Book I.

For GH, HC, are equal to EH, HC; and the bafes GC, EC2, equal: Therefore the angles GHC, EHC, are equal 1. ader. 15. Each of them are right angles, and HC perpendicular to AB. Which was required.

b

a def. 10.

b II.

c ax. I.

*Fig. to

prop. 9.

d 13. e 5.

fax. 3.

2. 13.

b. hyp.

W

PRO P. XIII. THE OR.

HEN a right line ftands upon a right line, making angles with it, thefe angles are either two right angles, or, together, equal to two right angles.

For, let a right line, AB, ftand upon the right line CD, making angles CBA, ABD; these angles fhall either be two right angles, or, together, equal to two right angles.

For, if CBA, ABD, be equal, they are right angles; if not, from the point B draw BE, at right angles, to DC: Therefore the two right angles CBE, EBD, are equal to the three angles ABC, ABE, EBD; but the two angles ABC, ABD, are equal to the fame three angles: Therefore the two angles ABC, ABD, are equal to the two angles CBE, EBD; that is, equal to two right angles. Wherefore, &c.

COR. If the two fides of an ifofceles triangle ADE, be pro-, duced to B, C, the angles below the bafe will be equal to one another; for the angles ADE, EDB*, are equal to two right angles, and AED, DEC, are equal to two right angles; but the angles ADE, AED, above the bafe, are proved equal: Therefore the remaining angles BDE, DEC, are equal f: There fore, in every ifofceles triangle, the angles above the base are equal to one another; and, if the fides be produced, the angles below the bafe are likewise equal to one another.

PRO P. XIV. THE OR.

F to any right line, and point therein, two right lines be drawn

I from oppofite points, making the adjacent angles together e

qual to two right angles, these two right lines will make one conti nued right line.

For, if to any right line AB, and point B therein, be drawn two right lines, CB, DB, from the oppofite points, C, D, making the angles ABC, ABD, equal to two right angles; then DBC. will be one right line. If not, let CBE be one right line; then the angles ABC, ABE, will be equal to two right angles; but ABC, ABD, are equal to two right angles; Therefore the

two angles ABC, ABD, are equal to the two angles ABC, Book I, ABE. Take ABC from both; then the angle ABE will be equal to the angle ABD, a part to the whole; which cannot be. Wherefore, &c.

COR. Hence two right lines CBD, CBE, cannot have a common fegment as CB; or BD, BE, cannot both be in a right line with CB.

PRO P. XV. THE O R.

F two right lines mutually cut each other, the oppofite angles are equal.

IF

a 13.

Let the right lines AB, CD, mutually cut each other in the point E, the angles AEC, DEB, will be equal; and likewife the angles CEB, AED, equal to one another. For, because the right line CE falls upon the right line AB, the angles AEC, CEB, are equal to two right angles. For the fame reason the angles AED, AEC, are equal to two right angles : Therefore the two angles AEC, CEB, are equal to the two angles AEC, AED. Take the common angle AEC from both, the remain-b Ax... ing angles CEB, AED, are equal". Again, because AEC, AED, are equal to two right angles, and AED, DEB, equal to two right angles, take the common angle AED from both, the remaining angles AEC, DEB, will be equal. Wherefore, &c.

COR. 1. Hence, two right lines cutting each other, the angles at the fection are equal to four right angles.

2. All the angles conftitute about any point are equal to four right angles.

Fone

PRO P. XVI. THE OR.

one fide of a triangle be produced, the outward angle will be greater than either of the inward oppofite angles.

Let ABC be a triangle, and one of its fides BC be produced to D, the outward angle ACD will be greater than the angle CBA, or BAC.

For, bifect AC in E; join BE, which produce to F; make a 10. EF equal to EB, and join FC; then the two fides AE, EB, are equal to the two fides FE, EC, and the angles AEB, FEC, equal: Therefore the bafes FC, AB, are equal; and the angles b 15,

ECF,

Book. I. ECF, EAB, likewife equal: Therefore the angle ACD is greater than the angle BAC. In like manner, if the fide BC is bifected in E, EF made equal to AE, and FC joined, the angle BCG, or ACD4, is greater than ABC; but ACD is likewife proved greater than BAC. Wherefore,

C 4.

d 15.

&c.

a 16.

b 131

a 3.

b 16.

WO angles of any triangle, however taken, are, together, lefs than two right angles.

Let ABC be the triangle, any two angles in it are less than two right angles.

For, produce BC both ways to D, E; then, becaufe the outward angle ACD is greater than ABC, add ACB to both; then the angles ACD, ACB, are greater than ABC, ACB, or, BAC, ACB; but ACD, ACB, are equal to two right angles : Therefore ABC, ACB, or ACB, BAC, are lefs than two right angles. For the fame reafon, ABE, ABC, are greater than BAC, ABC. Wherefore, &c.

COR. Hence, if a right line fall upon two right lines, making the inward angles on the fame fide lefs than two right angles, thefe lines will meet one another on that fide where the angles are lefs than right ones.

PRO P. XVIII. THEOR.

HE greater fide of every triangle fubtends the greater.

Tangle.

Let ABC be a triangle, and the fide AC greater than AB; then the angle ABC will be greater than the angle ACB.

For, from the greater AC cut off AD, equal to AB, join DB; then, because ADB is greater than ACB, ABD is likewife greater than ACB, and ABC much greater. Wherefore,

&c.

PRO P. XIX. THE OR.

HE greater angle of every triangle is fubtended by the greater fide.

TH

In the triangle ABC let the angle ABC be greater than the Book I angle BCA; the fide AC will be greater than AB. If not, let AČ be either equal or lefs than AB. If equal, then the angle ABC is equal to ACB; but it is not: Therefore AC is not a 5. equal to AB. If AC is lefs than AB, the angle ABC is lefs b Hyp. than ACB; but it is not: Therefore AC is not lefs than AB. c 18. It is therefore greater, fince it has been proved neither equal nor lefs. Wherefore, &c.

T

PRO P. XX. THE OR.

WO fides of any triangle, however taken, are greater
than the third.

In any triangle, ABC, two fides of it, however taken, are greater than the third, viz. AB, AC, greater than BC; AC, BC, greater than AB; or BC, AB, greater than AC. For, produce any fide, as BA, to D; make AD equal to ACa; and a 3. join DC: Then, because AD is equal to AC, the angles ADC, ACD, are equal; but the angle BCD is greater than ACD, 5. that is, than ADC: Therefore the fide BD is greater than BC; but BD is equal to BA, AC: Therefore the fides BA, AC, are greater than BC. Wherefore, &c.

PRO P. XXI. THE OR.

F two right lines be drawn from the extreme points of one fide of a triangle, to a point within the fame, thefe two right lines will be less than the fides of the triangle, but contain a greater angle.

From the extreme points of the right line BC, let the two right lines BD, CD, be drawn to the point D, within the fame; thefe lines fhall be less than the fides BA, AC; but the angle BDC will be greater than BAC.

For, produce BD to E; then the two fides BA, AE, are great

b

C 19.

er than the third fide BE; add EC to both; then BA, AČ, are a 20. greater than BE, EC. For the fame reason, BE, EC, are b, Ax.4; greater than BD, DC; but BA, AC, are greater than BE, EC; therefore much greater than BD, DC. But the angle BDC is greater than BAC; for the angle BDC is greater than BEC; and BEC is greater than BAČ: Therefore BDC is c 16. much greater than BAC. Wherefore, &c,

COB

BOOK I.

Def, 15.

$ 8.

COR. Hence BD, DC, are not equal to BA, AC, each to each. Wherefore, if in any case it is thought neceflary to prove that part of Prop. VII. when the one point falls within the triangle, it is evident from this.

T

PRO P. XXII. PRO B.

O make a triangle, whofe fides are equal to three given right lines, if any two of them, however taken, are greater than the third,

Let A, B, C, be the three given right lines, any two of which are greater than the third. Take any right line bounded at D, but not bounded at E, from which cut off DF equal to A, FG equal to B, and make GH equal to C; then, with the center F, and distance DF, defcribe the circle DKL; with the center G, and distance GH, describe the circle KLH; from the point K, where the circles cut each other, draw the right lines FK, KG; then FD is equal to FK*; but FD is equal to A; therefore FK is equal to A. For the fame reafon GK is equal to C, and FG is equal to B: Therefore the three sides FK, FG, GK, of the triangle FKG, are equal to the three given right lines, A, B, C. Wherefore there is conftitute, &c,

A

PRO P. XXIII. PRO B.

Ta given point, in any right line, to make an angle equal to a given right lined angle.

Let A be the given point in the right line AB; it is required to make an angle equal to the right lined angle DCE.

Affume any points D, E, in the right lines CD, CE, and join DE. At the point A, in the line AB, make a triangle AFG, whofe fides are equal to the three right lines CD, CE, DE2; then, because the two fides GA, AF, are equal to the two fides CE, CD, each to each, and the bafes GF, ED, equal, the angles GAF, ECD, are equal. Wherefore there is conftitute,' &c.

PRO P. XXIV. T HE OR.

I F two triangles have two fides of the one equal to two fides of the other, each to each, and the angle contained by the two fides of the one greater than the angle contained by the correfpond