Reasoning and proof in geometry: effects of a learning environment based on heuristic worked-out examples

Abstract

We analyze heuristic worked-out examples as a tool for learning argumentation and proof. Their use in the mathematics classroom was motivated by findings on traditional worked-out examples, which turned out to be efficient for learning algorithmic problem solving. The basic idea of heuristic worked-out examples is that they encourage explorative processes and thus reflect explicitly different phases while performing a proof. We tested the hypotheses that teaching with heuristic examples is more effective than usual classroom instruction in an experimental classroom study with 243 grade 8 students. The results suggest that heuristic worked-out examples were more effective than the usual mathematics instruction. In particular, students with an insufficient understanding of proof were able to benefit from this learning environment.

Appendix

A heuristic worked-out example: Opposing sides and angles of parallelograms (slightly shortened version)

• I. The problem

Nina and Tom have drawn and measured parallelograms. In doing so, they noticed that opposing sides were always of equal length. Moreover, opposing angles were always of equal size.

Tom: “We measured so many parallelograms: We have drawn all kinds of quadrangles and always we recognized that the opposing sides were of equal length and opposing angles were of equal size. I think, it has to be like this!”

Nina: “I think you are right, but I don’t know a reason. Maybe by chance, we have only drawn parallelograms for which the statement is correct? We cannot measure the angles and sides exactly. Perhaps they were only approximately of the same size.”

Tom: “So let’s try to prove our assumption like mathematicians would do!”

Tom and Nina try to prove the following mathematical proposition:

“In a parallelogram opposing sides are of equal length and opposing angles are commensurate!”

In the following we have a look at how they solved the mathematical problem. Please read their solution, but try to complete all steps on your own.

• II. Examination of the Problem

First we want to reproduce the measurement results of Tom and Nina. You will need a set square, paper, and pencils.

(a) Draw a parallelogram ABCD and mark the angles with α, β, γ, δ. Afterwards measure and note the sizes of its sides and angles.

(b) The experiments suggest that opposing sides and angles are of equal size in all parallelograms. You may remember that this characteristic is called congruence. In 7th grade you learned that congruent sides and angles can be transformed on one another by using congruency mappings.

Nina and Tom remember some facts:

Nina: “When did we hear of angles and sides that have the same size?”

Tom: “In the 7th grade.”

Nina: “Yes, when we learned about congruency mappings.”

What kind of congruency mappings do you remember?

Answer: ——

(c) The pictures on this page display several congruency mappings. Figure out what kind of congruency mappings are displayed and mark the congruent sides of the triangles.

Nina and Tom think about using the properties of congruency mappings for parallelograms.

Tom: “So far, by using congruency mappings we got new parallelograms.”

Nina: “Now, what can we do with the congruency mappings?”

Tom: “We could transform all the figures in congruent figures.”

Nina: “But we don’t want to construct new parallelograms. Rather we want to demonstrate that opposing sides and angles are of equal sizes. Therefore we need to transform the parallelogram on itself.”

Try to detect all the symmetry axes, rotation centers, and the center of the point of reflection of the parallelogram and mark them in a figure.

• III. Statement

So far it seems that Tom’s and Nina’s statement was correct. Try to rephrase this statement in a formula:

Prove what you wrote by filling in the gaps in the following text:

We know that a parallelogram is a ——, in which the —— sides are parallel.

We claim:

If A, B, C, D are the—of a parallelogram and α, β, γ, δ are its——, then you can say that:

— = —, — = —, — = —, — = — .

Compare your statement to the statement of Tom and Nina on the first page.

• IV. What do you know about quadrangles, parallel straight lines and congruency transformations

There are a lot of arguments that could be important for the proof of the conjecture. In particular, please try to remember the following facts:

• Congruency mappings map

  straight lines on ——,

  circles on ——,

  sections on ——with equal ——,

  angles on —— of equal ——.

• The point of reflection is a rotation of —— degrees.

• The sum of angles in a quadrangle always is —— degrees.

• In a point of reflection every point lies on a ——

  with its image point and the ——.

  The center is exactly ——

  between —— and ——

  ——.

• The point of reflection maps straight lines on——

• With the line of reflection straight lines that are orthographic to the axis of reflection are mapped on the ——

• With the line of reflection straight lines that are orthographic to the axis of reflection are mapped on the ——

• Straight lines that are parallel to the axis of the reflection are mapped on the ————.

• V The Proof Idea

The length of a side and the size of an angle remain unchanged when congruency mappings are applied. Thus, if we find a mapping, which maps each side of a parallelogram on the opposing side and each angle on the opposing angle we will know that they are of equal size. Have another look on section (d):

Where would the point of symmetry or the axis of reflection have to be?

Try to extend your proof idea to a proof.

• VI. Proving the Conjecture

ABCD is an arbitrary parallelogram with the angles α, β, γ, δ. We draw the diagonal [AC] and call the middle of it M, as in the figure.

Then A is mapped with a —— on M to A’ = —— and C to C’ = ——, because M is the—— between A and C. The straight-line AB is mapped to a —— straight-line, which goes trough A’ = ____, so to _______. Also the straight-line BC is mapped to a —— straight-line through C’ = ——, so to ——.

Now we find B’, the image point of B. To do this we use the fact that B is the point of intersection of the straight-lines —— and ——. The image point of B has to be the point of intersection of both image lines, that means the point of intersection of—— and ——. This point of intersection is —, that is why B’ = ——.

In the same way we conclude that D’ = ——. So the —— on M maps the parallelogram to ——. This means that the section [AB] is mapped to ——, and the section [BC] is mapped to ——. Because the —— maps the sections to ——, it can be concluded that —— = —— and —— = ——.

Furthermore it follows that α’ = —— and β’ = ——. Because the —— maps angles to —— of —— size, we conclude that —— = —— and —— = ——.