7 Geometrical Reasoning
Task 7.2.6 Step-by-Step: Pythagoras’ Theorem – Proof Four
Accompanying Figure 7.2h is a set of statements, referring to the lengths and triangles from the figure. The statements are set out in random order. Make a copy of these statements, cut them out, and re-arrange them to give a logical step-by- step proof of Pythagoras’ theorem.
You may feel that you need to insert more statements so that your proof flows with no sudden gaps. Use blank statement cards as necessary.
∆ABC has a right angle at C ∠DBC is common AC AD ––– = ––– a2= c2– b2 AB AC c – x a –––– = –– ∠ACB = ∠ADC = 90° a c
∆ACD and ∆ABC are similar Let AD = x
a2= c(c – x) ∠DAC is common
b x
–– = –– Drop the perpendicular from C to AB,
c b meeting AB at D
BD BC
–––– = –––– b2= cx
BC BA
a2= c2– cx ∆BDC and BCA are similar
∠BDC = ∠BCA = 90° DB = (c – x)
Note: when two triangles are written next to one another with the statement ‘are similar’, the letters have been arranged so as to correspond.
When you are satisfied that you have constructed a logical chain of reasoning, present it to somebody else and check if they can follow the steps you have set out. Allow them time to rearrange the state- ments or to introduce new ones that you perhaps considered to be superfluous.
Comment
It is quite likely that another person will produce a different ordering of these statements. It is also likely that both their order and your order will be valid logically. Reflecting on this can be productive, because it lies at the heart of a daily dilemma for mathematics teachers. You arrive to look over the shoulder of a pupil working on a problem that you know.The steps set out on the pupil’s page are not the steps that you took when you solved that problem. How do you enter the logical chain that the pupil has started to construct and, from a standing start, offer constructive ways of moving on from their ‘being stuck in the moment’ position?
Exploiting the Different Proofs
Pythagoras’ theorem states that ‘In a right-angled triangle … ’.Where has the fact of a right angle been used? Is it necessary?
Looking back at the basis of Proof Two, the two images in Figure 7.2i make it quite clear that if the triangle is not right-angled, then the hole in the middle is no longer a square and so the proof breaks down. In this visual proof the relationship only holds if the triangle is a right-angled triangle. There is an interactive version of the diagram ‘7b Pythagoras without a right-angle’ on the CD-ROM.
It is worth articulating to yourself what happens when the angle is larger than, or smaller than, 90°. Every theorem in mathematics hides a surprise (otherwise it would not be worth stating much less proving!). Here it is amazing that it is just 90° that yields the simple relationship between the areas and hence the sides of the triangle. It takes a bit of drawing and thinking to see
why Proof Three using pentagons breaks down when the angle is not 90°.
Proof One holds more surprises: the squares used in Proof One can all be deformed into parallelograms as in Figure 7.2j.
Unlike the other proofs, the basic dia- gram still seems to work! One triangle has been shaded in. Its edges can be seen to
determine the edges of the parallelograms. By gazing at the diagram, you may be able to convince yourself (and then a sceptic?) that the large parallelograms are equal in area to the sum of one each of the other two sizes of parallelograms.
This result seems to be due to Pappus of Alexandria, who lived and worked in Alexandria 600 years after Euclid (900 years after Pythagoras) and who produced a number of important theorems that still bear his name. Pappus’ area theorem gives a way of constructing a parallelogram on the third side of any triangle in such a way that its area will be the sum of the areas of two parallelograms constructed on the two smaller sides. It is a beautiful and astonishing result.
Proof Four not only justifies Pythagoras’ theorem but also yields an important generalisation, known as the law of cosines. Consider Figure 7.2k, (known as Vecten’s diagram) which is a development of the diagram you were invited to construct in Task 5.4.2.Three altitudes have been added to the diagram and these have each been extended to cut a square into two rectangles.You saw in Task 5.4.2 that the areas of the four triangles are equal in magnitude.
120 BLOCK 2
Figure 7.2i
Figure 7.2j
In addition, each pair of rectangles with the same density of grey has the same area.
One way to prove this result proceeds by finding that the area of each one of the pair is the product of the cosine of the angle at their shared vertex, and the sides of their respective squares. In other words, the area of each grey rectangle is of the form
ab× cosC, where C is the angle between the two sides a and b.
The area of each square is the sum of the other two squares minus the two extra, equal-shaded rectangles.Thus c2 = a2+ b2– 2ab× cosC and this is the law of cosines
that is the generalisation of Pythagoras’ theorem. In the case of a right-angled trian- gle, cos 90° is zero.
In order to appreciate the full scope of generality, it is also necessary to consider what happens to the diagram when one of the angles in the triangle is obtuse.
For more on Pythagoras and the theorem that bears his name see the website St Andrews (webref) about history of mathematics and the website ‘Cut-the-Knot’ (webref). (See References at the back of the book for web addresses.)
7.3 REASONING ABOUT IN-CIRCLES AND EX-CIRCLES
Pythagoras’ theorem is just the tip of the iceberg, albeit an important tip. There are thousands of surprising relationships that can be established in geometry.This section considers some circles connected with triangles.
The section involves thinking about the in-circle and the ex-circle of a triangle. The in-circle of a triangle is a circle drawn inside the triangle in such a way that it is tangential to all three sides.The ex-circle is a circle tangent to all three sides of a tri- angle, but two of those points are on sides
extended beyond the triangle. The in-circle and one of the three ex-circles of a triangle are shown in Figure 7.3a.
Three proofs of a formula for the radius of the in-circle are presented, in this section, followed by an opportunity to revisit the same ideas with ex- circles. One outcome is a formula for the area of a triangle known as Heron’s formula.
In-circles
As an introduction to in-circles, you are invited to start by imagining.
GEOMETRICAL REASONING 121
Reflection 7.2
What differences have you noticed in your degree of conviction and in what you had to do to understand the four proofs offered? Which ones offer the most opportunity for you to evoke surprise in learners with whom you work?
What changes are you aware of as you move from discerning specific details, to recognising relation- ships, to perceiving properties that hold in general, to reasoning on the basis of properties?
For Euclid, it was necessary to justify this result on the basis of already established properties; starting from the intersection of two of the angle bisectors, he dropped perpendiculars to the sides of the triangle and used the conditions for congruence of various triangles.What else would you need to do to convince a sceptic?
The remainder of this sub-section provides the outlines for three different proofs of a formula for the radius of the in-circle of a right-angled triangle. One issue for consid- eration is what facility each proof requires in the reader. Notice what you have to do in each case to discern what is relevant, recognise relationships and perceive these as properties so as to appreciate the generality.
First Proof
The first proof is algebraic in nature. Start by assuming the in-circle has already been drawn inside a right-angled triangle. Denote that which is wanted but not yet known, namely the radius, by a letter, say
r.The triangle can be split up into three triangles, each with a base on
a side of the original triangle, and height r, as shown in Figure 7.3b. 122 BLOCK 2