A Mathematician's Excuse Me (Er ... Don't Get Caught Short!)Author(s): Paul WakefieldSource: Mathematics in School, Vol. 30, No. 5 (Nov., 2001), pp. 25-27Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30215495 .

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A Mathematician's

Excuse Me

(Er ... don't get caught short!)

by Paul Wakefield

Background There has been much talk recently about the re- introduction of proof into the National Curriculum. In particular proofs in geometry are making a return and provide an interesting challenge for pupils and teachers. Having looked at what is expected by the new curriculum, I concluded that the result that seemed to provide the most challenge to budding mathematicians (Key Stage 4, Higher Tier) is the proof of: 'the angle subtended by an arc at the centre of a circle is twice the angle subtended by any point on the circumference'.

Imagine my surprise to find in conversations with those who should know better that there exists a proof of this result that is regarded as standard and indeed trivial. However, the proof is incomplete and does not in any way address the true situation. My flabber was ghasted further when I looked at several texts only to find either a simple statement of the result or, if a proof was provided, then it consisted of the incomplete one. Even (Waring, 2000) a book dealing with how to teach and learn the skills needed for proof provides the seemingly bog-standard incomplete one.

Surely the new GCSE texts will not commit such a heinous crime. However, in the event that they do, I have written this note explaining the full proof of this result in an effort to prevent fellow teachers being 'caught short' by bright young things pointing out the deficiencies in the seemingly well-accepted but nevertheless incorrect proof.

Analysis of the Situation

Figure 1 shows the bog-standard situation given and proven in most texts. However, this is only Stage 1 of the three stages needed for a complete proof.

D

C

A~ ~B

Fig. 1

We arrive at Stage 2 by thinking of angle Z ACB as being fixed and allow the line AD to move clockwise towards B. The proof given for Stage 1 is only valid until the points A, C and D lie on a straight line. Once the line AD travels past this point we are in a new and distinct situation (Fig 2). This is Stage 2 and a proof must be given for this situation. By the way, an argument based on angles in the same segment is not appropriate for this as the proof for angles in the same segment is based on the result we are currently attempting to prove!

D C

A~ ~B

Fig. 2

Now if we let AD continue on its journey clockwise towards B then the proof for Stage 2 will only be valid until AD passes the point B, once past B we are once again in a new and distinct situation to the previous two (Fig. 3). This is Stage 3. A proof must also be given for Stage 3. However, notice that the angles involved are ZADB and the reflex ZACB.

C

D

Fig. 3

Mathematics in School, November 2001 The MA web site www.m-a.org.uk 25

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The Proof

Stage 1

By drawing in the line CD (dashed) we form two isosceles triangles DCA and DCB with base angles P and y, respectively (Fig. 4). So the angle subtended at the circumference Z ADB = P3 + y. Also, as angles in a triangle sum to 7 we see that Z ACD = 7 - 2P and Z BCD = 7 - 2y. So, as angles at a point sum to 7C the angle subtended at the centre, a, is equal to:

2n - ZACD - Z BCD = 27u - (7[ - 20) - (7 - 2y) = 2p + 27 = 2(p + y) = 2 x Z ADB.

D

PIY

r

C

a r r

A~ ~B

Fig. 4

Stage 2

If we join point C to point D (see dashed line in Fig. 5) this forms two overlapping isosceles triangles DCA and DCB with base angles 0 and y, respectively (see Figs 6 and 7).

T

C D r ~a 6

O r

A~ ~B

Fig. 5

or D8 D

Fig. 6

C r D

E Y

r

B

Fig. 7

We let Z ADB = 86 and from this we can deduce that P + 6

= y. We also let Z DCB = e (Fig. 8).

C OE D

O

A VB

Fig. 8

Now we form three equations:

First of all Z DOB and Z COA are vertically opposite angles and so are equal hence:

I -U - 0 = --7 y a + P = 8 + 7 (equation 1)

Next we see that the angles in triangle A CD sum to 7; and so we have:

a + E = R - 2p (equation 2)

Also we see (Fig. 8) that the angles in triangle BCD sum to 7T and so we have:

7= n-y-P - 8 = Rt-- y-(P + 8) = r - - = - - 2y (equation 3)

We now put E = 7C - 2y into equation 2 to obtain:

a + E = 7C - 2P = a + [- 2y = 7C - 2P > a = 2y- 2P = 2(y- P).

Finally, we put a = 2(y - P) into equation 1 to get:

X + P = 8 + y 2(y- P) + P = 8 + y > 2y-2P + P - y7= 8

:8 =

= 1/2 U

i.e. ZACB = 2 x ZADB.

Stage 3

We form two isosceles triangles DCA and DCB with base angles y and P, respectively by joining point C to point D (see dashed line in Fig 9).

26 Mathematics in School, November 2001 The MA web site www.m-a.org.uk

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Ca

r r

r Y P

y I p ~B

D

Fig. 9

We let the reflex angle ZA CB be called ao and so we see that ZADB =

y+ P and Z ACD = C- 2y and ZDCB = C - 2p.

Using these values we have:

ao = 27C - (Z ACD + Z DCB) = 271 - (TE -2y + T - 2p) = 2y + 2p = 2(y + P) = 2 x Z ADB

i.e. (reflex) ZA CB = 2 x Z ADB.

Final Comments

I will leave the reader to decide whether the extra work involved in producing a correct proof is still within the target pupil's ability. Perhaps it is best to put our trust in those who have decided upon the content of the new curriculum as they must have considered what is required to learn and in particular teach the proof of such a result - mustn't they?

References

Waring, S. 2000 Can you Prove it? - Developing Concepts of Proof in Primary and Secondary Schools, The Mathematical Association.

Keywords: Proof; Angles in Circles.

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