By repeating this demonstration, we prove that if P is a point on LK such that

AP=m-AB,

where m is a positive integer, the line OP is perpendicular to LK at P and PO=AO. But only one perpendicular can be drawn at P to LK. Hence this perpendicular passes through 0. Now take D, so that

where n is a positive integer, and draw a line perpendicular to LK at D. If this perpendicular should intersect either BO or AO at a point O' in the segments BO or AO, then, by the demonstration just finished, BO and AO would also intersect at O', contrary to hypothesis. Hence this perpendicular passes through O and DO = AO.

It follows that if P is any point on LK such that

where m and n are positive integers, the perpendicular at P to LK passes through O and PO AO. Also, since by hypothesis, only one straight line can be drawn from P to O, the line PO is perpendicular to LK.

m

n

draw OP and OP1, and let pass through rational values

approaching as a limit.

4AP,O=lim

APO,

P10=lim PO.

But APO is always a right angle and PO is always equal to AO. Hence 4AP,O=rt. 4 and P10= AO.

Our theorem is therefore proved for the line LK. If L'K' is any other line, we may take A' and B' any two points on it,

and draw the perpendiculars A'O' and B'O', intersecting at O'. Take AB on LK so that AB=A'B'. The two triangies ABO and A'B'O' are congruent and A'O' 40. The distance AO is therefore independent of the line LK or of the position of the point A on the line. We will place OA=4.

A corollary of our theorem is that all straight lines are of constant length. For, from the proof we have used, it is evident that, if P is any point on AB,

АР ДАОР
АВДАОВ

Now if 44OP=2, the line OA coincides with OP, and AP becomes 7, the total length of the line. Then

15. All lines which pass through a point O meet again in a point O such that the distance O1 is constant.

Let 0 (Fig. 16) be any point and OA any line through 0. Take OA=1 (sec. 14) and draw LK perpendicular to AO. Let OB be any other line through O intersecting LK in B. Then OB is perpendicular to LK (sec. 14). Prolong AO to 01 so that AO1 = AO and draw 0,B. The triangles AOB and AO,B are congruent, since two sides and the included angle of one are congruent respectively to two sides and the included angle of the other. Hence

and

ABO1 = ABO=rt. 4,

O1B=OB=OA.

Therefore the line OBO1 is a straight line and

001=24.

Since all lines are of finite length (sec. 14) any line through O returns through 01 to 0. Two cases are usually considered.

First, the point O1 may coincide with O. The total length of a straight line is then 24 and any two lines have only one point in common.

Secondly, the point O1 may be distinct from O, but the lines 001 continued through O1 meet again in O. The total length of a line is then 44 and two lines meet in two points. The Riemannian geometry in this case is the same as the geometry on the surface of a sphere.

VI. THE SUM OF THE ANGLES OF A TRIANGLE

16. Consider any triangle ABC (Fig. 17). Take E, the middle point of AB, F the middle point of AC, and draw a straight line EF. From A, B, and C draw the lines AG, BK, and CL perpendicular to EF.

K

E

FL

G

B

C

FIG. 17.

=

In the right triangles AEG and EBK, EA EB and 4GEA=4BEK. Hence the two triangles are congruent and

BK= AG, 4KBE = 4GAE.

Similarly, the right triangles AGF and FLC are congruent and

If we define equivalent figures as those which may be divided into parts which are congruent in pairs, it appears that the triangle ABC is equivalent to the quadrilateral BCLK. Also, the sum of the angles of the triangle ABC is equal to the sum of the angles KBC and LCB of the quadrilateral BCLK.

This quadrilateral BCLK has two right angles, L and K, and two equal sides, KB and LC, adjacent to the right angles and opposite to each other. Such a figure we shall call an isosceles birectangular quadrilateral.

The study of the sum of the angles and of the area of a triangle is thus reduced to the study of an equivalent isosceles birectangular quadrilateral.

17. Let ABCD (Fig. 18) be an isosceles birectangular quadrilateral with right angles at A and B. For convenience, we

shall call AB the base, CD the summit, and C and D the summit angles of the quadrilateral.

Take L the middle point of the base and draw LK perpendicular to the base. Fold LBDK on LK as an axis. It is clear that the point D falls on C. Hence, the summit angles of an isosceles birectangular quadrilateral are equal. Also, LK is perpendicular to CD at its middle point K and the quadrilateral LBDK has three right angles.

Now through H, the middle point of LK, draw EF perpendicular to LK. Fold HFDK on HF as an axis. The point D will fall at B', B, or B" according as KD is less than, equal to,

or greater than LB. In these three cases 4D is greater than, equal to, or less than, B respectively. Hence:

Each summit angle of an isosceles birectangular quadrilateral is less than, equal to, or greater than, a right angle, according as the summit of the quadrilateral is greater than, equal to, or less than, the base.

18. In the Euclidean geometry each summit angle of an isosceles birectangular quadrilateral is equal to a right angle.

This is a familiar proposition of the Euclidean geometry and need not be proved here. We shall prove, however, the following theorem:

In the Lobachevskian and Riemannian geometries, a summitangle of an isosceles birectangular quadrilateral cannot equal a right angle.

Let ABCD (Fig. 19) be an isosceles birectangular quadrilateral with right angles at A and B. If possible, suppose 4C4D=rt. 4. Then (sec. 17) CD=AB. Take two points

M and N on AC and BD respectively so that CM = DN and draw MN.

Then ABMN is an isosceles birectangular quadrilateral with right angles at A and B. MNDC is an isosceles birectangular quadrilateral with right angles at C and D. Then MN must be perpendicular to AC and BD or we should have, by sec. 17, MN greater than one and less than the other of the two equal lines AB and CD, which is absurd.

Since M is any point between A and C it appears that the segments AC and BD are equidistant. By prolonging the lines AC and BD and considering congruent segments, it appears that the lines AC and BD are equidistant throughout their extent. Since this is impossible in the Lobachevskian and Riemannian geometries (secs. 11, 12, 13) the theorem is proved.

19. Each summit angle of an isosceles birectangular quadrilateral is less than a right angle in the Lobachevskian geometry and greater than a right angle in the Riemannian geometry.

In Fig. 18, the line CK measures the distance of the line AC from the line LK at the point C. In the Lobachevskian geometry, if the line AC is taken sufficiently long, CK>AL (sec. 12). If, therefore, CK were in any position less than AL, there would exist at least one other position in which CK = AL. This is impossible (sec. 18) and hence CK is always greater than AL and the angle C less than a right angle (sec. 17).

In the Riemannian geometry the lines AC and LK eventually intersect. Hence, if 4C is sufficiently long CK< AL, and therefore CK is always less than AL and the angle С greater than a right angle.

20. In the Euclidean, Lobachevskian, and Riemannian geometries respectively the sum of the angles of a triangle is equal to, less than, and greater than, two right angles.

We have seen in sec. 16 that the sum of the angles of a triangle is equal to that of the summit angles of an isosceles birectangular quadrilateral.

The theorem then follows from secs. 18, 19.