one of the equal angles A B C, A CB, is equal to two right angles (2.).

Cor. 3. The straight line which bisects the vertical angle of an isosceles triangle, bisects the base at right angles: and conversely, the straight line which bisects the base at right angles, passes through the vertex, and bisects the vertical angle.

Cor. 4. If there be two isosceles triangles upon the same base (whether they be upon the same side of it or upon different sides), the straight line which joins their vertices or summits, or that straight line produced, shall bisect the base at right angles. For the straight line which bisects the base at right angles passes through the vertex of each.

PROP. 7. (EUc. i. 8.)

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases. equal, the angle contained by the two sides of the one shall be equal to the angle contained by the two sides equal to them of the other.

Let ABC, DEF be two triangles, having the two sides of the one equal to two sides of the other, each to each, and likewise the base B C equal to the base EF: the angle BAC shall be equal to the angle ED F.

At the point E in the straight line E F, make the angle FE G equal to the angle A B C (Post. 6.): take E G equal to BA or ED, and join G F, GD. Then because the triangles A B C, GEF, have two sides of the one equal to two sides of the other, each to each, and the included angles equal to one another, the base G F is equal to A C (4.) that is, to DF. Again, because in the triangle FDG, the side FD is equal to FG, the angle FDG is equal to FGD (6.). For the like reason, the angle E D G is equal

to EGD. Therefore the angle EDF, which is the sum or difference of the two EDG, FD G, is equal to the angle EGF, which is the sum or difference of the two EGD, FGD (ax. 2., 3.). But EG F is equal to BA C, because (4.) the triangle GEF is equal to the triangle ABC in every respect: therefore, (ax. 1.) the angle EDF is equal to the angle BAC. When G D coincides with GE, GED is a straight line, and the angles at G and D are the angles at the base of the isosceles triangle F D G ; wherefore the latter is equal to the angle at G, that is, to the angle at A, as before. Therefore, &c.

Cor. The two triangles are equal in every respect. (4. & 4 Cor.)

PROP. 8. (EUC. i. 17.)

Any two angles of a triangle are together less than two right angles. Let A B C be any triangle: any two of its angles, ABC and ACB, shall be together less than two right angles.

Bisect B C in D (Post. 3.): join A D, and produce it to E, so that DE may be equal to AD; and join CE.

B D

Then, because the triangles ABD ECD have two sides of the one equal to the two sides of the other, each to each, and the included angles A D B, EDC, equal to one another (3.), the angle ECD is equal to the angle ABD or AB C. (4.). Therefore, the two angles A B C, AC B taken together are equal to the two angles ECD, ACB taken together (ax. 2.), that is, to the angle ACE. But (2. Cor. 2.) the angle A CE is less than two right angles. Therefore, the angles A B C, ACB together are less than two right angles.

Therefore, &c.

Cor. 1. (Euc. i. 16.) If one side of a triangle ABC, as BC, be produced to F, the exterior angle A C F shall be greater than either of the interior and opposite angles at A and B; for either of these angles taken with the angle ACB, is less than two right angles, but the angle A CF, taken with the same A CB, is equal to two right angles (2.).

Cor. 2. A triangle cannot have more than one right angle, or more than one obtuse angle.

PROP. 9. (EUC. i. 18. & 19.)

If one side of a triangle be greater

complete path from A to B, which is shorter than ADEB by the intermediate part D E.

Therefore, there is no path from A to B, the straight line A CB excepted, than which a shorter may not be found between A and B. But, since none of the paths from A to B can be less than of some certain length, there must be some, one or more, shorter than the others. Therefore, the shortest path is the straight line ACB. From this property the straight line which joins two points derives the name of the distance between them.

B

Hence, also, we may infer, that of any_two paths, ACB, AD B, leading from A to B, and everywhere concave towards the straight line AB, that which is enveloped by the other, as ADB, is the shortest. For of all the paths not lying between ADB and the straight line A B, there is none, A D B excepted, than which a shorter may not be found.

And this is the case whether the paths ACB and AD B be both of them curvilineal, or one of them, (ACB or ADB) rectilineal.

PROP. 11. (Euc. i. 24 & 25.)

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle which is contained by the two sides of the one greater than the angle which is contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other: and conversely.

B

CE

Let А В С, DE F, be two triangles having the two sides AB, AC of the one equal to the two sides DE, DF of the other, each to each, but the angle BAC greater than the angle EDF: the base BC shall be greater than the base E F.

At the point D in the straight line DE, make the angle EDG equal to the angle BAC (Post. 6.); take DG equal to AC, and join EG, G F. Then, because the triangles ABC, DEG have two sides of the one equal to two sides of the other, each to each, and the included angles BA C, EDG equal to one another, the bases BC, EG are equal to one another (4.).

Now, the line D F falls between DE and DG, because the angle EDG is equal to BA C, which is supposed to be greater than ED F. But the point F may fall 1° without the triangle DEG; or 2° upon the base E G; or 3° within the triangle D E G.

In the first case, because DG is equal to A C, that is, to D F, the angle DFG is equal to the angle DG F (6.). But the angle EFG is greater than DFG, and EGF is less than DG F. Therefore much more is the angle EFG greater than the angle EGF. Therefore, also, the side E G, that is B C, is greater than EF (9.).

In the second case, it is at once evident that EG, that is, B C, is greater than E F.

AA

EFG

produced to H and K. Then, because DF is equal to DG, the angles H FG, KGF, upon the other side of the base of the isosceles triangle D FG, are equal to

In the third case, let D F and D G be

D

G K

one another (6. Cor. 2.). But E F G is greater than H FG, and E G F is less than KG F. Therefore much more is the angle EFG greater than the angle EGF, as in the first case, and the side EG, that is B C, is greater than E F (9.).

Next, let the base B C be greater than the base EF; the angle BAC shall likewise be greater than the angle EDF. For B A C cannot be equal to EDF, because then (4.) the base BC would be equal to the base EF; neither can it be less than E D F, because then, by the former part of the proposition, the base BC would be less than the base EF. Therefore, the angle B A C cannot but be greater than the angle EDF. Therefore, &c.

PROP. 12.

A straight line may be drawn perpendicular to a given straight line of indefinite length, from any given point without it; but, from the same point, there cannot be drawn more than one perpendicular to the same straight line.

Let BC be a given straight line of indefinite length, and A any given point without it. A perpen

dicular may be drawn from the point A to the straight line B C.

In B C take any point D; join AD, and produce it to any point E, (Post. 1.). With the centre A and the radius AE describe a circle cutting B C in the points B and C upon each side of the point D, (Post. 2.). Bisect BC in F, and join AF, AB, AC, (Post. 3.). Then because AB is equal to A C, ABC is an isosceles triangle. Therefore AF, which is drawn from the vertex A to the middle point of the base BC, is perpendicular to the base (6. Cor. 3.); that is, a straight line AF may be drawn from the point A perpendicular to the straight line B C. But, from the same point A there cannot be drawn more than one perpendicular to the same straight line B C. For, if any other straight line AD were perpendicular to B C, the two angles, ÁDF and AFD, of the triangle AD F, would be together equal to two right angles, which (8.) is impossible.

Therefore, &c.

Cor. 1. If from any point A to a straight line B C, there be drawn a straight line A B which is not at right angles to B C, a second straight line AC may be drawn from A to B C, which shall be equal to AB; for, the perpendicular AF being drawn, and FC being taken equal to FB, it may easily be shown (4.) that A C is equal to A B.

Cor. 2. Of straight lines A B, AD, which are drawn from A to B C upon the same side of the perpendicular AF, that which is nearer to the perpendicular, as AD, is less than the other, which is more remote. For, the angle ABD or ABF being (8. Cor. 1.) less than the exterior right angle AFC, and again AFD or AFC less than the exterior angle ADB, much more is the angle ABD less than AD B, and therefore also the side AD less than AB (9.).

Cor. 3. In the same manner it may be shown that the perpendicular AF is the least of all straight lines which can be drawn from A to B C. For, if A B be any other straight line, the angle A B F being less than the exterior angle AFC, that is than AFB (def. 10.), the side A F is a so less than the side A B (9.). For this reason, the perpendicular

AF is called also the distance of the point A from the line B C.

Cor. 4. Hence, if, from the centre A, a circle be described with a radius less than the perpendicular A F, it will not meet the straight line BC; if with a radius equal to A F, it will meet BC in one point only, which is the foot of the perpendicular; and if with a radius greater than A F, it will meet B C in two points, which are at equal distances from the foot of the perpendicular, upon either side of it.

and side A B of the one be equal to the hypotenuse DF and side DE of the other. The triangles ABC, DEF, shall be equal to one another in every respect.

For, if the side A B be made to coincide with DE, which is equal to it, the right angle ABC will also coincide with the right angle DEF (1. and ax. 11). Therefore, if AC do not coincide with DF, but fall otherwise, as DG, there will be drawn from the point D to the line E F, upon the same side of the perpendicular, two straight lines that are equal to one another, which is impossible (12. Cor. 2.). Therefore, A C coincides with D F, and the triangle_ A B C coincides with the triangle DEF, that is, (ax. 11.) the triangles ABC and DEF are equal in every respect.

Next, let the hypotenuse AC and angle ACB of the one triangle be equal to the hypotenuse DF and angle DFE of the other. In this case, also, the triangles shall be equal in every respect.

For if the hypotenuse A C be made to coincide with D F, which is equal to it, the angle A CB will also coincide with DFE, which is equal to it. Therefore, if A B do not coincide with DE, but fall otherwise, as DG there will be

line EF, 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 AB.

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 GF. Through E draw lm, 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 H K, 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,)

な

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 G H.

Hence, it appears, that A B, 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 E F, 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,