THE

ELEMENTS OF EUCLID.

APPENDIX A.

PROFESSOR PLAYFAIR, in his "Elements of Geometry" (Glasgow, 1795), substituted the axiom that "two straight lines, which cut one another, cannot be each of them parallel to the same straight line," for Bk. I. Ax. 12, as given in the text. If this method be adopted, the above may be made Axiom A. of the 1st Book, and the assertion of Euclid's 12th Axiom proved by means of it; this will form Bk. I. Prop. A., and may be placed conveniently after Prop. XXVIII. The props then in which Ax. 12 is introduced may either be proved as in the text, reference being made to Prop. A. instead of to Ax. 12; or their proofs may be so modified as to depend immediately on Playfair's Axiom. We shall enunciate Ax. A., state and prove Prop. A., and prove Bk. I. Prop. XXIX. immediately by help of Ax. A., according to Playfair.

AXIOM A.

Two straight lines, which cut one another, cannot be each of them parallel to the same straight line.

BK. I. PROP. A.

If a straight line cut two straight lines, so as to make the two interior angles on the same side of it together less than two right angles: then these two straight lines, being continually produced, shall at length meet on that side of the cutting line on which are the angles that are together less than two right angles.

Let the straight line EFGH cut AB, CD in F, G, so as to make the two interior angles BFG, FGD on the same side of EH (viz. that side towards B, D,) together less than two right angles. Then AB, CD, being continually produced, shall at length meet in some point on the side of EH towards B, D.

E

A

F

L

B

K

H

H

G

D

For if they do not meet on the side of EH towards в, D; they must either meet if produced on the side of EH towards A, C, or else be parallel.

I. Let, if possible, AB, CD meet if produced on the side of EH towards A, C.

Then the angles GFA, FGC are two angles of a triangle, and therefore are together less (i. 17) than two right angles. But since GF makes with AB the adjacent angles AFG, BFG, these two angles are equal to two right angles; and for like reason the two angles CGF, FGD are equal to two right angles: hence, adding equals to equals, the four angles AFG, FGC, FGD, GFB are equal to four right angles. But two of them, AFG, FGC have been shewn to be less than two right angles, therefore the other two BFG, fgd are greater than two right angles; and by hyps they are also less which is impossible. Therefore AB, CD do not, if produced, meet on the side of EH towards A, C.

II. Let, if possible, AB, CD be parallel.

At the point F in the straight line GF make the angle GFL equal to the angle FGC; and produce LF to K.

Then, because FG makes with CD the adjacent angles FGC, FGD, these two angles are equal to two right angles; and the angle FGC is equal to the angle GFL by const": therefore the two angles GFL, FGD are equal to two right angles. But the two angles GFB, FGD are by hyps less than two right angles; therefore the two angles GFL, FGD are greater than the two angles BFG, FGD. From each of these unequals take away the common angle FGD; therefore the remaining angle GFL is greater than the remaining angle GFB, and FL does not coincide with FB; that is, AB, KL are two straight lines, cutting one another in F. But since HE, cutting the two straight lines KL, CD in G, F makes the alternate angles CGF, GFL equal, KL is parallel (i. 27) to CD; and AB is supposed parallel to CD. That is, the two straight lines AB, CD, which cut one another in F, are each of them parallel to the same straight line CD: which by the axiom (Ax. A) is impossible. Therefore AB, CD are not parallel.

Hence, since it has been shewn, that AB, CD neither, if produced, meet on the side of EH towards A, C, nor are parallel, AB, CD must, if continually produced, meet on the side of EH towards B, D. Which was to be proved.

The following is the proof of Bk. I. Prop. XXIX. as given by Playfair, and depending immediately on Axiom A.

Then by const" the angle HGK is unequal to the angle HGA; therefore GK and GA do not coincide, and AB, KL are two straight lines cutting one another in G.

Now

because EF cutting KL, CD in G, H makes the alternate angles KGH, GHD equal, therefore KL is parallel (i. 28) to CD; and AB is by hyps parallel to CD. That is, the two

straight lines AB, KL, which cut one another in & are each of them parallel to the same straight line CD: which by the axiom (Ax. A) is impossible. Therefore the angles AGH, GHD are not unequal, that is, the alternate angles AGH, GHD are equal. In like manner it may be shewn that the alternate angles BGH, GHC are equal. Which

was to be proved.

The proofs of Parts II. and III. are the same as those in the text.

Playfair deduces what is given above as Prop. A from this prop", to which he makes it a Corollary.

APPENDIX B.

If we lay down the following as an Axiom:

If the distance of a point from the centre of a circle be less than the radius of the circle, the point is within the circle; and if the distance of a point from the centre of a circle is greater than the radius, the point is without the circle:

The 2nd and 18th Props of Bk. III. may be proved by a direct proof, instead of the ex absurdo demonstrations given in the text, as we shall now shew; and the above axiom will be referred to as Bk. III. Ax. A. These methods are due to Commandine, who published the fifteen books of Euclid's Elements in Latin, at Pisauri, fol., in the year 1572.

BK. III. PROP. II.

Let ABC be a circle and A, B any two points in the circumference. Then the straight line AB, which joins them, shall fall within the circle.

In AB take any point E; find D the centre of the circle (iii. 1); and join ᎠᎪ, ᎠᎬ, ᎠᏴ,

Because DA is equal to DB by the def" of a circle, the angle DAB is equal (i. 5) to the angle DBA; and because the side AE of the triangle DAE is produced to B, the exterior angle DEB is greater than the interior and opposite

angle DAB (i. 16): therefore the angle DEB is likewise greater than the angle DBE. But the greater angle of a triangle is subtended by the greater side (i. 19); therefore DB is greater than DE. That is, DE the distance of the point E from the centre D is less than the radius of the circle ABC; therefore by the axiom (iii. Ax. A) the point E is within the circle ABC. In like manner it may be shewn that every other point in AB is within the circle ABC, excepting A and B which are points in the circumference: therefore the straight line AB, which joins A and B, falls within the circle. Which was to be proved.

Let ABC be a circle, of which D is the centre and DA any radius; and let the straight line AE be drawn at right angles to DA from its extremity a.

Then :

I. AE shall fall without the circle ABC. In AE take any point F, and join DF. Because the angle DAF is a right angle by const", the angle DFA must be less than a right angle; for if it were either equal to or greater than a right angle, the two angles DAF, DFA of the triangle DAF would be not less than two right angles: which is impossible (i. 17). Hence the angle DAF is greater than the angle DFA; and the greater angle is subtended by the

:

E

F

Ө

B

D

A

greater side (i. 19): therefore DF is greater than DA. That is, DF the distance of the point F from the centre D is greater than the radius of the circle ABC: therefore

D d