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and EF will meet HQ since LK is parallel to PQ. Now take

=

DG EF and draw BG. Then the two triangles BEF and DBG are congruent and therefore

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Hence the line BC meets LK at some point between E and G. But BC is any line through B in the angle opening QBA and LK and BQ do not meet. Therefore BQ is parallel to LK. 5. If two lines are parallel to a third, they are parallel to each other.

We distinguish two cases according as the third line lies between the two lines or not.

In the first case, let AK and DQ (Fig. 4) be each parallel to ML. We wish to prove that AK is parallel to DQ. Draw

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AC any line through A in the angle opening DAK. AC will meet ML in some point F since AK is parallel to ML. CF produced will also meet DQ, since ML and DQ are parallel.

Hence any line through A in the angle opening DAK meets DQ. On the other hand AK cannot meet DQ since it cannot meet ML. Hence AK and DQ are parallel.

In the second case, let AK and DQ (Fig. 5) be each parallel to ML. We wish to prove that AK is parallel to DQ. Draw through A the line AK' parallel to DQ. Then by the first case AK' is parallel to ML and hence coincides with AK.

III. THE EUCLIDEAN ASSUMPTION

6. We may replace Postulate 5 of Euclid or Assumption XIII of Veblen by the following assumption while retaining all the other assumptions of either author.

Through any point in the plane there goes one and only one line parallel to a given line.

That one parallel exists, is, in fact, proved in the twentyeighth proposition of the first book of Euclid (Veblen, VIII). To assume that only one parallel exists is equivalent to assuming that in Fig. 1 the lines AL and AK form one and the same straight line. Hence

4HAL=&HAK=rt. 4.

Take now M (Fig. 6), the middle point of AB and draw MD perpendicular to PQ and intersecting AL in C. Then as

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just shown,

DCK is a right angle. The two right triangles

AMC and BMD are congruent and 4CAB=4ABD.

Therefore

4QBA+4BAK=2 rt. 4s.

By our definition of parallels, any line through A in the angle BAK meets PQ.

Hence our assumption is equivalent to Euclid's Postulate 5. From this follows the Euclidean geometry.

IV. THE LOBACHEVSKIAN ASSUMPTION

7. While retaining all the other assumptions of the Euclidean geometry, we will replace Postulate 5 of Euclid or Assumption XIII of Veblen by the following assumption due to Lobachevsky. Through any point in the plane there go two lines parallel to a given line.

It follows that, in Fig. 6,

4QBA+4BAK< 2 rt. 4s.

For if the sum of the angles QBA and BAK were greater than two right angles we could draw through A in the angle BAK a line not meeting BQ by Euclid I, 28. This is contrary to our assumption that AK and BQ are parallel.

On the other hand, if the sum of the angles were equal to two right angles we should have the Euclidean assumption.

8. The following theorems are of vital importance in subsequent proofs.

Theorem I. Let AB and CD be two parallel lines cut by a third line C and let A'B' and C'D' be two other parallel lines cut by a line A'C', and let 4DCA= 4D'C'A'; then

(1) Iƒ A'C' = AC, 4C'A'B' = 4CAB
(2) If A'C'< AC, ¿C'A'B' >¿CAB

(3) If A'C' > AC, 4C'A'B'< 4CAB

Consider first the case A'C' AC (Fig. 7).

If

=

C'A'B' were less than 4CAB draw AK so that

CAK

= 4C'A'B'. Then AK meets CD in some point K. Take K' on

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= 4C'A'B'. This is impossible, since A'B' does not meet C'D'. Hence 4C'A'B' cannot be less than 4CAB. Similarly 4CAB cannot be less than 4C'A'B' and hence 4C'A'B' = 4CAB.

Consider secondly the case A'C'< AC (Fig. 8).

On CA take CA" equal to C'A' and draw A"B" parallel to CD. Then &CA′′B"=4C'A'B', as just shown, and AB and A"B" are parallel (sec. 5). Therefore, by sec. 7,

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The third case A'C' >AC, is handled like the second case. Theorem II. Let AB and CD be two parallel lines cut by a third line AC and let A'B' and C'D' be two other parallel lines cut by a line A'C', and let 4CAB= 4C'A'B' and ACD = 4 A'C'D', then AC A'C'.

For each of the suppositions AC< A'C' and AC>A'C' contradicts Theorem I.

If in Theorem I we take 4DCA = 4D'C'A' = rt. 4, the angles CAB and C'A'B' are the angles of parallelism for the distance AC and A'C' respectively (sec. 2). Theorem I includes then, as a special case, the following:

Theorem III. The angle of parallelism is fixed for a fixed distance and decreases as the distance increases.

If we denote the distance AH (Fig. 1) by p, the angle of parallelism HAL is denoted in Lobachevsky's notation by II (p). Theorem III asserts that II (p) is a decreasing function of p. The exact determination of II (p) will be given in sec. 33. may note, however, that II (p) is always less than a right angle. In other words,

We

Theorem IV. If two lines have a common perpendicular they neither intersect nor are parallel.

The converse of IV is also true, as we shall now show.

9. Two straight lines which neither intersect nor are parallel have a common perpendicular.*

Let LM and EF (Fig. 9) be two straight lines which neither intersect nor are parallel. We wish to show that they have a common perpendicular. Take A and B any two points on LM and draw AH and BK perpendicular to EF. If AH=BK the existence of a common perpendicular to LM and EF follows quickly, as shown below. Suppose then that

BK< AH.

Draw KS parallel to LM. Place † the rt. FKB on the rt. FHA so that K falls on H, KF takes the direction HF

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and KB takes the direction HA. The point B falls at B' between H and A, BM takes the position B'M' and KS the position HS', parallel to B'M'.

Since 4FKS= 4FHS' a line parallel to KS (and hence to LM) drawn through H lies in the angle opening FHS' (sec. 7). Hence HS' intersects LM and therefore B'M' intersects LM at some point P (Veblen, Theorem 17).

Draw PR perpendicular to EF. Place the right angle FHB' on the right angle FKB. Then the line PR takes the position QT, where QT = PR and QT is perpendicular to EF.

Take now W halfway between R and T and draw WV

*The proofs in this and the following section are due to Hilbert, Neue Begründung der Bolyai-Lobatschefkyschen Geometrie, Math. Ann., Vol. LVII.

† Here and subsequently, we use the principle of superposition to abbreviate the proof. The theorems on congruence may of course be employed without the aid of any idea of mechanical motion.

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