Geometry, Plane, Solid, and Spherical, in Six Books

drawn from the point D to the line EF, two perpendiculars, which is impossible (12). Therefore, A B coincides with D E, and the triangle ABC coincides with the triangle DEF, that is, (ax. 11.) the triangles ABC and D E F are equal in every respect. Therefore, &c.

SECTION 3. Parallels.

PROP. 14.

Straight lines which are at right angles to the same straight line, are parallel: and, conversely, parallel straight lines are at right angles to the same straight line; that is, if a straight line be drawn through the two at right angles to one of them, it shall be at right angles to the other likewise.

The first part of the proposition is manifest; for if A B and C D be each of them at right angles to the straight

line E F, and be not parallel, they must meet one another; in which case, there will be two perpendiculars drawn to the same straight line EF from the same point, viz. the point of concourse. But (12.) this is impossible. Therefore, AB cannot meet CD, though produced ever so far both ways, that is (def. 12.) A B is parallel to CD.

In the next place, let A B be parallel to CD, and from any point E of A Blet EF be drawn at right angles to CD: EF shall also be at right angles to A B.

Through E let any straight line L M be drawn which is not at right angles to EF. Produce FE to G, so that EG may be equal to E F, and from G draw GH perpendicular to G F. Through E draw Im, making the angle G Em equal to the angle FEM. In FD take any point K: make G H equal to F K, and join HK, cutting the lines L M, Im in the points M, m. Then it may easily be shown (by doubling over the figure, and applying the straight line E F upon EG, so that the point F may coincide with the point G, and therefore the straight line FK with GH, and E M with Em,)

that E M is equal to E m, and M K to m H.

Now, it seems sufficiently evident, that, the straight lines G H and FK being at right angles to the same straight line, and therefore (by the first part of the proposition) never meeting one another, the distance H K, of any two corresponding points in them, neither increases nor diminishes, but remains always equal to FG; while, on the other hand, the lines EM and Em cutting one another in E, the distance M m, of any two corresponding points in them, continually increases with the distance from E, and may, by sufficiently producing EM, Em, be made greater than any assigned distance, as FG. Therefore, the straight lines E M, Em may be produced, so that Mm may become greater than HK. But, because M K is always equal to m H, Mm cannot become greater than H K, unless the straight line E M cuts the line F D, and Em the line G H.

Therefore EM and Em may be produced, until they respectively meet the lines FD and GH.

Hence, it appears, that AB, which never meets CD, cannot but be at right angles to EF: for, it has been shown, that any straight line which passes through E, and is not at right angles to EF, may be produced to meet C D. Therefore, &c.

Cor. 1. Any point E being given, a straight line A B may be drawn through that point, which shall be parallel to a given straight line CD (12. and Post. 5.).

Cor. 2. Through the same given point, there cannot be drawn more than one parallel to the same given straight line.

Cor. 3. If a straight line cut one of two parallels, it may be produced to cut the other likewise.

Scholium.

The second part of this proposition is not supported by that cogency of demonstration which is said, and with truth, to characterize every other part of Geometry. Of the two particulars which have been assumed, one indeed, viz., that which regards the unlimited divergency of cutting lines, seems almost axiomatic or self-evident. The other is not equally so. It may be illustrated by observing that, at equal intervals, upon either side of E F, the distances of corresponding points are equal to one another; and thence arguing, that,

from one of these to the other, the distance can neither have been increasing nor diminishing, for that, had either been the case, since the lines are straight, the distance of their corresponding points would have continued to increase or to diminish. Should this, however, as we can easily imagine, fail to satisfy the student, we must refer him to measurement for such a degree of conviction as it can afford.

In fact, although it may be shown without difficulty, that certain straight lines will never meet one another; the converse, viz., that straight lines, which never meet one another, must have certain properties, has never been strictly demonstrated. It is agreed by Geometrical writers that some assumption is indispensable.*

The following is, perhaps, as simple as any that can be proposed, while it has the advantage also of not being many steps distant from the proposition in question.

"If from two points of one straight line to another, there fall two unequal perpendiculars, the straight lines will meet one another, if produced, upon the side of the lesser perpendicular."

Hence, if two straight lines be parallel, the perpendiculars drawn from the points of the one to the other, must all

That of Euclid (the famous twelfth axiom, see 15. Cor. 4.) is the converse of our eighth proposition, and asserts that, if the two interior angles made by two straight lines with a third be together less than two right angles, the two straight lines will meet one another, and, with the third, form a triangle,

if produced far enough.

Simson's demonstration of this axiom rests upon an assumption, which is scarcely more evident than that of the text, viz., that if the perpendiculars which are drawn from two points of one straight line to another be equal, any other perpendicular, drawn from a point of the first to the other, shall be equal to

either of them.

of them be equal; and hence, if a straight line be drawn at right angles to one of two parallels, it may easily be shown to cut the other at angles, which are equal to one another, that is, at right angles.

It is demonstrated in Prop. 16., that, if two straight lines be parallel, the perpendiculars drawn from the points of the one to the other must all of them be equal but that demonstration itself rests upon the converse part of Prop. 14., which is here in question. The reader must not imagine, therefore, that the above assumption is at ail assisted by that demonstration.

PROP. 15. (Euc. i. 27, 28 and 29.) Straight lines which make equal angles with the same straight line, towards the same parts, are parallel: and, conversely, if two parallel straight lines be cut by the same straight line, they shall make equal angles with it towards the same parts.

Let the straight lines A B, CD make equal angles BEG, DFG, with the same straight line E F, towards the same parts: AB shall be parallel to CD.

Bisect E F in H (Post. 3.), from H draw HK perpendicular to CD (12.), and produce KH to meet AB in L. Then because the angle HE L is equal to BE G (3.), and that B E G is equal to DFG or HFK, the angle HEL is equal to the angle H F K (ax. 1.). The vertical angles EH L, FHK are likeA modern geometer of great celebrity, M. Legendre, wise equal to one another (3.). Therehas, after more than one alteration, suggested, fore, the triangles HEL and HFK finally, an experimental proof not very different from that which is here adopted, as best suited to having two angles of the one equal to an Elementary Treatise. He is, notwithstanding, of two angles of the other, each to each, opinion, that the grand truth with which it is so intiand their sides H E, H F, which lie bemately connected, viz., that "the three angles of a triangle are together equal to two right angles," tween the equal angles, also equal to may be referred to the general principle of homoone another, are equal in every respect geneity. Of this it is unnecessary to say more in this place, than that it teaches us in the present (5.). Therefore, the angle HLE is instance, that the angles of a triangle depend, not equal to the angle H K F, that is, to a upon the absolute magnitude of its sides, but upon their relative magnitude; so that a triangle whose right angle. But straight lines which sides are 3, 4, and 5 times some given line will are at right angles to the same straight have the same angles whether the given line be line KL are parallel (14.). Therefore, A B is parallel to C D.

an inch or a mile; a truth which, indeed, seems to be nearly related to the more simple truth, that an angle is not increased by producing the sides which contain it, and leads directly to the theory of parallel straight lines. But, though the principle be of extensive application, the reasoning by which it is established has been shown to be incomplete, and such as, if great circumspection be not used, may even lead to fallacies,

Next let A B be parallel to C D, and let them be cut by the same straight line GEF; the angles BEG, D FG, which are towards the same parts, shall be equal to one another.

Bisect E F, and draw the straight line I. K, as before. Then, because LK is at right angles to C D, and that A B is parallel to CD, L K is at right angles also to AB (14.). And, because, in the right-angled triangles H EL, H FK, the hypotenuse HE, and adjacent angle EHL of the one, are equal to the hypotenuse HF, and adjacent angle FHK (3.) of the other, they are equal in every respect (13.). Therefore, the angle HEL is equal to the angle HFK, and B E G, which is equal to H E L (3.), is equal to D F G.* Therefore, &c.

Cor. 1. When two straight lines A B, C D, are cut by a third, E F, the angle BE G, is called the exterior angle, and the angle D FG, the interior and opposite angle upon the same side of the line. Therefore, if one straight line fall upon two other straight lines, so as to make the exterior angle equal to the interior and opposite upon the same side, those two straight lines shall be parallel: and conversely.

Cor. 2. The angles A E F, E FD, are called alternate angles. And AEF is always equal to BEG (3.). Therefore, if one straight line fall upon two other straight lines so as to make the two alternate angles equal to one another, those two straight lines shall be parallel and conversely.

Cor. 3. The angles B E F, and DFE, are called interior angles on the same side of the line. Now, when BEG is equal to DFG, or DFE, the angles BEF and DFE are together equal to two right angles, because BEF and BEG, are together equal to two right angles (2.). Therefore, if one straight line, falling upon two other straight lines, make the two interior angles, on the same side, together equal to two right angles, the two straight lines shall be parallel: and conversely.

Cor. 4. (Euc. i. ax. 12.) If one straight line, falling upon two other straight lines, make the two interior angles, on the same side, together less than two right angles,

A shorter demonstration may be had, by considering that if the angles BEG, DFG, be equal to one another, the two interior angles upon each side of E F will be together equal to two right angles; and, therefore, cannot be two angles of a triangle (8.), that is A B, CD cannot meet upon either side of EF; and hence the converse, because (14. Cor. 2.) only one parallel can be drawn through the same point to the same straight line. That which is given in the text, however, seems preferable, as pointing out the connexion of the proposition with Prop. 14., which immediately precedes it.

the two straight lines shall meet upon that side, if produced far enough.

PROP. 16.

Parallel straight lines are every where equidistant; that is, if from any two points of the one, perpendiculars be drawn to the other, those perpendiculars shall be equal to one another. Let AB, C D be two parallel straight lines; and from the points A, B, of AB, let AC, BD be drawn perpendicular to CD (12.): AC shall be equal to B D.

Join B C. Then because AC and B D are perpendicular to the same straight line CD, they are parallel (14.). Therefore, the alternate angles AC B, D B C, are equal to one another (15. Cor. 2.). Again, because A B is parallel to CD, the alternate angles ABC, DCB are equal to one another (15.Cor. 2.). Therefore the triangles A B C, D C B, having two angles of the one equal to two angles of the other, each to each, and the same line, BC, lying between the equal angles, are equal in every respect (5.). Therefore, AC is equal to B D. Therefore, &c.

Cor. It appears from the demonstration, that if AC be only parallel to BD, AC and B D will be equal to one another. Therefore, the parts of parallel straight lines, which are intercepted by parallel straight lines, are equal to one another.

PROP. 17. (Euc. i. 30.)

Straight lines, which are parallel to the same straight line, are parallel to one another.

Let the straight lines AB, CD be each of them parallel to EF: A A B shall be pa- c rallel to C D.

For, if the straight line G H be at right angles to E F, it will be at right angles to A B, because A B is parallel to EF (14.): and, for the like reason, it will be at right angles to CD: therefore, A B and CD being at right angles to the same straight line, G H, are parallel to one another (14.).

Therefore, &c.

Cor. Hence it appears that the quadrilaterals into which the parallelogram ABCD is divided in def. 23., by lines drawn parallel to two adjacent sides

are likewise themselves parallelograms the exterior angle shall be equal to the (def. 17.)

PROP. 18.

If, of two angles in the same plane, the sides of the one be parallel to the sides of the other, or perpendicular to the sides of the other, in the same order, the two angles.shall be equal.

Let AB C, DEF be two angles in the same plane, having the sides AB, BC of the one M parallel to the sides DE, EF of the other, each to each; and let the sides AB, DE, lie in the same direction from the sides BC, EF: the angle ABC shall be equal to the angle DEF.

Join B E, and produce it to G. Then, because AB, D E are parallel, and G B falls upon them, the exterior angle DEG is equal (15. Cor. 1.) to the interior and opposite angle ABE: and, for the like reason, FEG is equal to CBE: therefore, if these two equals be taken respectively from the former two, (ax. 3.) the remaining angle ABC will be equal

to DEF.

Secondly, let H KL be an angle, the sides of which are perpendicular to those of the angle ABC, each to each, viz., HK to AB, and K L to BC: the angle HKL shall be equal to ABC. Draw BM perpendicular to BA (Post. 5.), and therefore (14.) parallel to K H, and BN perpendicular to BC, and therefore parallel to K L. Then, by the first part of the proposition, the angle M BN is equal to HK L. But, because the right angle M B A is equal to the right angle NBC (1.), and the part NAB common to both, the remaining angle MBN is equal to ABC (ax. 3.). Therefore, (ax. 1.) ABC is equal to H K L.

Therefore, &c.

Cor. The demonstration of the second case, viz. that in which the sides of the one angle are perpendicular to the sides of the other, requires only that the angle M B A be equal to N B C.

Therefore, if, of two angles, the sides of the one make equal angles with the sides of the other, respectively, in the same order, and towards the same parts, the two angles shall be equal.

PROP. 19. (EUc. i. 32.)
If one side of a triangle be produced,

two interior and opposite angles; and the three angles of every triangle are together equal to two right angles.

Let the side BC of the triangle A B C be produced to D: the exterior angle ACD shall be equal to the two interior and opposite angles at A and B; and the three angles of the triangle ABC shall be together equal to two right angles.

Draw CE parallel to BA (14. Cor.1.). Then, because AC meets these parallels, the alternate angles ACE and A are equal; and because B D falls upon the same parallels, the angles ECD and B are equal (15. Cor. 1. and 2.). Therefore, the whole angle A CD, which is made up of the two angles A CE, ECD together, is equal to the angles at A and B together (ax. 2.).

To each of these equals add the angle of the triangle A B C are together equal ACB: therefore, (ax. 2.) the three angles to the angles A CD, AC B, that is, to two right angles (2.).

Therefore, &c.

Cor. 1. If two triangles have two angles of the one equal to two angles of the other, their third angles will likewise be equal to one another.

Cor. 2. (Euc. i. 26, second part of.) Hence, if two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, the equal sides being opposite to equal angles in each, the two triangles shall be equal in every respect (5.).

Cor. 3. In a right-angled triangle, the right angle is equal to the sum of the other two angles: and, conversely, if one angle of a triangle be equal to the sum of the other two, that angle shall be a right angle.

Cor. 4. In a right-angled triangle, the right angle to the middle of the opposite straight line, which is drawn from the side, is equal to half that side: and, conversely, if this be the case in any triangle, the angle from which the straight line is drawn shall be a right angle.

For, if the right angle be divided into parts equal respectively to the two acute into two isosceles triangles. And, in angles, the triangle will be divided (6.) the converse, the triangle being made up of two isosceles triangles, one of its angles is equal to the other two.

PROP. 20. (EUc. i. 32. Corr.1. and 2.) All the exterior angles of any rectilineal figure are together equal to four right angles: and all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides.

For, if from any point in the same plane, straight lines be drawn, one after the other, parallel to the sides of the figure, the angles contained by these straight lines about that point, will be equal to the exterior angles of the figure (18.), each to each, because their sides are parallel to the sides of the figure. Thus, the angles a, b, c, d, e, are respectively equal to the exterior angles A, B, C, D, E. But the former angles are together (3. Cor.) equal to four right angles; therefore, all the exterior angles of the figure are together equal to four right angles (ax. 1.).

Again, since every interior angle, together with its adjacent exterior angle, is equal to two right angles (2.); all the interior angies, together with all the exterior angles, are equal to twice as many right angles as the figure has angles. But all the exterior angles are, by the former part of the proposition, equal to four right angles; and the figure has as many angles as sides: therefore all the interior angles together with four right angles are equal to twice as many right angles as the figure has sides.

Therefore, &c.

Cor. The four angles of a quadrilateral are together equal to four right angles.

CBD equal to one another, for they are
alternate angles (15.), the side AB is
equal to CD, and the angle ABD to
CDB (4.); and, hence, because A BD,
C D B are alternate angles (15.), AB is
also parallel to CD.
Therefore, &c.

PROP. 22. (EUc. i. 34, first part of.)

The opposite sides and angles of a parallelogram are equal, and its diagonals bisect one another: and, conversely, if, in any quadrilateral figure, the opposite sides be equal; or if the opposite angles be equal; or if the diagonals bisect one another; that quadrilateral shall be a parallelogram.

Let A B C D be a parallelogram (see the last figure), and let its diagonals AC, BD cut one another in the point E: the sides AD, BC, as also, A B, CD, shall be equal to one another; the angles A and C, as also B and D shall be equal; and the diagonals A C, BD shall be bisected in E.

For, in the first place, that the opposite sides, as AD and B C, are equal, is evident, because they are parts of parallels intercepted by parallels (16. Cor.). Also, the opposite angles are equal, as at D and B; for. the angle at D is equal to the vertical angle formed by CD, AD produced (3.), and the latter to the angle B (18.).

Lastly, with regard to the bisection of the diagonals: because AD is parallel to B C, the two triangles E AD, ECB have the two angles E AD, EDA of the one equal to the two angles EC B, EBC of the other, each to each (15.); and it has been shown, that the interjacent sides A D, B C are equal to one another; therefore, (5.) EA is equal to EC, and ED to E B, that is, AC, BD are bisected in E.

Next, let the opposite sides of the another: it shall be a parallelogram. quadrilateral ABCD be equal to one For, in the triangles A B D, CD B, the three sides of the one are equal to the three sides of the other, each to each; therefore, the angle ABD is equal to CDB (7.), and (15.) A B is parallel to CD. And, for the like reason, A D is parallel to B C.

Or, let the opposite angles be equal: then, because the angles at A and B together are equal to the angles at C and D together, and that the four angles of the quadrilateral (20. Cor.) are equal to four right angles, the angles at A and B are together equal to two right angles, and

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Geometry, Plane, Solid, and Spherical, in Six Books: To which is Added, in ...