Proof. Select in the ordinary plane geometry (the possibility of which has already been demonstrated in Section 9, pp. §–§) any two straight lines perpendicular to each other as the axes of $x$ and $y$. Construct about the origin $O$ of this system of co-ordinates an ellipse having the major and minor axes equal to 1 and $\frac{1}{2}$, respectively. Finally, let $F$ denote the point situated upon the positive $x$-axis at the distance $\frac{3}{2}$ from $O$. Consider all of the circles which cut the ellipse in four real points. These points may be either distinct or in any way coincident. Of all the points situated upon these circles, we shall attempt to determine the one which lies upon the $x$-axis farthest from the origin. For this purpose, let us begin with an arbitrary circle cutting the ellipse in four distinct points and intersecting the positive $x$-axis in the point $C$. Suppose this circle then turned about the point $C$ in such a manner that two or more of the four points of intersection with the ellipse finally coincide in a single point $A$, while the rest of them remain real. Increase now the resulting tangent circle in such a way that $A$ always remains a point of tangency with the ellipse. In this way we obtain a circle which is either tangent to the ellipse in also a second point $B$, or which has with the ellipse a four-point contact in $A$. Moreover, this circle cuts the positive $x$-axis in a point more remote than $C$. The desired farthest point will accordingly be found among those points of intersection of the positive $x$-axis by circles lying exterior to the ellipse and being doubly tangent to it. All such circles must lie, as we can easily see, symmetrically with respect to the $y$-axis. Let $a$, $b$ be the co-ordinates of any point on the ellipse. Then an easy calculation shows that the circle, which is symmetrical with respect to $y$-axis and tangent to the ellipse at this point, cuts off from the positive $x$-axis the segment

The greatest possible value of this expression occurs for $b=\frac{1}{2}$ and, hence, is equal to $\frac{1}{2}\mid \sqrt{7}\mid$. Since the point on the $x$-axis which we have denoted by $F$ has for its abscissa the value $\frac{3}{2}>\frac{1}{2}\mid \sqrt{7}\mid$, it follows that among the circles cutting the ellipse four times there is certainly none which passes through the point $F$.

We will now construct a new plane geometry in the following manner. As points in this new geometry, let us take the points of the -plane. We will define a straight line of our new geometry in the following manner. Every straight line of the -plane which is either tangent to the fixed ellipse, or does not cut it at all, is taken unchanged as a straight line of the new geometry. However, when any straight line $g$ of the -plane cuts the ellipse, say in the points $P$ and $Q$, we will then define the corresponding

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straight line of the new geometry as follows. Construct a circle passing through the points $P$ and $Q$ and the fixed point $F$. From what has just been said, this circle will have no other point in common with the ellipse. We will now take the broken line, consisting of the arc $PQ$ just mentioned and the two parts of the straight line $g$ extending outward indefinitely from the points $P$ and $Q$, as the required straight line in our new geometry. Let us suppose all of the broken lines constructed which correspond to straight lines of the -plane. We have then a system of broken lines which, considered as straight lines of our new geometry, evidently satisfy the axioms I, 1–2 and III. By a convention as to the actual arrangement of the points and the straight lines in our new geometry, we have also the axioms II fulfilled.

Moreover, we will call two segments $AB$ and $A\prime B\prime$ congruent in this new geometry, if the broken line extending between $A$ and $B$ has equal length, in the ordinary sense of the word, with the broken line extending from $A\prime$ to $B\prime$.

Finally, we need a convention concerning the congruence of angles. So long as neither of the vertices of the angles to be compared lies upon the ellipse, we call the two angles congruent to each other, if they are equal in the ordinary sense. In all other cases we make the following convention. Let $A,B,C$ be points which follow one another in this order upon a straight line of our new geometry, and let $A\prime ,B\prime ,C\prime$ be also points which lie in this order upon another straight line of our new geometry. Let $D$ be a point lying outside of the straight line $ABC$ and $D\prime$ be a point outside of the straight $A\prime B\prime C\prime$. We will now say that, in our new geometry, the angles between these straight lines fulfill the congruences

whenever the natural angles between the corresponding broken lines of the ordinary geometry fulfill the proportion These conventions render the axioms IV, 1–5 and V valid.

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In order to see that Desargues’s theorem does not hold for our new geometry, let us consider the following three ordinary straight lines of the -plane; viz., the axis of $x$, the axis of $y$, and the straight line joining the two points of the ellipse and . Since these three ordinary straight lines pass through the origin, we can easily construct two triangles so that their vertices shall lie respectively upon these three straight lines and their homologous sides shall be parallel and all three sides shall lie exterior to the ellipse. As we may see from figure 40, or show by an easy calculation, the broken lines arising from the three straight lines in question do not intersect in a common point. Hence, it follows that Desargues’s theorem certainly does not hold for this particular plane geometry in which we have constructed the two triangles just considered.

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This new geometry serves at the same time as an example of a plane geometry in which the axioms I, 1–2, II–III, IV, 1–5, V all hold, but which cannot be considered as a part of a geometry of space.

21. Introduction of an Algebra of Segments based upon Desargues’s Theorem and Independent of the Axioms of Congruence²³

In order to see fully the significance of Desargues’s theorem (Theorem 32), let us take as the basis of our consideration a plane geometry where all of the axioms I 1–2, II–III are valid, that is to say, where all of the plane axioms of the first three groups hold, and then introduce into this geometry, in the following manner, a new algebra of segments independent of the axioms of congruence.

Take in the plane two fixed straight lines intersecting in $O$, and consider only such segments as have $O$ for their origin and their other extremity in one of the fixed lines. We will regard the point $O$ itself as a segment and call it the segment $O$. We will indicate this fact by writing

Let $E$ and $E\prime$ be two definite points situated respectively upon the two fixed straight lines through $O$. Then, define the two segments $OE$ and $OE\prime$ as the segment 1 and write accordingly

$$OE=OE\prime =1\text{ or }1=OE=OE\prime .$$

(5.1)

We will call the straight line $EE\prime$, for brevity, the unit-line. If, furthermore, $A$ and $A\prime$ are points upon

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the straight lines $OE$ and $OE\prime$, respectively, and, if the straight line $AA\prime$ joining them is parallel to $EE\prime$, then we will say that the segments OA and OA’ are equal to one another, and write

In order now to define the sum of the segments $a=OA$ and $b=OB$, we construct $AA\prime$ parallel to the unit-line $EE\prime$ and draw through $A\prime$ a parallel to $OE$ and through $B$ a parallel to $OE\prime$. Let these two parallels intersect in $A\prime \prime$. Finally, draw through $A\prime \prime$ a straight line parallel to the unit-line $EE\prime$. Let this parallel cut the two fixed lines $OE$ and $OE\prime$ in $C$ and $C\prime$ respectively. Then $c=OC=OC\prime$ is called the sum of the segments $a=OA$ and $b=OB$. We indicate this by writing In order to define the product of a segment $a=OA$ by a segment $b=OB$, we make use of exactly the same construction as employed in Section 14, except that, in place of the sides of a right angle, we make use here of the straight lines $OE$ and $OE\prime$.

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The construction is consequently as follows. Determine upon $OE\prime$ a point $A\prime$ so that $AA\prime$ is parallel to the unit-line $EE\prime$, and join $E$ with $A\prime$. Then draw through $B$ a straight line parallel to $EA\prime$. This parallel will intersect the fixed straight line $OE\prime$ in the point $C\prime$, and we call $c=OC\prime$ the product of the segment $a=OA$ by the segment $b=OB$. We indicate this relation by writing

22. The Commutative and the Associative Law of Addition for Our new Algebra of Segments

In this section, we shall investigate the laws of operation, as enumerated in Section 13, in order to see which of these hold for our new algebra of segments, when we base our considerations upon a plane geometry in which axioms I 1–2, II–III are all fulfilled, and, moreover, in which Desargues’s theorem also holds.

First of all, we shall show that, for the addition of segments as defined in Section 21, the commutative law

holds. Let

$$\begin{array}{rcll}& a=OA=OA\prime & & \\ & b=OB=OB\prime & & \end{array}$$

Hence, $AA\prime$ and $BB\prime$ are, according to our convention, parallel to the unit-line. Construct the points $A\prime \prime$ and $B\prime \prime$ by drawing $A\prime A\prime \prime$ and $B\prime B\prime \prime$ parallel to $OA$ and also $AB\prime \prime$ and $BA\prime$ parallel to $OA$. We see at once that the line $A\prime \prime B\prime \prime$ is parallel to $AA\prime$ as the commutative law requires. We shall show the validity of this statement by the aid of Desargues’s theorem in the following manner. Denote the point of intersection of $AB\prime \prime$ and $A\prime A\prime \prime$ by $F$ and that of $BA\prime \prime$ and $B\prime B\prime \prime$ by $D$. Then, in the triangles $AA\prime F$ and $BB\prime D$, the homologous sides are parallel to each other. By Desargues’s theorem, it follows that the three points $O$, $F$, $D$ lie in a straight line. In consequence of this condition, the two triangles $OAA\prime$ and $DB\prime \prime A\prime \prime$ lie in such a way that the lines joining the corresponding vertices pass through the same point $F$, and since the homologous sides $OA$ and $DB\prime \prime$, as also $OA\prime$ and $DA\prime \prime$, are parallel to each other, then, according to the second part of Desargues’s theorem (Theorem 32), the third sides $AA\prime$ and $B\prime \prime A\prime \prime$ are parallel to each other.

To prove the associative law of addition

we shall make use of figure 44. In consequence of the commutative law of addition just demonstrated, the above formula states that the straight line $A\prime \prime B\prime \prime$ must be parallel to the unit-line.

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The validity of this statement is evident, since the shaded part of figure 44 corresponds exactly with figure 43.

23. The Associative Law of Multiplication and the Two Distributive Laws for the New Algebra of Segments

The associative law of multiplication

has also a place in our new algebra of segments.

Let there be given upon the first of the two fixed straight lines through $O$ the segments

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and upon the second of these straight lines, the segments

In order to construct the segments

in accordance with Section 21, draw $A\prime B\prime$ parallel to $AB$, $B\prime C\prime$ parallel to $BC$, $CD$ parallel to $AG$, and $A\prime D\prime$ parallel to $AD$. We see at once that the given law amounts to the same as saying that $CD$ must also be parallel to $C\prime D\prime$. Denote the point of intersection of the straight lines $A\prime D\prime$ and $B\prime C\prime$ by $F\prime$ and that of the straight lines $AD$ and $BC$ by $F$. Then the triangles $ABF$ and $A\prime B\prime F\prime$ have their homologous sides parallel to each other, and, according to Desargues’s theorem, the three points $O$, $F$, $F\prime$ must lie in a straight line. Because of these conditions, we can apply the second part of Desargues’s theorem to the two triangle $CDF$ and $C\prime D\prime F\prime$, and hence show that, in fact, $CD$ is parallel to $C\prime D\prime$.

Finally, upon the basis of Desargues’s theorem, we shall show that the two distributive laws

and hold for our algebra of segments.

In the proof of the first one of these laws, we shall make use of figure 46.²⁴ In this figure, we have

In the same figure, $B\prime \prime D_2$ is parallel to $C\prime \prime D_1$ which is parallel to the fixed straight line $OA\prime$, and $B\prime D_1$ is parallel to $C\prime D_2$, which is parallel to the fixed straight line $OA\prime \prime$. Moreover, we have $A\prime A\prime \prime$ parallel to $C\prime C\prime \prime$, and $A\prime B\prime \prime$ parallel to $B\prime A\prime \prime$, parallel to $F\prime D_2$, parallel to $F\prime \prime D_1$.

Our proposition amounts to asserting that we must necessarily have also

We construct the following auxiliary lines:

Let us denote the points of intersection of the straight lines $C\prime \prime D_1$ and $C\prime D_2$, $C\prime \prime D_1$ and $F\prime J$, $C\prime D_2$ and $F\prime \prime J$ by $G$, $H_1$, $H_2$, respectively. Finally, we obtain the other auxiliary lines indicated in the figure by joining the points already constructed.

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In the two triangles $A\prime B\prime \prime C\prime \prime$ and $F\prime D_{2}G$, the straight lines joining homologous vertices are parallel to each other. According to the second part of Desargues’s theorem, it follows, therefore, that

In the two triangles $A\prime C\prime \prime F\prime \prime$ and $F\prime G{H}_2$, the straight lines joining the homologous vertices are also parallel to each other. From the properties already demonstrated, it follows by virtue of the second part of Desargues’s theorem that we must have

Since in the two horizontally shaded triangles $OA\prime F\prime$ and $J{H}_{2}F\prime$ the homologous sides are parallel, Desargues’s theorem shows that the three straight lines joining the homologous vertices, viz.:

$$OJ,A\prime H_2,F\prime \prime F\prime$$

(5.2)

all intersect in one and the same point, say in $P$.

In the same way, we have necessarily

and since, in the two obliquely shaded triangles $OA\prime \prime F\prime$ and $J{H}_{1}F\prime \prime$, the homologous sides are parallel, then, in consequence of Desargues’s theorem, the three straight lines joining the homologous vertices, viz.: all intersect likewise in the same point, namely, in point $P$.

Moreover, in the triangles $OA\prime A\prime \prime$ and $J{H}_2{H}_1$, the straight lines joining the homologous vertices all pass through this same point $P$, and, consequently, it follows that we have

and, therefore, Finally, let us consider the figure $F\prime \prime H_{2}C\prime C\prime \prime H_{1}F\prime F\prime \prime$. Since, in this figure, we have

$$\begin{array}{rcll}F\prime \prime H_2\text{ parallel to}& C\prime F\prime ,& \text{ parallel to }C\prime \prime H_1,& \\ G{H}_{2}C\prime \text{ “ “}& F\prime \prime C\prime \prime ,& \text{ “ “}{H}_{1}F\prime ,& \\ C\prime C\prime \prime \text{ “ “}& H_1{H}_2,& & \end{array}$$

we recognize here again 43, which we have already made use of in Section 22 to prove the commutative law of addition. The conclusions, analogous to those which we reached there, show that we must have

and, consequently, we must have also which result concludes our demonstration.

To prove the second formula of the distributive law, we make use of an entirely different figure,—-47. In this figure, we have

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and, furthermore, we have

We have also

$$\begin{array}{rcll}AB& \text{ parallel to}& A\prime B\prime & \\ BD& \text{ “ “}& B\prime D\prime & \\ DG& \text{ “ “}& D\prime G\prime & \\ HJ& \text{ “ “}& H\prime J\prime .& \end{array}$$

That which we are to prove amounts, then, to demonstrating that

Denote the points in which $BD$ and $GD$ intersect the straight line $AH$ by $C$ and $F$, respectively, and the points in which $B\prime D\prime$ and $G\prime D\prime$ intersect the straight line $A\prime H\prime$ by $C\prime$ and $F\prime$, respectively. Finally, draw the auxiliary lines $FJ$ and $F\prime J\prime$, indicated in the figure by dotted lines.

In the triangles $ABC$ and $A\prime B\prime C\prime$, the homologous sides are parallel and, consequently, by Desargues’s theorem the three points $O$, $C$, $C\prime$ lie on a straight line. Then, by considering in the same way the triangles $CDF$ and $CD\prime F\prime$, it follows that the points $O$, $F$, $F\prime$ lie upon the same straight line and likewise, from a consideration of the triangles $FGH$ and $F\prime G\prime H\prime$, we find the points $O$, $H$, $H\prime$ to be situated on a straight line. Now, in the triangles $FHJ$ and $F\prime H\prime J\prime$, the straight lines joining the homologous vertices all pass through the same point $O$, and, hence, as a consequence of the second part of Desargues’s theorem, the straight lines $FJ$ and $F\prime J\prime$ must also be parallel to each other. Finally, a consideration of the triangles $DFJ$ and $D\prime F\prime J\prime$ shows that the straight lines $DJ$ and $D\prime J\prime$ are parallel to each other and with this our proof is completed. □

Proof. Suppose that the straight line $l$ passes through the origin $O$. Furthermore, let $C$ be a definite point upon $l$ different from $O$, and $P$ any arbitrary point of $l$. Let $OA$ and $OB$ be the co-ordinates of $C$ and $x$, $y$ be the co-ordinates of $P$. We will denote the straight line joining the extremities of the segments $x$, $y$ by $g$. Finally, through the extremity of the segment $1$, laid off on the $x$-axis, draw a straight line $h$ parallel to $AB$. This parallel cuts off upon the $y$-axis the segment $e$. From the second part of

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Desargues’s theorem, it follows that the straight line $g$ is also always parallel to $AB$. Since $g$ is always parallel to $h$, it follows that the co-ordinates $x$, $y$ of the point $P$ must satisfy the equation

$$ex=g.$$

Moreover, in 49 let $l\prime$ be any arbitrary straight line in our plane. This straight line will cut off on the $x$-axis the segment $c=OO\prime$. Now, in the same figure, draw through $O$ the straight line $l$ parallel to $l\prime$. Let $P\prime$ be an arbitrary point on the line $l\prime$. The straight line through $P\prime$, parallel to the $x$-axis, intersects the straight line $l$ in $P$ and cuts off upon the $y$-axis the segment $y=OB$. Finally, through $P$ and $P\prime$ let parallels to the $y$-axis cut off on the $x$-axis the segments $x=OA$ and $x\prime =OA\prime$.

We shall now undertake to show that the equation

is fulfilled by the segments in question. For this purpose, draw $O\prime C$ parallel to the unit-line and likewise $CD$ parallel to the $x$-axis and $AD$ parallel to the $y$-axis.

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Then, to prove our proposition amounts to showing that we must have necessarily

Let $D\prime$ be the point of intersection of the straight lines $CD$ and $A\prime P\prime$ and draw $O\prime C\prime$ parallel to the $y$-axis.

Since, in the triangles $OCP$ and $O\prime C\prime P\prime$, the straight lines joining the homologous vertices are parallel, it follows, by virtue of the second part of Desargues’s theorem, that we must have

In a similar way, a consideration of the triangles $ACP$ and $A\prime C\prime P\prime$ shows that we must have Since, in the triangles $ACD$ and $C\prime A\prime O\prime$, the homologous sides are parallel to each other, it follows that the straight lines $AC\prime$, $CA\prime$ and $DO\prime$ intersect in a common point. A consideration of the triangles $C\prime A\prime D$ and $ACO\prime$ then shows that $A\prime D$ and $CO\prime$ are parallel to each other. From the two equations already obtained, viz.:

$$ex=y\text{ and }x\prime =x+c,$$

(5.3)

follows at once the equation

$$ex\prime =y+ec.$$

(5.4)

If we denote, finally, by $n$ the segment which added to the segment 1 gives the segment 0, then, from this last equation, we may easily deduce the following

$$ex\prime +ny+nec=0,$$

(5.5)

and this equation is of the form required by theorem 34.

We can now show that the second part of the theorem is equally true; for, every linear equation

$$ax+by+c=0$$

(5.6)

may evidently be brought into the required form

$$ex+ny+nec=0$$

(5.7)

by a left-sided multiplication by a properly chosen segment.

It must be expressly stated, however, that, by our hypothesis, an equation of segments of the form

$$xa+yb+c=0,$$

(5.8)

where the segments $a$, $b$ stand to the right of the co-ordinates $x$, $y$ does not, in general, represent a straight line. □