PRO P. X. PRO B.

Book I.

To cut

O cut a given finite right line into two equal parts.

Let AB be a given right line, required to be cut into two equal parts; upon it defcribe an equilateral triangle ABC; bifect the angle ACB by the right line CD; then is the right a gi line AB bifected in D.

For, because AC is equal to CB, and CD common, and the angles ACD, BCD, equal, the bafes AD, BD, are equal:b 4. Therefore the right line AB is bisected in D. Which was required.

PRO P. XI. PR O B.

O draw a line at right angles to a given right line from a
point given in the fame.

T

Let AB be the given right line, and C the given point in it, from which it is required to draw a right line, at right angles to the given right line AB.

b

Affume any point D in AC, and make CE equal to CD"; a 3. DE describe an equilateral triangle DEF, and join FC, upon which will be at right angles to AB. For, because DC, CE, are equal, and CF common, the two fides DC, CF, are equal c. const, to the two fides EC, CF, and the bases FD, FE, equal ", the angles FCD, FCE, are likewife equald: Therefore each of them d 8. is a right angle, and FC perpendicular to AB. Which was e Def. 10j required.

b

T

PRO P. XII. PRO B.

O draw a right line perpendicular to a given indefinite right
line from a given point out of it.

Let AB be the given right line, and C the point given out of it, from which its required to let fall a perpendicular upon the indefinite given right line AB.

Affume any point D, on the oppofite fide of the right line AB; about the center C, with the distance CD, describe a circle EDG; bisect EG in H, join CG, CH, CE; then CH is the perpendicular required.

B

For

Book I.

For GH, HC, are equal to EH, HC; and the bases GC, EC, equal: Therefore the angles GHC, EHC, are equal. a def. 15. Each of them are right angles, and HC perpendicular to AB. Which was required.

b 8.

c def. 10.

a def ro.

b II.

c ax. I.

* Fig, to

prop. 9.

d 13. e 5.

f ax. 3.

2. 13.

b. hyp.

W

PRO P. XIII. THE OR.

HEN a right line ftands upon a right line, making angles with it, thele 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 angl、 s CBA, ABD; these angles fhall either be two right angles, or, together, equal to two right angles.

a

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 produced 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, ALD, above the bafe, are proved equal: Therefore the remaining angles BDE, DEC, are equal f: Therefore, in every ifofceles triangie, the angles above the base are equal to one another; and, if the fides be produced, the angles below the base are likewife equal to one another.

PRO P. XIV. THE OR.

F to any right line, and point therein, two right lines be drawn from oppofite points, making the adjacent angles together equal to two right angles, these two right lines will make one continued 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 2; but ABC, ABD, are equal to two right angles : Therefore the

b

a

two

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 OR.

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

I'

Let the right lines AB, CD, mutually cut each other in the point E, the angles AEC, DEB, will be equal; and likewise 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 reafon the a 13. angles AED, AEC, are equal to two right angles: Therefore the two angles AEC, CEB, are equal to the two angles AEC, AED b. Take the common angle AEC from both, the remaining angles CEB, AED, are equal. Again, because AEC, Ax. 3. AED, are equal to two right angles 2, 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 section are equal to four right angles.

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

PRO P. XVI. THEOR.

Fone 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.

b Ax. f. ›

For, bifect AC in E2; 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 bx51

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.

C 4. d 15.

&c.

Wherefore,

a 16.

b 130

a 3.

b 16.

PRO P. XVII. THE OR.

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

The

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.

b

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 less than right ones.

T

PRO P. XVIII. THE OR.

HE greater fide of every triangle fubtends the greater angle.

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

T

PRO P. XIX. THE O R.

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

In

In the triangle ABC let the angle ABC be greater than the Book L. angle BCA; the fide AC will be greater than AB. If not, let AC be either equal or lefs than AB. If equal, then the angle ABC is equal to ACB; but it is notb: 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.

WV O 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 AC; 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; 19、 but BD is equal to BA, AC: Therefore the fides BA, AC, are greater than BC. Wherefore, &c.

I'

PRO P. XXI. T H E O R.

F two right lines be drawn from the extreme points of one fide of a triangle, to a point within the fame, these 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 lefs 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

с

er than the third fide BEa; add EC to both; then BA, AČ, are a 10. greater than BE, EC. For the fame reafon, 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 BAC : Therefore BDC is 1ếu much greater than BAC. Wherefore, &c.

COR.