If M and N should meet, as at X, we would have two parallels to AB through the same point X, which is absurd.

Ax. b

[Through one point there is one and only one straight line parallel to a given straight line.]

Therefore M and N cannot meet, and, lying in the same plane, must be parallel.

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Q. E. D.

39. Defs. When two straight lines are cut by a third straight line, of the eight angles formed—

BA

a/b

ba A'B'

a, b, a', b', are interior angles.

A, B, A', B', are exterior angles.

a and a', or b and b', are alternate-interior angles. A and A', or B and B', are alternate-exterior angles. A and a', b and B', B and b', or a and A', are corresponding angles.

Question. Of the eight angles, which are always equal, and why? Question.-If A= A', what other angles are also equal to A, and why Are the remaining angles all equal, and if so, why?

Question.-If A= A' and also AB, what angles are equal, and why?

40. Defs.-Two figures are symmetrical with respect to a point called the centre of symmetry when, if one of them is revolved half way round on this point as a pivot, it will coincide with the other.

A single figure is said to be symmetrical with respect to a point called the centre of symmetry if, when the figure is turned half way round on this point as a pivot, each portion of the figure will take the position previously occupied by another part.

[A figure is said to be turned half way round a point when a line through the point turns through two right angles.]

5.5

TWO FIGURES SYMMETRICAL

WITH RESPECT TO O

A SINGLE FIGURE SYMMETRICAL
WITH RESPECT TO O

41. When two straight lines are cut by a third straight line, if the two interior angles on the same side of the cutting line are together equal to two right angles, then the two straight lines are parallel.

N'

M'

P

b'a

N

GIVEN-PQ cutting QM and PN so that a and b on the same side of PQ are together equal to two right angles.

About O, the middle point of PQ, as a pivot, revolve the figure QMXNP half way round to the symmetrical position. PM'X'N'Q, so that P and Q exchange places.

The angle a is the supplement of b.

Hyp.

Hence, when a takes the position a', PM' must be the prolongation of PN.

$29

(If two adjacent angles equal two right angles, their exterior sides form the same straight line.]

Likewise QN' is the prolongation of QM.

Now if these lines should meet on the right of PQ, as at X, they would also meet on the left, at X'.

$40

And we would have two straight lines between the two points, X and X', which is absurd.

Ax. a

$40

If they do not meet on the right of PQ, neither can they meet on the left of it. Hence QM and PN do not meet, and, being in the same plane, are parallel.

Q. E. D.

It may be observed that the preceding proposition rests on only two of the three geometric axioms stated in § 10, viz.: the superposition axiom, assumed in turning the figure unchanged about O, and the straight-line axiom, used to prove that there cannot be two straight lines between X and X'. The parallel axiom (viz.: that through a point only one straight line can be drawn parallel to a given straight line) has only been used so far in Propositions VII. and VIII. Mathematicians have tried to dispense with the parallel axiom entirely, but have not succeeded. In fact, Lobatchewsky in 1829 proved that we can never get rid of the parallel axiom without assuming the space in which we live to be very different from what we know it to be through experience. Lobatchewsky tried to imagine a different sort of universe in which the parallel axiom would not be true. This imaginary kind of space is called non-Euclidean space, whereas the space in which we really live is called Euclidean, because Euclid (about 300 B.C.) first wrote a systematic geometry of our space. In Lobatchewsky's space, Proposition IX. would be true, but Propositions VII. and VIII. would not be true, nor would §§ 47, 48, 49, 51, 58, etc., in Book I., and §§ 284, 327, 329, etc., in Book III.

To bisect a given straight line, AB

42. CONSTRUCTION.

B'

B

FIG. I

First method (Fig. 1).—At A and B erect Aa and Bb equal perpendiculars on opposite sides of AB. Join ab cutting. AB at O. O is the required middle point.

Proof. Suppose the middle point of AB is not O, but some other point as X.

Then turn the whole figure about X until AX coincides with its equal BX, A falling on B (call this position of A, A'), and B on A (call this position of B, B'). And O will assume the position O' on the opposite side of X.

§ 18

Then the perpendicular Aa will fall along Bb. [From a point in a straight line only one perpendicular can be drawn.] And a will fall on b (call this position of a, a').

[Since Aa is equal to Bb.]

Likewise b will fall on a (call this position of b, b').

Then the straight line aOb takes the position a'O'b'. That is, through two points, a and b, there would be two straight lines, which is absurd.

Ax. a

Hence the supposition that O is not the middle point is false, and is the middle point.

Q. E. D.

Second method (Fig. 2).—From A and B as centres with the same radius describe arcs intersecting at X and Y. Join XY intersecting AB at O, the required middle point.

[This method can be proved correct after reaching § 104.]

43. If two straight lines are cut by a third straight line, making the alternate-interior angles equal, the lines are parallel.

[When two straight lines are cut by a third straight line, if the two interior angles on the same side of the cutting line are together equal to two right angles, then the two straight lines are parallel.]

Q. E. D.