Book I.

a IO. I.

PROP. XVI. THEOR.

[F one fide of a triangle be produced, the exterior angle is greater than either of the interior 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 ACin E, joinBE and produce it to F, and make EF equal to BE; join alfo FC,and produce AC to

G.

C

B

A

F

E

D

G

Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle b 15. 2. AEB is equalb to the angle CEF, because they are oppofite vertical angles; there4. 1. fore the base AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles ta 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 ACD is greater than BAE: In the fame manner, if the fide BC be bifected, it may be demonftrated that 15. 1. the angle BCG, that is d, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D.

A

PROP. XVII, THEOR.

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

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

Produce BC to D; and because ACD is the exterior angle of the triangle ABC, ACD is a 16. 1. greater a than the interior and oppofite angle ABC; to each of

A

B

C

D

thefe

these add the angle ACB; therefore the angles ACD,ACB are Book. I. greater than the angles ABC, ACB; but ACD, ACB are together equal b to two right angles; therefore the angles ABC, b 13. 1. BCA are lefs than two right angles. In like manner, it may demonftrated, that BAC, ACB, as alfo CAB,ABC, are less than two right angles. Therefore any two angles, &c. Q. E. D.

PROP. XVIII. THEOR.

be

'HE greater fide of every triangle is oppofite to the
greater angle.

TH

the interior and oppofite angle DCB; but ADB is equal e to c 5. 1, ABD, because 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. There fore the greater fide, &c. Q. E. D.

THE

PROP, XIX. TH FOR.

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 likewise 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

a 5. I.

ABC

Book I. ABC would be lefs b than the angle ACB; but it is not; there fore 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.

b 18. I.

See N.

A

PROP. XX. THEOR.

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

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

Produce BA to the point D, a 3. 1. and make a AD equal to AC; and join DC.

b5. 1.

Because DA is equal to AC, the angle ADCis likewife equal b to ACD; but the angle BCD is greater than the angle ACD; therefore the angle BCDisgreater than the angle ADC; and be

C

B

A

D

cause the angle BCD of the triangle DCB is greater than its €19. 1. angle BDC, and that the greater fide is oppofite to the greater angle; therefore the fide DB is greater than the fide BC; but DB is equal to BA and AC; therefore the fides BA, AC are greater than BC. In the fame manner it may be demonftrated, that the fides AB, BC are greater than CA,and BC, CA great er than AB. Therefore any two fides, &c. Q. E. D.

IFa

PROP. XXI. THEOR.

See N. TF, from the ends of the fide of a triangle, there be drawn two ftraight lines to a point within the triangle, these shall be lefs than the other two fides of the triangle, but shall contain a greater angle.

Let the two ftraight lines BD, CD be drawn from B, C, the ends of the fide BC of the triangle ABC, to the point D within it; BD and DC are less than the other two fides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC.

Produce BD to E; and because two fides of a triangle are greater than the third fide, the two fides BA, AE of the tri

angle

then are BA, AC greater than BD, DC.

Again, because the exterior angle of a triangle is greater than the interior and oppofite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the fame reason, the exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonftrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D.

PROP. XXII. PRO B.

O make a triangle of which the fides fhall be equal See N. to three given straight lines, but any two whatever

a

of these must be greater than the third a.

Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B ; and B and C than A. It is required to make a triangle of which the fides fhall be equal to A, B, C, each to each.

Take a ftraight line DE terminated at the point D, but unlimited towards E, and

a 20. I.

make a DF equal to A,

FG to B, and GH equal
to C; and from the cen-
tre F, at the distance,
FD, defcribe bthe circle D
DKL; and from the cen-

tre G, at the distance
GH, defcribe b another

circle HLK; and joinKF, KG; the triangle KFG

has its fides equal to the three ftraight lines, A, Because the point F is the centre of the circle

Book I. Lig

C. 15. Def.

с

equal to FK; but FD is equal to the ftraight line A; therefore FK is equal to A: Again, because G is the centre of the circle LKH, GH is equal to GK; but GH is equal to C; therefore alfo GK is equal to C; and FG is equal to B; therefore the three straight lines KF, FG, GK, are equal to the three A, B. C: And therefore the triangle KFG has its three fides KF, FG, GK equal to the three given straight lines, A, B, C. Which was to be done.

PROP. XXIII. PRO B.

Ta given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle; it is required to make an angle at the given point A in the given ftraight line AB, that fhall be equal to the given rectilineal angle DCE.

Take in CD, CE any points D, E, and a 22. 1. join DE; and make a the triangle AFG the fides of which fhall be equal to the three

ДД

B

G

ftraight lines CD, DE, CE, so that CD be equal to AF, CE to AG, and DE to FG; and because DC, CE are equal to FA, AG, each to each, and the base DE to the base FG; the angle b 8. 1. DCE is equal b to the angle FAG. Therefore, at the given point A in the given ftraight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done.

See N.

IF

PROP. XXIV. THEOR.

F two triangles have two fides of the one equal to two fides of the other, each to each, but the angle contained by the two fides of one of them greater than the angle contained by the two fides equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other.

Let