A History of Mathematics From Mesopotamia to Modernity

Geometries andSpace 211

P AQ

BC

Fig. 14Proof of the ‘angles of a triangle’ theorem.

AB

CD

Fig. 15Figure for Exercise 1(b).

B

C

A

D

Fig. 16The figure for Exercise 2; AB is 1 cm.

point is that CD must be parallel to AB. In fact, if the angles∠BAC,∠ACD are greater than two

right angles, then CD must meet AB on the left, say at E, and by symmetry it must also meet on

the right at F. But then there are two straight lines joining E,F.

Accordingly, CD does not meet AB, so∠BAC,∠ACD are together equal to two right angles;

since∠BAC is a right angle, this means that∠ACD is one, and similarly for the fourth angle.

Once we have this, it is easy to use congruent triangles to show that AB=CD.

2. See Fig. 16. We have: distance to meeting point is length of CD, or 0.5 tanα=0.5/tan( 10 −^10 )
   For such a small angle, we can approximate tanxbyx, arriving at CD= 0.5× 1010 as
   stated.


3. (a) (See Fig. 8) What is meant is that the angle G is a (decreasing) function of the length AB,
   and so to any angle G there corresponds a unique length AB of the side of the quadrilateral.
   As a result, you can define a measure of length not by arbitrary choice, as is done in Euclidean
   geometry, for example, by giving a fixed value to the length of a metre or ‘Paris foot’; but by
   stating that your unit will be that length AB which corresponds, say, to 80◦for the angle G.
   (b) This follows from the angle-sum theorem. In fact, the area of ABGD is twice the area
   of triangle ABG. By the angle-sum theorem, this is 2C(π−∠GAB−∠ABG−∠BGA) =
   2C(π−π/ 4 −π/ 2 −α/ 2 )=C(π/ 2 −α), where C is the relevant constant for the geometry
   andαis the angle at G. It follows that the angle G is a decreasing function of the area ABGD,
   and so of AB.


4. Let AD be perpendicular to BC, and A′D′to B′C′, and suppose that AD=A′D′. Now let AF
   be a cutting line from A to BC, and construct F′on B′C′so that D′F′=DF. Then it is easy
   to see that ADF, A′D′F′arecongruenttriangles (equal sides and included right angle), and so