PROP. 12. B. I.

The given line must not be finite, for if it were, the problem might be impossible, as is evident from the figure if C be the given point and the line finite AE.

PROP. 13. B. I.

Euclid says makes angles with it, because a line drawn to the extremity of another stands upon it, but makes only one angle with it.

PROP. 20. B. I.

Some writers contend that this proposition should be placed among the axioms, and that such was the opinion of Archimedes, who assumes that a right line is the shortest line between two points; but there he is comparing right lines with curves: Archimedes would never have assumed what Euclid had demonstrated; nor is every proposition whose truth is very evident, to be placed among the axioms, our aim should be to have the principles as few as possible.

This demonstration is not Euclid's, but some ancient geometer's.

PROP. 22. B. I.

In this proposition, as well as in the first, Euclid assumes that the circles intersect; but this is easily demonstrated from the conditions of the problem, that any two of the given lines must be greater than the third; for since the sum of the radii DG and EF is greater than DE, part of each circle must be within the other; and since the sum of DE and either of the others is greater than the third, part of each circle must be outside the other.

PROP. 24. B. I.

In the construction of this proposition the Greek editors have omitted the words which is not the greater,

but these are absolutely necessary to prevent a diversity of cases, the point G falling sometimes above, sometimes below, and sometimes upon the line BC: when the construction here given is made, Clavius has shewn that the point G must fall below BC, and it can be demonstrated in the following manner, which is not unlike his.

Let the side AC be greater than AB, and, if it be Fig. 4. possible, let the point G fall upon the line BC; since the side AC is greater than AB, the angle B is greater than the angle C; but the angle AGC is greater than the internal angle B, and therefore greater than the angle C; therefore the side AC is greater than AG, but by hypothesis they are equal, which is absurd.

Now if it be possible, let the point G fall above BC, Fig. 5. and produce AG to E; it can be demonstrated as in the former case that the line AE is less than AC, and therefore that AG is less than AC; but by hypothesis they are equal, which is absurd.

If the sides AB and AC be equal, the demonstration is the same, except that the angles B and C are equal.

If the angle BAG be constructed with the greater side, and the point G fall either upon or above BC, the proposition can be demonstrated in the following

manner:

First, let the point G fall upon BC, and it is evident Fig. 6. that BC is greater than BG.

2. Let G fall above BC; since the sum of the lines Fig. 7. AG and GB is less than the sum of AC and CB, but AG is equal to AC, BG must be less than BC.

This latter case can be demonstrated otherwise thus, Fig. 8. draw GC and produce AG and AC; since the angles

EGC and FCG are equal (1), and BGC greater than (1) Prop. 5. one of them, and BCG less than the other, BGC is B. 1. greater than BCG, and therefore the side BC greater than BG.

PROP. 26. B. I.

The line equal to the side DE ought to be cut off, conterminous to the side BC.

Fig. 9.

B. 1.

PROP. 27. B. I.

If the definition of parallel lines be changed, and an axiom laid down, which Euclid made use of, this proposition, the two following, and the 12th axiom, can be demonstrated in the following manner.

Definition.

Parallel right lines are those which however produced are always equidistant.

Axiom.

Any part of a finite right line can be taken so often as to exceed the whole.

Lemma I.

If two right lines AB and CD be parallel, any right line MN, which is perpendicular to one of them AB, is also perpendicular to the other CD.

Take any point A, and make NB equal to NA, draw AC and BD perpendicular to AB, and draw AM and BM.

In the triangles ANM and BNM the angles at N are equal, the sides NA and NB are also equal, and NM is common to both, therefore the angles NMA and NMB are equal, the sides MA and MB, and the angles NAM (1) Prop. 4. and NBM (1): Take away the equal angles NAM and NBM from the equals NAC and NBD, the remainders MAC and MBD are equal, the sides MA and AC are equal to MB and BD, therefore the angles AMC and BMD are equal (1), but NMA and NMB are also (2) Def. 11, equal, therefore the angles NMC and NMD are equal, and therefore NM is perpendicular to CD (2).

B. 1:

Fig. 10.

Lemma II.

If two right lines AB and CD be perpendicular to the same line NM, they are parallel.

For, if it be possible, let CD not be parallel to AB, and through M draw PR parallel to AB, this line is

perpendicular to NM (1), therefore the angle RMN (1) Lemma is right, but DMN is also a right angle, which is

absurd.

Lemma 3.

If two right lines AB and CD be parallel, and AC Fig, 11. and BD be drawn perpendicular to them, the intercepted parts AB and CD are equal.

2.

For since the right lines AC and BD are perpendi- (1) Lemma cular to the same line AB, they are parallel (1), and therefore the lines AB and CD are equal (2).

PROP. 29. B. 1.

(2) Defin.

parall.

The right line EF, cutting the parallel right lines Fig. 12. AB and CD, makes the alternate angles, AGH and GHD, CHG and HGB, equal; and the external angle equal to the internal and opposite at the same side, EGA to GHC and EGB to GHD; and the internal angles at the same side, AGH and CHG, BGH and DHG, equal to two right angles.

If the line EF be perpendicular to the parallels, the proposition is evident.

But if not, draw HK and GL perpendicular to CD. In the triangles LHG and KGH, the sides LG and HK (1), LH and GK (2) are equal, and GH (1) Defin. common to both, therefore the angles LHG and KGH parall are equal, and therefore the angles CHG and BGH, which make up with them two right angles.

Since the vertically opposite angles EGA and BGH are equal, and BGH is equal to the alternate angle CHG, the external angle EGA is equal to the internal GHC; and in the same manner it can be demonstrated that EGB and GHD are equal.

Since the angles CHG and DHG are equal to two right angles, and BGH is equal to the alternate angle CHG, DHG and BGH are equal to two right angles; and in the same manner it can be demonstrated that GHC and AGH are equal to two right angles.

(2) Lem. 3.

Fig. 13.

Fig. 14.

(1) Prop. prec.

Fig. 15.

B. 1.

PROP. 27. B.I.

If a right line EF cutting two right lines, AB and CD, make the alternate angles equal, AEF and EFD, those lines are parallel.

For, if it be possible, let the line CD not be parallel to AB, and through E draw XZ parallel to CD, the alternate angles XEF and EFD are equal, but AEF and EFD are equal, therefore XEF and AEF are equal, which is absurd.

PROP. 28. B. I.

If a right line EF intersect two right lines, AB and CD, and make the external angle equal to the internal and opposite upon the same side of the line, EGA to GHC, and EGB to GHD; or make the internal angles at the same side, AGH and CHG, or BGH and DHG, equal to two right angles, the two right lines are parallel to one another.

1. Let EGA and GHC be equal, and since EGA and BGH are equal, the alternate angles GHC and BGH are equal, and therefore the lines AB and CD are parallel (1); in the same manner it can be demonstrated if the angles EGB and GHD be given equal.

2. Let AGH and CHG be equal to two right angles, and since AGH and BGH are also equal to two right angles, the alternate angles CHG and BGH are equal, and therefore AB and CD are parallel; in the same manner it can be demonstrated if BGH and DHG be given equal to two right angles.

Lemma 4.

If any right line NP cut another line BD, it can be produced so as to cut any line XZ parallel to BD.

Take any point O in the line NP, make OS equal (1) Prop. 3. to OA (1), draw OM and SE perpendicular to BD, and OC perpendicular to SE, which must fall between S and E because the angle ASE is acute, since the angle AES is right (2).

(2) Cor.

Prop. 17.
B. 1.