Theory of Parallels

A straight line fits upon itself in all its positions. By this I mean that during the revolution of the surface containing it the straight line does not change its place if it goes through two unmoving points in the surface: (i. e., if we turn the surface containing it about two points of the line, the line does not move.)

Two straight lines can not intersect in two points.

A straight line sufficiently produced both ways must go out beyond all bounds, and in such way cuts a bounded plain into two parts.

Two straight lines perpendicular to a third never intersect, how far soever they be produced.

A straight line always cuts another in going from one side of it over to the other side: (i. e., one straight line must cut another if it has points on both sides of it.)

Vertical angles, where the sides of one are productions of the sides of the other, are equal. This holds of plane rectilineal angles among themselves, as also of plane surface angles: (i. e., dihedral angles.)

Two straight lines can not intersect, if a third cuts them at the same angle.

In a rectilineal triangle equal sides lie opposite equal angles, and inversely.

In a rectilineal triangle, a greater side lies opposite a greater angle. In a right-angled triangle the hypothenuse is greater than either of the other sides, and the two angles adjacent to it are acute.

Rectilineal triangles are congruent if they have a side and two angles equal, or two sides and the included angle equal, or two sides and the angle opposite the greater equal, or three sides equal.

A straight line which stands at right angles upon two other straight lines not in one plane with it is perpendicular to all straight lines drawn through the common intersection point in the plane of those two.

The intersection of a sphere with a plane is a circle.

A straight line at right angles to the intersection of two perpendicular planes, and in one, is perpendicular to the other.

In a spherical triangle equal sides lie opposite equal angles, and inversely.

Spherical triangles are congruent (or symmetrical) if they have two sides and the included angle equal, or a side and the adjacent angles equal.

From here follow the other theorems with their explanations and proofs.

All straight lines which in a plane go out from a point can, with reference to a given straight line in the same plane, be divided into two classes—into cutting and not-cutting.

The boundary lines of the one and the other class of those lines will be called parallel to the given line.

From the point A (Fig. 1) let fall upon the line BC the perpendicular AD, to which again draw the perpendicular AE.

In the right angle EAD either will all straight lines which go out from the point A meet the line DC, as for example AF, or some of them, like the perpendicular AE, will not meet the line DC. In the uncertainty whether the perpendicular AE is the only line which does not meet DC, we will assume it may be possible that there are still other lines, for example AG, which do not cut DC, how far soever they may be prolonged. In passing over from the cutting lines, as AF, to the not-cutting lines, as AG, we must come upon a line AH, parallel to DC, a boundary line, upon one side of which all lines AG are such as do not meet the line DC, while upon the other side every straight line AF cuts the line DC.

The angle HAD between the parallel HA and the perpendicular AD is called the parallel angle (angle of parallelism), which we will here designate by $\Pi (p)$ for AD = p.

If $\Pi (p)$ is a right angle, so will the prolongation AE′ of the perpendicular AE likewise be parallel to the prolongation DB of the line DC, in addition to which we remark that in regard to the four right angles, which are made at the point A by the perpendiculars AE and AD, and their prolongations AE′ and AD′, every straight line which goes out from the point A, either itself or at least its prolongation, lies in one of the two right angles which are turned toward BC, so that except the parallel EE′ all others, if they are sufficiently produced both ways, must intersect the line BC.

If $\Pi (p)<\frac{1}{2}\pi$, then upon the other side of AD, making the same angle DAK = $\Pi (p)$ will lie also a line AK, parallel to the prolongation DB of the line DC, so that under this assumption we must also make a distinction of sides in parallelism.

All remaining lines or their prolongations within the two right angles turned toward BC pertain to those that intersect, if they lie within the angle HAK = $2\Pi (p)$ between the parallels; they pertain on the other hand to the non-intersecting AG, if they lie upon the other sides of the parallels AH and AK, in the opening of the two angles EAH = $\frac{1}{2}\pi -\Pi (p)$, E′AK = $\frac{1}{2}\pi -\Pi (p)$, between the parallels and EE′ the perpendicular to AD. Upon the other side of the perpendicular EE′ will in like manner the prolongations AH′ and AK′ of the parallels AH and AK likewise be parallel to BC; the remaining lines pertain, if in the angle K′AH′, to the intersecting, but if in the angles K′AE, H′AE′ to the non-intersecting.

In accordance with this, for the assumption $\Pi (p)=\frac{1}{2}\pi$ the lines can be only intersecting or parallel; but if we assume that $\Pi (p)<\frac{1}{2}\pi$, then we must allow two parallels, one on the one and one on the other side; in addition we must distinguish the remaining lines into non-intersecting and intersecting.

For both assumptions it serves as the mark of parallelism that the line becomes intersecting for the smallest deviation toward the side where lies the parallel, so that if AH is parallel to DC, every line AF cuts DC, how small soever the angle HAF may be.

A straight line maintains the characteristic of parallelism at all its points.

Given AB (Fig. 2) parallel to CD, to which latter AC is perpendicular. We will consider two points taken at random on the line AB and its production beyond the perpendicular.

Let the point E lie on that side of the perpendicular on which AB is looked upon as parallel to CD.

Let fall from the point E a perpendicular EK on CD and so draw EF that it falls within the angle BEK.

Connect the points A and F by a straight line, whose production then (by Theorem 16) must cut CD somewhere in G. Thus we get a triangle ACG, into which the line EF goes; now since this latter, from the construction, can not cut AC, and can not cut AG or EK a second time (Theorem 2), therefore it must meet CD somewhere at H (Theorem 3).

Now let E′ be a point on the production of AB and E′K′ perpendicular to the production of the line CD; draw the line E′F′ making so small an angle AE′F′ that it cuts AC somewhere in F′; making the same angle with AB, draw also from A the line AF, whose production will cut CD in G (Theorem 16.)

Thus we get a triangle AGC, into which goes the production of the line E′F′; since now this line can not cut AC a second time, and also can not cut AG, since the angle BAG = BE′G′, (Theorem 7), therefore must it meet CD somewhere in G′.

Therefore from whatever points E and E′ the lines EF and E′F′ go out, and however little they may diverge from the line AB, yet will they always cut CD, to which AB is parallel.

Two lines are always mutually parallel.

Let AC be a perpendicular on CD, to which AB is parallel if we draw from C the line CE making any acute angle ECD with CD, and let fall from A the perpendicular AF upon CE, we obtain a right-angled triangle ACF, in which AC, being the hypothenuse, is greater than the side AF (Theorem 9.)

Make AG = AF, and slide the figure EFAB until AF coincides with AG, when AB and FE will take the position AK and GH, such that the angle BAK = FAC, consequently AK must cut the line DC somewhere in K (Theorem 16), thus forming a triangle AKC, on one side of which the perpendicular GH intersects the line AK in L (Theorem 3), and thus determines the distance AL of the intersection point of the lines AB and CE on the line AB from the point A.

Hence it follows that CE will always intersect AB, how small soever may be the angle ECD, consequently CD is parallel to AB (Theorem 16.)

In a rectilineal triangle the sum of the three angles can not be greater than two right angles.

Suppose in the triangle ABC (Fig. 4) the sum of the three angles is equal to $\pi +a$; then choose in case of the inequality of the sides the smallest BC, halve it in D, draw from A through D the line AD and make the prolongation of it, DE, equal to AD, then join the point E to the point C by the straight line EC. In the congruent triangles ADB and CDE, the angle ABD = DCE, and BAD = DEC (Theorems 6 and 10); whence follows that also in the triangle ACE the sum of the three angles must be equal to $\pi +a$; but also the smallest angle BAC (Theorem 9) of the triangle ABC in passing over into the new triangle ACE has been cut up into the two parts EAC and AEC. Continuing this process, continually

halving the side opposite the smallest angle, we must finally attain to a triangle in which the sum of the three angles is $\pi +a$, but wherein are two angles, each of which in absolute magnitude is less than $\frac{1}{2}a$; since now, however, the third angle can not be greater than $\pi$, so must $a$ be either null or negative.

If in any rectilineal triangle the sum of the three angles is equal to two right angles, so is this also the case for every other triangle.

If in the rectilineal triangle ABC (Fig. 5) the sum of the three angles = $\pi$, then must at least two of its angles, A and C, be acute. Let fall from the vertex of the third angle B upon the opposite side AC the perpendicular p. This will cut the triangle into two right-angled triangles, in each of which the sum of the three angles must also be $\pi$, since it can not in either be greater than $\pi$, and in their combination not less than $\pi$.

So we obtain a right-angled triangle with the perpendicular sides p and q, and from this a quadrilateral whose opposite sides are equal and whose adjacent sides p and q are at right angles (Fig. 6.)

By repetition of this quadrilateral we can make another with sides np and q, and finally a quadrilateral ABCD with sides at right angles to each other, such that AB = np, AD = mq, DC = np, BC = mq, where m and n are any whole numbers. Such a quadrilateral is divided by the diagonal DB into two congruent right-angled triangles, BAD and BCD, in each of which the sum of the three angles = $\pi$.

The numbers n and m can be taken sufficiently great for the right-angled triangle ABC (Fig. 7) whose perpendicular sides AB = np, BC = mq, to enclose within itself another given (right-angled) triangle BDE as soon as the right-angles fit each other.

Drawing the line DC, we obtain right-angled triangles of which every successive two have a side in common.

The triangle ABC is formed by the union of the two triangles ACD and DCB, in neither of which can the sum of the angles be greater than $\pi$; consequently it must be equal to $\pi$, in order that the sum in the compound triangle may be equal to $\pi$.

In the same way the triangle BDC consists of the two triangles DEC and DBE, consequently must in DBE the sum of the three angles be equal to $\pi$, and in general this must be true for every triangle, since each can be cut into two right-angled triangles.

From this it follows that only two hypotheses are allowable: Either is the sum of the three angles in all rectilineal triangles equal to $\pi$, or this sum is in all less than $\pi$.

From a given point we can always draw a straight line that shall make with a given straight line an angle as small as we choose.

Let fall from the given point A (Fig. 8) upon the given line BC the perpendicular AB; take upon BC at random the point D; draw the line AD; make DE = AD, and draw AE.

In the right-angled triangle ABD let the angle ADB = $a$; then must in the isosceles triangle ADE the angle AED be either $\frac{1}{2}a$ or less (Theorems 8 and 20). Continuing thus we finally attain to such an angle, AEB, as is less than any given angle.

If two perpendiculars to the same straight line are parallel to each other, then the sum of the three angles in a rectilineal triangle is equal to two right angles.

Let the lines AB and CD (Fig. 9) be parallel to each other and perpendicular to AC.

Draw from A the lines AE and AF to the points E and F, which are taken on the line CD at any distances FC > EC from the point C.

Suppose in the right-angled triangle ACE the sum of the three angles is equal to $\pi -\alpha$, in the triangle AEF equal to $\pi -\beta$, then must it in triangle ACF equal $\pi -\alpha -\beta$, where $\alpha$ and $\beta$ can not be negative.

Further, let the angle BAF = a, AFC = b, so is $\alpha +\beta$ = a − b; now by revolving the line AF away from the perpendicular AC we can make the angle a between AF and the parallel AB as small as we choose; so also can we lessen the angle b, consequently the two angles $\alpha$ and $\beta$ can have no other magnitude than $\alpha =0$ and $\beta =0$.

It follows that in all rectilineal triangles the sum of the three angles is either $\pi$ and at the same time also the parallel angle $\Pi (p)=\frac{1}{2}\pi$ for every line p, or for all triangles this sum is $<\pi$ and at the same time also $\Pi (p)<\frac{1}{2}\pi$.

The first assumption serves as foundation for the ordinary geometry and plane trigonometry.

The second assumption can likewise be admitted without leading to any contradiction in the results, and founds a new geometric science, to which I have given the name Imaginary Geometry, and which I intend here to expound as far as the development of the equations between the sides and angles of the rectilineal and spherical triangle.

For every given angle $\alpha$ there is a line p such that $\Pi (p)=\alpha$.

Let AB and AC (Fig. 10) be two straight lines which at the intersection point A make the acute angle $\alpha$; take at random on AB a point

B′; from this point drop B′A′ at right angles to AC; make A′A″ = AA′; erect at A″ the perpendicular A″B″; and so continue until a perpendicular CD is attained, which no longer intersects AB. This must of necessity happen, for if in the triangle AA′B′ the sum of all three angles is equal to $\pi$ − a, then in the triangle AB′A″ it equals $\pi$ − 2a, in triangle AA″B″ less than $\pi$ − 2a (Theorem 20), and so forth, until it finally becomes negative and thereby shows the impossibility of constructing the triangle.

The perpendicular CD may be the very one nearer than which to the point A all others cut AB; at least in the passing over from those that cut to those not cutting such a perpendicular FG must exist.

Draw now from the point F the line FH, which makes with FG the acute angle HFG, on that side where lies the point A. From any point H of the line FH let fall upon AC the perpendicular HK, whose prolongation consequently must cut AB somewhere in B, and so makes a triangle AKB, into which the prolongation of the line FH enters, and therefore must meet the hypothenuse AB somewhere in M. Since the angle GFH is arbitrary and can be taken as small as we wish, therefore FG is parallel to AB and AF = p. (Theorems 16 and 18.)

One easily sees that with the lessening of p the angle $\alpha$ increases, while, for p = 0, it approaches the value $\frac{1}{2}\pi$; with the growth of p the angle $\alpha$ decreases, while it continually approaches zero for p = $\infty$.

Since we are wholly at liberty to choose what angle we will under-

stand by the symbol $\Pi (p)$ when the line p is expressed by a negative number, so we will assume

an equation which shall hold for all values of p, positive as well as negative, and for p = 0.

The farther parallel lines are prolonged on the side of their parallelism, the more they approach one another.

If to the line AB (Fig. 11) two perpendiculars AC = BD are erected and their end-points C and D joined by a straight line, then will the quadrilateral CABD have two right angles at A and B, but two acute angles at C and D (Theorem 22) which are equal to one another, as we can easily see by thinking the quadrilateral superimposed upon itself so that the line BD falls upon AC and AC upon BD.

Halve AB and erect at the mid-point E the line EF perpendicular to AB. This line must also be perpendicular to CE, since the quadrilaterals CAEF and FDBE fit one another if we so place one on the other that the line EF remains in the same position. Hence the line CD can not be parallel to AB, but the parallel to AB for the point C, namely CG, must incline toward AB (Theorem 16) and cut from the perpendicular BD a part BG < CA.

Since C is a random point in the line CG, it follows that CG itself nears AB the more the farther it is prolonged.

Two straight lines which are parallel to a third are also parallel to each other.

We will first assume that the three lines AB, CD, EF (Fig. 12) lie in one plane. If two of them in order, AB and CD, are parallel to the outmost one, EF, so are AB and CD parallel to each other. In order to prove this, let fall from any point A of the outer line AB upon the other outer line FE, the perpendicular AE, which will cut the middle line CD in some point C (Theorem 3), at an angle DCE $<\frac{1}{2}\pi$ on the side toward EF, the parallel to CD (Theorem 22).

A perpendicular AG let fall upon CD from the same point, A, must fall within the opening of the acute angle ACG (Theorem 9); every other line AH from A drawn within the angle BAC must cut EF, the parallel to AB, somewhere in H, how small soever the angle BAH may be; consequently will CD in the triangle AEH cut the line AH somewhere in K, since it is impossible that it should meet EF. If AH from the point A went out within the angle CAG, then must it cut the prolongation of CD between the points C and G in the triangle CAG. Hence follows that AB and CD are parallel (Theorems 16 and 18).

Were both the outer lines AB and EF assumed parallel to the middle line CD, so would every line AK from the point A, drawn within the angle BAE, cut the line CD somewhere in the point K, how small soever the angle BAK might be.

Upon the prolongation of AK take at random a point L and join it

with C by the line CL, which must cut EF somewhere in M, thus making a triangle MCE.

The prolongation of the line AL within the triangle MCE can cut neither AC nor CM a second time, consequently it must meet EF somewhere in H; therefore AB and EF are mutually parallel.

Now let the parallels AB and CD (Fig. 13) lie in two planes whose intersection line is EF. From a random point E of this latter let fall a perpendicular EA upon one of the two parallels, e. g., upon AB, then from A, the foot of the perpendicular EA, let fall a new perpendicular AC upon the other parallel CD and join the end-points E and C of the two perpendiculars by the line EC. The angle BAC must be acute (Theorem 22), consequently a perpendicular CG from C let fall upon AB meets it in the point G upon that side of CA on which the lines AB and CD are considered as parallel.

Every line EH [in the plane FEAB], however little it diverges from EF, pertains with the line EC to a plane which must cut the plane of the two parallels AB and CD along some line CH. This latter line cuts AB somewhere, and in fact in the very point H which is common to all three planes, through which necessarily also the line EH goes; consequently EF is parallel to AB.

In the same way we may show the parallelism of EF and CD.

Therefore the hypothesis that a line EF is parallel to one of two other parallels, AB and CD, is the same as considering EF as the intersection of two planes in which two parallels, AB, CD, lie.

Consequently two lines are parallel to one another if they are parallel to a third line, though the three be not co-planar.

The last theorem can be thus expressed:

Three planes intersect in lines which are all parallel to each other if the parallelism of two is pre-supposed.

Triangles standing opposite to one another on the sphere are equivalent in surface.

By opposite triangles we here understand such as are made on both sides of the center by the intersections of the sphere with planes; in such triangles, therefore, the sides and angles are in contrary order.

In the opposite triangles ABC and A′B′C′ (Fig. 14, where one of them must be looked upon as represented turned about), we have the sides AB = A′B′, BC = B′C′, CA = C′A′, and the corresponding angles at the points A, B, C are likewise equal to those in the other triangle at the points A′, B′, C′.

Through the three points A, B, C, suppose a plane passed, and upon it from the center of the sphere a perpendicular dropped whose prolongations both ways cut both opposite triangles in the points D and D′ of the sphere. The distances of the first D from the points ABC, in arcs of great circles on the sphere, must be equal (Theorem 12) as well to each other as also to the distances D′A′, D′B′, D′C′, on the other triangle (Theorem 6), consequently the isosceles triangles about the points D and D′ in the two spherical triangles ABC and A′B′C′ are congruent.

In order to judge of the equivalence of any two surfaces in general, I take the following theorem as fundamental:

Two surfaces are equivalent when they arise from the mating or separating of equal parts.

A three-sided solid angle equals the half sum of the surface angles less a right-angle.

In the spherical triangle ABC (Fig. 15), where each side $<\pi$, designate the angles by A, B, C; prolong the side AB so that a whole circle ABA′B′A is produced; this divides the sphere into two equal parts.

In that half in which is the triangle ABC, prolong now the other two sides through their common intersection point C until they meet the circle in A′ and B′.

In this way the hemisphere is divided into four triangles, ABC, ACB′, B′CA′, A′CB, whose size may be designated by P, X, Y, Z. It is evident that here $\mathrm{P}+\mathrm{X}=\mathrm{B}$, $\mathrm{P}+\mathrm{Z}=\mathrm{A}$.

The size of the spherical triangle Y equals that of the opposite triangle ABC′, having a side AB in common with the triangle P, and whose third angle C′ lies at the end-point of the diameter of the sphere which goes from C through the center D of the sphere (Theorem 26). Hence it follows that

$\mathrm{P}+\mathrm{Y}=\mathrm{C}$, and since $\mathrm{P}+\mathrm{X}+\mathrm{Y}+\mathrm{Z}=\pi$, therefore we have also

We may attain to the same conclusion in another way, based solely upon the theorem about the equivalence of surfaces given above. (Theorem 26.)

In the spherical triangle ABC (Fig. 16), halve the sides AB and BC, and through the mid-points D and E draw a great circle; upon this let fall from A, B, C the perpendiculars AF, BH, and CG. If the perpendicular from B falls at H between D and E, then will of the triangles so made BDH = AFD, and BHE = EGC (Theorems 6 and 15), whence follows that the surface of the triangle ABC equals that of the quadrilateral AFGC (Theorem 26).

If the point H coincides with the middle point E of the side BC (Fig. 17), only two equal right-angled triangles, ADF and BDE, are made, by whose interchange the equivalence of the surfaces of the triangle ABC and the quadrilateral AFEC is established.

If, finally, the point H falls outside the triangle ABC (Fig. 18), the perpendicular CG goes, in consequence, through the triangle, and so we go over from the triangle ABC to the quadrilateral AFGC by adding the triangle FAD = DBH, and then taking away the triangle CGE = EBH.

Supposing in the spherical quadrilateral AFGC a great circle passed through the points A and G, as also through F and C, then will their arcs between AG and FC equal one another (Theorem 15), consequently also the triangles FAC and ACG be congruent (Theorem 15), and the angle FAC equal the angle ACG.

Hence follows, that in all the preceding cases, the sum of all three angles of the spherical triangle equals the sum of the two equal angles in the quadrilateral which are not the right angles.

Therefore we can, for every spherical triangle, in which the sum of the three angles is S, find a quadrilateral with equivalent surface, in which are two right angles and two equal perpendicular sides, and where the two other angles are each $\frac{1}{2}\mathrm{S}$.

Let now ABCD (Fig. 19) be the spherical quadrilateral, where the sides AB = DC are perpendicular to BC, and the angles A and D each $\frac{1}{2}\mathrm{S}$.

Prolong the sides AD and BC until they cut one another in E, and further beyond E, make DE = EF and let fall upon the prolongation of BC the perpendicular FG. Bisect the whole arc BG and join the mid-point H by great-circle-arcs with A and F.

The triangles EFG and DCE are congruent (Theorem 15), so FG = DC = AB.

The triangles ABH and HGF are likewise congruent, since they are right angled and have equal perpendicular sides, consequently AH and AF pertain to one circle, the arc AHF = $\pi$, ADEF likewise = $\pi$, the angle $\mathrm{HAD}=\mathrm{HFE}=\frac{1}{2}\mathrm{S}-\mathrm{BAH}=\frac{1}{2}\mathrm{S}-\mathrm{HFG}=\frac{1}{2}\mathrm{S}-\mathrm{HFE}-\mathrm{EFG}=\frac{1}{2}\mathrm{S}-\mathrm{HAD}-\pi +\frac{1}{2}\mathrm{S}$; consequently, angle $\mathrm{HFE}=\frac{1}{2}(\mathrm{S}-\pi )$; or what is the same, this equals the size of the lune AHFDA, which again is equal to the quadrilateral ABCD, as we easily see if we pass over from the one to the other by first adding the triangle EFG and then BAH and thereupon taking away the triangles equal to them DCE and HFG.

Therefore $\frac{1}{2}(\mathrm{S}-\pi )$ is the size of the quadrilateral ABCD and at the same time also that of the spherical triangle in which the sum of the three angles is equal to S.

If three planes cut each other in parallel lines, then the sum of the three surface angles equals two right angles.

Let AA′, BB′ CC′ (Fig. 20) be three parallels made by the intersection of planes (Theorem 25). Take upon them at random three points A, B, C, and suppose through these a plane passed, which consequently will cut the planes of the parallels along the straight lines AB, AC, and BC. Further, pass through the line AC and any point D on the BB′, another plane, whose intersection with the two planes of the parallels AA′ and BB′, CC′ and BB′ produces the two lines AD and DC, and whose inclination to the third plane of the parallels AA′ and CC′ we will designate by $w$.

The angles between the three planes in which the parallels lie will be designated by X, Y, Z, respectively at the lines AA′, BB′, CC′; finally call the linear angles BDC = a, ADC = b, ADB = c.

About A as center suppose a sphere described, upon which the intersections of the straight lines AC, AD AA′ with it determine a spherical triangle, with the sides p, q, and r. Call its size $\alpha$. Opposite the side q lies the angle $w$, opposite r lies X, and consequently opposite p lies the angle $\pi +2\alpha -w-\mathrm{X}$, (Theorem 27).

In like manner CA, CD, CC′ cut a sphere about the center C, and determine a triangle of size $\beta$, with the sides p′, q′, r′, and the angles, $w$ opposite q′, Z opposite r′, and consequently $\pi +2\beta -w-\mathrm{Z}$ opposite p′.

Finally is determined by the intersection of a sphere about D with the lines DA, DB, DC, a spherical triangle, whose sides are l, m, n, and the angles opposite them $w+\mathrm{Z}-2\beta$, $w+\mathrm{X}-2\alpha$, and Y. Consequently its size $\delta =\frac{1}{2}(\mathrm{X}+\mathrm{Y}+\mathrm{Z}-\pi )-\alpha -\beta +w$.

Decreasing $w$ lessens also the size of the triangles $\alpha$ and $\beta$, so that $\alpha +\beta -w$ can be made smaller than any given number.

In the triangle $\delta$ can likewise the sides l and m be lessened even to vanishing (Theorem 21), consequently the triangle $\delta$ can be placed with one of its sides l or m upon a great circle of the sphere as often as you choose without thereby filling up the half of the sphere, hence $\delta$ vanishes together with $w$; whence follows that necessarily we must have

In a rectilineal triangle, the perpendiculars erected at the mid-points of the sides either do not meet, or they all three cut each other in one point.

Having pre-supposed in the triangle ABC (Fig. 21), that the two perpendiculars ED and DF, which are erected upon the sides AB and BC at their mid points E and F, intersect in the point D, then draw within the angles of the triangle the lines DA, DB, DC.

In the congruent triangles ADE and BDE (Theorem 10), we have AD = BD, thus follows also that BD = CD; the triangle ADC is hence isosceles, consequently the perpendicular dropped from the vertex D upon the base AC falls upon G the mid-point of the base.

The proof remains unchanged also in the case when the intersection point D of the two perpendiculars ED and FD falls in the line AC itself, or falls without the triangle.

In case we therefore pre-suppose that two of those perpendiculars do not intersect, then also the third can not meet with them.

The perpendiculars which are erected upon the sides of a rectilineal triangle at their mid-points, must all three be parallal to each other, so soon as the parallelism of two of them is pre-supposed.

In the triangle ABC (Fig. 22) let the lines DE, FG, HK, be erected perpendicular upon the sides at their mid-points D, F, H. We will in the first place assume that the two perpendiculars DE and FG are parallel, cutting the line AB in L and M, and that the perpendicular HK lies between them. Within the angle BLE draw from the point L, at random, a straight line LG, which must cut FG somewhere in G, how small soever the angle of deviation GLE may be. (Theorem 16).

Since in the triangle LGM the perpendicular HK can not meet with MG (Theorem 29), therefore it must cut LG somewhere in P, whence follows, that HK is parallel to DE (Theorem 16), and to MG (Theorems 18 and 25).

Put the side BC = 2a, AC = 2b, AB = 2c, and designate the angles opposite these sides by A, B, C, then we have in the case just considered

as one may easily show with help of the lines AA′, BB′, CC′, which are drawn from the points A, B, C, parallel to the perpendicular HK and consequently to both the other perpendiculars DE and FG (Theorems 23 and 25).

Let now the two perpendiculars HK and FG be parallel, then can the third DE not cut them (Theorem 29), hence is it either parallel to them, or it cuts AA′.

The last assumption is not other than that the angle

If we lessen this angle, so that it becomes equal to $\Pi (a)+\Pi (b)$, while we in that way give the line AC the new position CQ, (Fig. 23), and designate the size of the third side BQ by 2c′, then must the angle CBQ at the point B, which is increased, in accordance with what is proved above, be equal to

whence follows c′ > c (Theorem 23).

In the triangle ACQ are, however, the angles at A and Q equal, hence in the triangle ABQ must the angle at Q be greater than that at the point A, consequently is AB > BQ, (Theorem 9); that is c > c′.

We call boundary line (oricycle) that curve lying in a plane for which all perpendiculars erected at the mid-points of chords are parallel to each other.

In conformity with this definition we can represent the generation of a boundary line, if we draw to a given line AB (Fig. 24) from a given point A in it, making different angles CAB = $\Pi (a)$, chords AC = 2a; the end C of such a chord will lie on the boundary line, whose points we can thus gradually determine.

The perpendicular DE erected upon the chord AC at its mid-point D will be parallel to the line AB, which we will call the Axis of the boundary line. In like manner will also each perpendicular FG erected at the mid-point of any chord AH, be parallel to AB, consequently must this peculiarity also pertain to every perpendicular KL in general which is erected at the mid-point K of any chord CH, between whatever points C and H of the boundary line this may be drawn (Theorem 30). Such perpendiculars must therefore likewise, without distinction from AB, be called Axes of the boundary line.

A circle with continually increasing radius merges into the boundary line.

Given AB (Fig. 25) a chord of the boundary line; draw from the end-points A and B of the chord two axes AC and BF, which consequently will make with the chord two equal angles BAC = ABF = $\alpha$ (Theorem 31).

Upon one of these axes AC, take anywhere the point E as center of a circle, and draw the arc AF from the initial point A of the axis AC to its intersection point F with the other axis BF.

The radius of the circle, FE, corresponding to the point F will make on the one side with the chord AF an angle AFE = $\beta$, and on the

other side with the axis BF, the angle EFD = $\gamma$. It follows that the angle between the two chords BAF = $\alpha -\beta <\beta +\gamma -\alpha$ (Theorem 22); whence follows, $\alpha -\beta <\frac{1}{2}\gamma$.

Since now however the angle $\gamma$ approaches the limit zero, as well in consequence of a moving of the center E in the direction AC, when F remains unchanged, (Theorem 21), as also in consequence of an approach of F to B on the axis BF, when the center E remains in its position (Theorem 22), so it follows, that with such a lessening of the angle $\gamma$, also the angle $\alpha -\beta$, or the mutual inclination of the two chords AB and AF, and hence also the distance of the point B on the boundary line from the point F on the circle, tends to vanish.

Consequently one may also call the boundary-line a circle with infinitely great radius.

Let AA′ = BB′ = $x$ (Figure 26). be two lines parallel toward the side from A to A′, which parallels serve as axes for the two boundary arcs (arcs on two boundary lines) AB = $s$, A′B′ = $s^{\prime }$, then is

where $e$ is independent of the arcs $s$, $s^{\prime }$ and of the straight line $x$, the distance of the arc $s^{\prime }$ from $s$.

In order to prove this, assume that the ratio of the arc $s$ to $s^{\prime }$ is equal to the ratio of the two whole numbers $n$ and $m$.

Between the two axes AA′, BB′ draw yet a third axis CC′, which so cuts off from the arc AB a part AC = $t$ and from the arc A′B′ on the same side, a part A′C′ = $t^{\prime }$. Assume the ratio of $t$ to $s$ equal to that of the whole numbers $p$ and $q$, so that

Divide now $s$ by axes into $nq$ equal parts, then will there be $mq$ such parts on $s^{\prime }$ and $np$ on $t$.

However there correspond to these equal parts on $s$ and $t$ likewise equal parts on $s^{\prime }$ and $t^{\prime }$, consequently we have

Hence also wherever the two arcs $t$ and $t^{\prime }$ may be taken between the two axes AA′ and BB′, the ratio of $t$ to $t^{\prime }$ remains always the same, as

long as the distance $x$ between them remains the same. If we therefore for $x=1$, put $s=e{s}^{\prime }$, then we must have for every $x$

Since $e$ is an unknown number only subjected to the condition $e>1$, and further the linear unit for $x$ may be taken at will, therefore we may, for the simplification of reckoning, so choose it that by $e$ is to be understood the base of Napierian logarithms.

We may here remark, that $s^{\prime }=0$ for $x=\infty$, hence not only does the distance between two parallels decrease (Theorem 24), but with the prolongation of the parallels toward the side of the parallelism this at last wholly vanishes. Parallel lines have therefore the character of asymptotes.

Boundary surface (orisphere) we call that surface which arises from the revolution of the boundary line about one of its axes, which, together with all other axes of the boundary-line, will be also an axis of the boundary-surface.

A chord is inclined at equal angles to such axes drawn through its end-points, wheresoever these two end-points may be taken on the boundary-surface.

Let A, B, C, (Fig. 27), be three points on the boundary-surface; AA′, the axis of revolution, BB′ and CC′ two other axes, hence AB and AC chords to which the axes are inclined at equal angles A′AB = B′BA, A′AC = C′CA (Theorem 31.)

Two axes BB′, CC′, drawn through the end-points of the third chord BC, are likewise parallel and lie in one plane, (Theorem 25).

A perpendicular DD′ erected at the mid-point D of the chord AB and in the plane of the two parallels AA′, BB′, must be parallel to the three axes AA′, BB′, CC′, (Theorems 23 and 25); just such a perpendicular EE′ upon the chord AC in the plane of the parallels AA′, CC′ will be parallel to the three axes AA′, BB′, CC′, and the perpendicular DD′. Let now the angle between the plane in which the parallels AA′ and BB′ lie, and the plane of the triangle ABC be designated by $\Pi (a)$, where $a$ may be positive, negative or null. If $a$ is positive, then erect FD = $a$ within the triangle ABC, and in its plane, perpendicular upon the chord AB at its mid-point D.

Were $a$ a negative number, then must FD = $a$ be drawn outside the triangle on the other side of the chord AB; when $a=0$, the point F coincides with D.

In all cases arise two congruent right-angled triangles AFD and DFB, consequently we have FA = FB.

Erect now at F the line FF′ perpendicular to the plane of the triangle ABC.

Since the angle D′DF = $\Pi (a)$, and DF = $a$, so FF′ is parallel to DD′ and the line EE′, with which also it lies in one plane perpendicular to the plane of the triangle ABC.

Suppose now in the plane of the parallels EE′, FF′ upon EF the perpendicular EK erected, then will this be also at right angles to the plane of the triangle ABC, (Theorem 13), and to the line AE lying in this plane, (Theorem 11); and consequently must AE, which is perpendicular to EK and EE′, be also at the same time perpendicular to FE, (Theorem 11). The triangles AEF and FEC are congruent, since they are right-angled and have the sides about the right angles equal, hence is

AF = FC = FB.

A perpendicular from the vertex F of the isosceles triangle BFC let fall upon the base BC, goes through its mid-point G; a plane passed through this perpendicular FG and the line FF′ must be perpendicular to the plane of the triangle ABC, and cuts the plane of the parallels BB′, CC′, along the line GG′, which is likewise parallel to BB′ and CC′, (Theorem 25); since now CG is at right angles to FG, and hence at the same time also to GG′, so consequently is the angle C′CG = B′BG, (Theorem 23).

Hence follows, that for the boundary-surface each of the axes may be considered as axis of revolution.

Principal-plane we will call each plane passed through an axis of the boundary surface.

Accordingly every Principal-plane cuts the boundary-surface in the boundary line, while for another position of the cutting plane this intersection is a circle.

Three principal planes which mutually cut each other, make with each other angles whose sum is $\pi$, (Theorem 28).

These angles we will consider as angles in the boundary-triangle whose sides are arcs of the boundary-line, which are made on the boundary surface by the intersections with the three principal planes. Consequently the same interdependence of the angles and sides pertains to the boundary-triangles, that is proved in the ordinary geometry for the rectilineal triangle.

In what follows, we will designate the size of a line by a letter with an accent added, e. g. $x^{\prime }$, in order to indicate that this has a relation to that of another line, which is represented by the same letter without accent $x$, which relation is given by the equation

Let now ABC (Fig. 28) be a rectilineal right-angled triangle, where the hypothenuse AB = c, the other sides AC = b, BC = a, and the angles opposite them are

At the point A erect the line AA′ at right angles to the plane of the triangle ABC, and from the points B and C draw BB′ and CC′ parallel to AA′.

The planes in which these three parallels lie make with each other the angles: $\Pi (\alpha )$ at AA′, a right angle at CC′ (Theorems 11 and 13), consequently $\Pi ({\alpha }^{\prime })$ at BB′ (Theorem 28).

The intersections of the lines BA, BC, BB′ with a sphere described about the point B as center, determine a spherical triangle mnk, in which the sides are mn = $\Pi (c)$, kn = $\Pi (\beta )$, mk = $\Pi (a)$ and the opposite angles are $\Pi (b)$, $\Pi ({\alpha }^{\prime })$, $\frac{1}{2}\pi$.

Therefore we must, with the existence of a rectilineal triangle whose sides are a, b, c and the opposite angles $\Pi (\alpha )$, $\Pi (\beta )$ $\frac{1}{2}\pi$, also admit the existence of a spherical triangle (Fig. 29) with the sides $\Pi (c)$, $\Pi (\beta )$, $\Pi (a)$ and the opposite angles $\Pi (b)$, $\Pi ({\alpha }^{\prime })$, $\frac{1}{2}\pi$.

Of these two triangles, however, also inversely the existence of the spherical triangle necessitates anew that of a rectilineal, which in consequence, also can have the sides a, ${\alpha }^{\prime }$, $\beta$, and the oppsite angles $\Pi (b^{\prime })$, $\Pi (c)$, $\frac{1}{2}\pi$.

Hence we may pass over from a, b, c, $\alpha$, $\beta$, to b, a, c, $\beta$, $\alpha$, and also to a, ${\alpha }^{\prime }$, $\beta$, b′, c.

Suppose through the point A (Fig. 28) with AA′ as axis, a boundary-surface passed, which cuts the two other axes BB′, CC′, in B″ and C″, and whose intersections with the planes the parallels form a boundary-triangle, whose sides are B″C″ = $p$, C″A = $q$, B″A = $r$, and the angles opposite them $\Pi (\alpha )$, $\Pi ({\alpha }^{\prime })$, $\frac{1}{2}\pi$, and where consequently (Theorem 34):

Now break the connection of the three principal-planes along the line BB′, and turn them out from each other so that they with all the lines lying in them come to lie in one plane, where consequently the arcs $p$, $q$, $r$ will unite to a single arc of a boundary-line, which goes through the

point A and has AA′ for axis, in such a manner that (Fig. 30) on the one side will lie, the arcs $q$ and $p$, the side b of the triangle, which is perpendicular to AA′ at A, the axis CC′ going from the end of b parallel to AA′ and through C″ the union point of $p$ and $q$, the side a perpendicular to CC′ at the point C, and from the end-point of a the axis BB′ parallel to AA′ which goes through the end-point B″ of the arc $p$.

On the other side of AA′ will lie, the side c perpendicular to AA′ at the point A, and the axis BB′ parallel to AA′, and going through the end-point B″ of the arc $r$ remote from the end point of b.

The size of the line CC″ depends upon b, which dependence we will express by CC″ = $f(b)$.

In like manner we will have BB″ = $f(c)$.

If we describe, taking CC′ as axis, a new boundary line from the point C to its intersection D with the axis BB′ and designate the arc CD by $t$, then is BD = $f(a)$.

BB″ = BD + DB″ = BD + CC″, consequently

Moreover, we perceive, that (Theorem 33)

If the perpendicular to the plane of the triangle ABC (Fig. 28) were erected at B instead of at the point A, then would the lines c and $r$ remain the same, the arcs $q$ and $t$ would change to $t$ and $q$, the straight lines a

and b into b and a, and the angle $\Pi (\alpha )$ into $\Pi (\beta )$, consequently we would have

whence follows by substituting the value of $q$,

and if we change $\alpha$ and $\beta$ into b′ and c,

further, by multiplication with $e^{f(b)}$

Hence follows also

Since now, however, the straight lines a and b are independent of one another, and moreover, for b = 0, $f(b)=0$, $\Pi (b)=\frac{1}{2}\pi$, so we have for every straight line a

Therefore,

Hence we obtain besides by mutation of the letters

If we designate in the right-angled spherical triangle (Fig. 29) the sides $\Pi (c)$, $\Pi (\beta )$, $\Pi (a)$, with the opposite angles $\Pi (b)$, $\Pi ({\alpha }^{\prime })$, by the letters a, b, c, A, B, then the obtained equations take on the form of those which we know as proved in spherical trigonometry for the right-angled triangle, namely,

from which equations we can pass over to those for all spherical triangles in general.

Hence spherical trigonometry is not dependent upon whether in a

rectilineal triangle the sum of the three angles is equal to two right angles or not.

We will now consider anew the right-angled rectilineal triangle ABC (Fig. 31), in which the sides are a, b, c, and the opposite angles $\Pi (\alpha )$, $\Pi (\beta )$, $\frac{1}{2}\pi$.

Prolong the hypothenuse c through the point B, and make BD = $\beta$; at the point D erect upon BD the perpendicular DD′, which consequently will be parallel to BB′, the prolongation of the side a beyond the point B. Parallel to DD′ from the point A draw AA′, which is at the same time also parallel to CB′, (Theorem 25), therefore is the angle

If from B we lay off $\beta$ on the hypothenuse c, then at the end point D, (Fig. 32), within the triangle erect upon AB the perpendicular DD′, and from the point A parallel to DD′ draw AA′, so will BC with its prolongation CC′ be the third parallel; then is, angle CAA′ = $\Pi (b)$, DAA′ = $\Pi (c-\beta )$, consequently $\Pi (c-\beta )=\Pi (\alpha )+\Pi (b)$. The last equation is then also still valid, when c = $\beta$, or c $<\beta$.

If c = $\beta$ (Fig. 33), then the perpendicular AA′ erected upon AB at the point A

is parallel to the side BC = a, with its prolongation, CC′, consequently we have $\Pi (\alpha )+\Pi (b)=\frac{1}{2}\pi$, whilst also $\Pi (c-\beta )=\frac{1}{2}\pi$, (Theorem 23).

If c $<\beta$, then the end of $\beta$ falls beyond the point A at D (Fig. 34) upon the prolongation of the hypothenuse AB. Here the perpendicular DD′ erected upon AD, and the line AA′ parallel to it from A, will likewise be parallel to the side BC = a, with its prolongation CC′.

Here we have the angle DAA′ = $\Pi (\beta -c)$, consequently

(Theorem 23).

The combination of the two equations found gives,

whence follows

Substituting here the value, (Theorem 35)

we have

Since here $\beta$ is an arbitrary number, as the angle $\Pi (\beta )$ at the one

side of c may be chosen at will between the limits 0 and $\frac{1}{2}\pi$, consequently $\beta$ between the limits 0 and $\infty$, so we may deduce by taking consecutively $\beta$ = c, 2c, 3c, &c., that for every positive number $n$, ${\left[\tan \frac{1}{2}\Pi (c)\right]}^n=\tan \frac{1}{2}\Pi (nc)$.

If we consider $n$ as the ratio of two lines x and c, and assume that $\cot \frac{1}{2}\Pi (c)=e^c$, then we find for every line x in general, whether it be positive or negative,

where $e$ may be any arbitrary number, which is greater than unity, since $\Pi (x)=0$ for x = $\infty$.

Since the unit by which the lines are measured is arbitrary, so we may also understand by $e$ the base of the Napierian Logarithms.

Of the equations found above in Theorem 35 it is sufficient to know the two following,

applying the latter to both the sides a and b about the right angle, in order from the combination to deduce the remaining two of Theorem 35, without ambiguity of the algebraic sign, since here all angles are acute.

In a similar manner we attain the two equations

We will now consider a rectilineal triangle whose sides are a, b, c, (Fig. 35) and the opposite angles A, B, C.

If A and B are acute angles, then the perpendicular p from the vertex of the angle C falls within the triangle and cuts the side c into two parts, x on the side of the angle A and c − x on the side of the angle B. Thus arise two right-angled triangles, for which we obtain, by application of equation (1),

which equations remain unchanged also when one of the angles, e. g. B, is a right angle (Fig. 36) or and obtuse angle (Fig. 37).

Therefore we have universally for every triangle

For a triangle with acute angles A, B, (Fig. 35) we have also (Equation 2),

which equations also relate to triangles, in which one of the angles A or B is a right angle or an obtuse angle.

As example, for B = $\frac{1}{2}\pi$ (Fig. 36) we must take x = c, the first equation then goes over into that which we have found above as Equation 2, the other, however, is self-sufficing.

For B $>\frac{1}{2}\pi$ (Fig. 37) the first equation remains unchanged, instead of the second, however, we must write correspondingly

but we have

(Theorem 23), and also $\cos (\pi -\mathrm{B})=-\cos \mathrm{B}$.

If A is a right or an obtuse angle, then must c − x and x be put for x and c − x, in order to carry back this case upon the preceding.

In order to eliminate x from both equations, we notice that (Theorem 36)

If we substitute here the expression for $\cos \Pi (x)$, $\cos \Pi (c-x)$, we obtain

whence follows

and finally

In the same way we must also have

From these three equations we find

Hence follows without ambiguity of sign,

If we substitute here the value of $\sin \Pi (c)$ corresponding to equation (3.)

then we obtain

but by substituting this expression for $\cos \Pi (c)$ in equation (4),

By elimination of $\sin \Pi (b)$ with help of the equation (3) comes

In the meantime the equation (6) gives by changing the letters,

From the last two equations follows,

All four equations for the interdependence of the sides a, b, c, and the opposite angles A, B, C, in the rectilineal triangle will therefore be, [Equations (3), (5), (6), (7).]

If the sides a, b, c, of the triangle are very small, we may content ourselves with the approximate determinations. (Theorem 36.)

and in like manner also for the other sides b and c.

The equations 8 pass over for such triangles into the following:

Of these equations the first two are assumed in the ordinary geometry; the last two lead, with the help of the first, to the conclusion

Therefore the imaginary geometry passes over into the ordinary, when we suppose that the sides of a rectilineal triangle are very small.

I have, in the scientific bulletins of the University of Kasan, published certain researches in regard to the measurement of curved lines, of plane figures, of the surfaces and the volumes of solids, as well as in relation to the application of imaginary geometry to analysis.

The equations (8) attain for themselves already a sufficient foundation for considering the assumption of imaginary geometry as possible. Hence there is no means, other than astronomical observations, to use

for judging of the exactitude which pertains to the calculations of the ordinary geometry.

This exactitude is very far-reaching, as I have shown in one of my investigations, so that, for example, in triangles whose sides are attainable for our measurement, the sum of the three angles is not indeed different from two right-angles by the hundredth part of a second.

In addition, it is worthy of notice that the four equations (8) of plane geometry pass over into the equations for spherical triangles, if we put $a\sqrt{-1}$, $b\sqrt{-1}$, $c\sqrt{-1}$, instead of the sides a, b, c; with this change, however, we must also put

and similarly also for the sides b and c.

In this manner we pass over from equations (8) to the following: