Book I. For, if the angle CBA be equal to ABD, each of them is

b

a

a def. 7. a right angle ; but, if not, from the point B draw BE at b. I. right angles to CD; therefore the angles CBE, EBD are two right angles. Now, the angle CBE is equal to the two angles CBA, ABE together; add the angle EBD to each of thefe cquals, and the two angles CBE, EBD will be 2. Az equal to the three CBA, ABE, EBD. Again, the angle DBA is equal to the two angles DBE, EBA; add to each of thefe equals the angle ABC; then will the two angles DBA, ABC be equal to the three angles DBE, EBA, ABC: but the angles CBE, EBD have been demonftrated to be equal to the fame three angles; and things that are equal to the fame dr. Ax. are equal to one another; therefore the angles CBE, EBD are equal to the angles DBA, ABC; but CBE, EBD are two right angles; therefore DBA, ABC are together equal to two right angles. Wherefore, when a ftraight line, &c. Q. E. D.

I'

d

PROP. XIV. THEOR.

F, at a point in a ftraight line, two other ftraight lines, upon the oppofite fides of it, make the adjacent angles together equal to two right angles, thefe two ftraight lines fhall be in one and the fame straight line.

At the point B in the straight line AB, let the two ftraight lines BC, BD upon the oppofite fides of AB, make the adjacent angles ABC, ABD equal together to two right angles. BD is in the fame ftraight line with CB.

For, if BD be not in the fame ftraight line with CB,

A

E

B

D

let

let BE be in the fame ftraight line with it; therefore, be-, Book I. cause the ftraight line AB makes angles with the straight line CBE, upon one fide of it, the angles ABC, ABE are together equal a to two right angles; but the angles ABC, a 13. 1. ABD are likewife together equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD: Take away the common angle ABC, and the remaining angle ABE is equal to the remaining angle ABD, the lefs to the greater, which is impoffible; therefore BE is not in the fame straight line with BC. And in like manner, it may be demonstrated, that no other can be in the fame ftraight line with it but BD, which therefore is in the fame ftraight line with CB. Wherefore, if at a point, &c. Q. E. D.

IF

b

PROP. XV. THEOR.

F two ftraight lines cut one another, the vertical, or oppofite angles fhall be equal.

Let the two ftraight lines AB, CD cut one another in the point E; the angle AEC fhall be equal to the angle DEB, and CEB to AED.

For the angles CEA, AED, which the straight line AE makes with the straight line CD, are to- Agether equal a to two right angles and the angles AED, DEB, which the ftraight line DE makes with the ftraight line AB,

a

3. As.

are alfo together equal to two right angles; therefore the two angles CEA, AED are equal to the two AED, DEB. Take away the common angle AED, and the remaining angle CEA is equalb to the remaining angle DEB. In 1 3. the fame manner it can be demonstrated that the angles CEB, AED are equal. Therefore, if two ftraight lines, &c. Q. E. D.

COR. 1. From this it is manifeft, that, if two ftraight lines cut one another, the angles which they make at the point of their interfection, are together equal to four right angles.

COR. 2. And hence, all the angles made by any number of ftraight lines meeting in one point, are together equal to four right angles.

C 2

PROP.

Ал.

Book I.

a Io. I.

c

€ 4. I.

one fide of a triangle be produced, the exterior angle is greater than either of the interior, and oppofite angles.

Let ABC be a triangle, and let its fide BC be produced to D, the exterior angle ACD is greater than either of the interior oppofite angles CBA,

BAC.

Bifecta AC in E, join BE and produce it to F, and make EF equal to BE; join alfo FC, and produce AC to G.

Because AE is equal to EC,

b

A

C

D

and BE to EF; AE, EB are equal to CE, EF, each to B each; and the angle AEB is b15. 1. equal to the angle CEF, because they are vertical angles; therefore the bafe AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal fides are oppofite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore the angle ECD, that is ACD, is greater than BAE: In the fame manner, if the fide BC be bifected, it may be demonstrated that the angle BCG, that is b, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D. Q.E.

A

NY two angles of a triangle are together lefs than two right angles.

Let

Let ABC be any triangle; any two of its angles together are lefs than two right angles.

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is greater a than the interior and oppofite angle ABC; to each of thefe add the angle ACB; therefore

the angles ACD, ACB are greater than the angles ABC, ACB; but ACD, ACB are together equal b to two right b 13. 1. angles; therefore the angles ABC, BCA are less than two right angles. In like manner, it may be demonstrated, that BAC, ACB, as also, CAB, ABC, are less than two right angles. Therefore, any two angles, &c. Q. E. D.

PROP. XVIII. THEOR.

HE greater fide of every triangle has the great-
er angle oppofite to it.

TH

Let ABC be a 'triangle, of which the fide AC is greater than the fide AB; the angle ABC is alfo greater than the angle BCA.

From AC, which is greater than
AB, cut off a AD equal to AB, B

C

and join BD; and because ADB is the exterior angle of the triangle BDC, it is greater b than the interior and oppofite angle DCB; but ADB is equal to ABD, becaufe the fide AB is equal to the fide AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than ACB, Therefore the greater fide, &c. Q. E. D.

a 3. I.

b 16. 1.

c 5.1.

22

Book I.

a 5. r.

TH

PROP. XIX. THEOR.

HE greater angle of every triangle is fubtended by the greater fide, or has the greater fide oppofite to it.

Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the fide AC is likewife greater than the fide AB.

For, if it be not greater, AC muft either be equal to AB, or lefs than it; it is not equal, because then the angle ABC would be equal a to the angle ACB; but it is not; therefore AC is not equal to AB; neither is it lefs; because then the angle

B

A

bis. 1. ABC would be lefs b than the angle ACB; but it is not; therefore the fide AC is not lefs than AB; and it has been fhewn that it is not equal to AB; therefore AC is greater. than AB. Wherefore the greater angle, &c. Q. E. D.

a 3: I.

b 5. I.

A

PROP. XX. THEO R.

NY two fides of a triangle are together greater than the third fide.

greater

Let ABC be a triangle; any two fides of it together are greater than the third fide, viz. the fides BA, AC than the fide BC; and AB, BC greater than AC; and BC, CA greater than AB,

Produce BA to the point

D, and make a AD equal
to AC; and join DC.

Because DA is equal to
AC, the angle ADC is like-
wife equalb to ACD; but
the angle BCD is greater
than the angle ACD; there-

D

fore the angle BCD is greater than the angle ADC; and be.

caufe