OMITTED]

If we add (1), (2), (3), and (4), we have

Area([DELTA]BFG) + Area([DELTA]DEH) + Area([DELTA]AEF) + Area([DELTA]CHG)

= 1/4 Area([DELTA]BAC) + 1/4 Area([DELTA]DAC) + 1/4 Area([DELTA]ADB) + 1/4 Area([DELTA]CDB)

= 1/4 [Area([DELTA]BAC) + Area([DELTA]DAC) + Area([DELTA]ADB) + Area([DELTA]CDB)]

= 1/4 [Area(ABCD) + Area(ABCD)] = 1/2[2x Area(ABCD)]

= 1/2 Area(ABCD).

Therefore, Area (EFGH) = Area (ABCD)--[Area([DELTA]BFG) + Area([DELTA]DEH) + Area([DELTA]AEF) + Area([DELTA]CHG)]

= Area(ABCD) - 1/2 Area(ABCD) = 1/2 Area(ABCD). That is, the area of the midpoint quadrilateral of a convex quadrilateral is a half the area of the quadrilateral.

Case 2: Concave Quadrilateral ABCD

Poof: Construct diagonals [bar.AC] and [bar.BD] (Figure 10). Since E is the midpoint of [bar.AD] and F is the midpoint of [bar.AB], by Lemma 1,

Area([DELTA]AEF) = 1/4Area([DELTA]ADB) ... (1).

Use the same reasoning, we can prove:

Area([DELTA]CHG) = 1/4 Area([DELTA]CDB) ... (2),

Area([DELTA]DEH) = 1/4 Area([DELTA]DAC) ... (3), and

Area([DELTA]BFG) = 1/4 Area([DELTA]BAC) ... (4).

[FIGURE 10 OMITTED]

(1) + (2) yields

Area([DELTA]AEF) + Area([DELTA]CHG) = 1/4 Area([DELTA]ADB) + 1/4 Area([DELTA]CDB) = 1/4 Area([DELTA]BCD) ... (5)

(3) - (4) yields Area([DELTA]DEH)--Area([DELTA]BFG) = 1/4 Area([DELTA]DAC) - 1/4 Area([DELTA]BAC) =

1/4 [Area(ADAC)--Area([DELTA]BAC)] = 1/4 A rea(ABCD)

(5) + (6) yields

Area([DELTA]AEF) + Area(ACHG) + Area([DELTA]DEH)--Area([DELTA]BFG) = Area([DELTA]BCD) + 1/4 Area(ABCD) = 1/2 Area(ABCD) [??]

Area (EFBGH) = Area (ABCD)--[Area([DELTA]AEF) + Area([DELTA]CHG) + Area([DELTA]DEH)]

= Area (ABCD)- 1/2 Area(ABCD) - Area(ABFG).

Therefore, we have

Area (EFBGH) + Area([DELTA]BFG) = 1/2 Area(ABCD) [??]

Area(EFGH) = 1/2 Area(ABCD) That is, the area of the midpoint quadrilateral of a concave quadrilateral is a half the area of the quadrilateral.

Case 3: Crossed Quadrilateral

Proof: For the sake of simplicity, in the following figure (Figure 11), we use labels to express areas (for example, use yl to express the area of CHG). K is not one of the vertices of quadrilateral ABCD. However, when we look at this crossed quadrilateral as two triangles, K is the shared vertex of them (ABAK and ACDK). This will be used in the proof.

[FIGURE 11 OMITTED]

From Lemmal, yl = (l/4)*(yl + y2 + gl + g4 + w2) [??] 3yl = y2 + gl + g4 + w2 ... (1) Use the same reasoning, we can prove:

3(y3 + g3 + g4) = y4 + w2 [??] 3y3 = y4 + w2--3g3--3g4 ... (2)

3y2 = y1 + g1 + g2 + w1 ... (3)

3(y4 + g3 + g2) = y3 + wl [??] 3y4 = y3 + wl - 3g3 - 3g2 ... (4)

(1) - (2): 3(y1 - y3) = y2 - y4 + gl + 3g3 + 4g4 ... (5)

(3) - (4): 3(y2 - y4) = y1 - y3 + g1 + 3g3 + 4g2 ... (6)

(5) + (6): 3(y1--y3) + 3(y2--y4) = (y1--y3) + (y2--y4) + 2g1 + 4g2 + 6g3 + 4g4

[??] (y1 - y3) + ( y2--y4) = g1 + 2g2 + 3g3 + 2g4

[??] (y1 + y2 + g1)--(y3 + y4 + g3) = 2(g1 + g2 + g3 + g4),

which is Area([DELTA]CDK)--Area([DELTA]BAK) = 2*Area(EFGH).

According to the explanation for the area of a crossed quadrilateral described above, the area of midpoint quadrilateral EFGH is half the area of the original quadrilateral ABCD.

The proofs (for Cases 1, 2, and 3) described above were done by one group of students with some help from us who finally presented the proofs to the rest of the class. These proofs were developed separately. In order to help students see the connections among these cases, at the end of the investigations described in this article, we introduced them to Bretschneider's Formula (http://home.att.net/~numericana/answer/ formula.htm#bretschneider). Bretschneider's Formula does hold for convex and concave quadrilaterals, and it's also good for (the signed areas of) crossed quadrilaterals. Expressed as [(4A).sup.2] = 4[p.sup.2][q.sup.2]-[([a.sup.2]-[b.sup.2] + [c.sup.2] - [d.sup.2]).sup.2], the formula gives the simplest way to express the area (A) of a quadrilateral in terms of its sides (a, b, c, d) and diagonals (p, q) without any restrictions.

Conclusion

Our experiences working with the pre-service secondary school teachers who participated in this activity using GSP indicate that pre-service teachers can benefit from mathematical explorations and investigations with dynamic geometry software as a tool. The activities shown in the article indicate that giving students the opportunity to pose and prove/disprove conjectures with the aid of GSP can be a great learning tool. The three cases of quadrilateral (concave, convex and crossed) are easily investigated by virtue of the existence of those features of GSP. (Please also see  for a thorough discussion in this aspect.] The dynamic features enable one to change convex quadrilateral to concave and then to crossed and back to convex by simply dragging one point of the quadrilateral. With the aid of measurement feature of GSP, students can make conjectures in terms of what they see as the quadrilateral changes from one case to another.

Due to the dynamics and measurement feature of GSP, it can construct visual representations easily and accurately as opposed to static figures that can be done with paper and pencil. These features in GSP make it possible for students to concentrate more on the problem than just struggling with a visual representation to model a given situation.

References

 Choi-koh, S. S. (1999). A student's learning of geometry using the computer, The Journal of Educational Research, 92, 301-11.

 Dixon, J. K. (1997). Computer use and visualization in students' construction of reflection and rotation concepts, School Science and Mathematics, 97, 352-8.

 de Villiers, M. (1998). A Sketchpad discovery involving triangles and quadrilaterals. KZN Mathematics Journal, Vol 3, No 1, 11-18.

 Hollebrands, K. (2007). The Role of a Dynamic Software Program for Geometry in the Strategies High School Mathematics Students Employ. Journal for Research in Mathematics Education, 38(2), 164-192.

 Jackiw, N. (2001). The geometer's sketchpad 4.06 [computer software]. Berkeley, CA: Key Curriculum Press.

 National Council of Teachers of Mathematics. (2000). Principles and standards for school mathematics: Reston, VA: Author.

 Ponte, J. P. (2001). Investigating mathematics and learning to teach mathematics. In F. Lin & T. J. Cooney (Eds.), Making sense of mathematics teacher education (pp. 53-72). Boston: Kluwer.

Supplementary Files

 GSP files for Figures 1, 2, 3,4,5,6,7,8,9,10, and 11.

Samuel Obara

e-mail: so16@txstate.edu

Department of Mathematics, Texas State University

San Marcos, Texas, USA

Zhonghong Jiang

e-mail: zj10@txstate.edu

Department of Mathematics, Texas State University

San Marcos, Texas, USA

(1) Convex quadrilateral is any quadrilateral with no diagonal falling outside the figure.

(2) Concave quadrilateral is any quadrilateral with one diagonal falling outside the figure.

(3) Crossed quadrilateral is any quadrilateral with both diagonals falling outside the figure. (4) This theorem states that the segment connecting the midpoints of two sides of a triangle (midsegment) is parallel to and one half as long as the third side.
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