Book I.

a 10. I.

b 15.1.

PROP. XVI. THEOR.

F fide of a triangle be produced, the exterior Tangle is greater than either of the interior oppo

fite angles.

Let ABC be a triangle, and let its fides BC be produced to D, the exterior angle ACD is greater than either of the

interior opposite 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.

A

10

C

C

G

F

1

D

Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; and the angle AEB is B equal b to the angle CEF, because they are opposite vertical angles; therefore the base C 4.1.. 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 bisected, it may be demonftrated that the angle BCG, that is d, the angle ACD, is greater than the angle ABC. Therefore, if one fide, &c. Q. E. D.

d15 1.

A

PROP. XVII. THEOR.

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

Le

the angle ACB; therefore B

C

а 16. 1.

D

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 demonftrated, that BAC, ACB, as alfo CAB, ABC, are less than two right angles. Therefore, any two angles, &c. Q. E. D.

T

HE greater side of every triangle has the great-
er angle opposite to it.

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

Because AC is greater than AB, make a AD equal to AB, В

:

1)

C

a 3. 1.

and join BD; and because ADB is the exterior angle of the Criangle BDC, it is greater b than the interior and opposite b 16. 1. angle DCB; but ADB is equal e to ABD, because the fide c 5. 1. AB is equal to the fide AD; therefore the angle ABD is ikewife greater than the angle ACB; wherefore much more Is the angle ABC greater than ACB. Therefore the greater fide, &c. Q. E. D.

:

T

PROP. XIX. THEOR.

HE greater angle of every triangle is fubtend ed by the greater fide, or has the greater fide

opposite to it.

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

A

a 5. 1.

For, if it be not greater, AC must either be equal to AB, or less 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 B

C

b 18. 1.

less; because then the angle
ABC would be less b than the angle ACB; but it is not,
therefore the side AC is not less than AB; and it has been
shewn that it is not equal to AB; therefore AC is greater
than AB. Wherefore the greater angle, &c. Q. E. D.

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

a 3.1. D, and make a AD equal to AC; and join DC.

D

A

Because DA is equal to

b 5. 1.

AC, the angle ADC is like-
wise equal b to ACD; but
the angle BCD is greater

than the angle ACD there- B

C

fore the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle. DCB is greater than its - Book I angle BDC, and that the greater c fide is opposite 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 demonstrated, that the fides AB, BC are greater than CA, and BC,

с 19.1.

CA greater than AB. Therefore any two fides, &c.

Q. E. D.

F from the ends of the fide of a

f a triangle, there be See N.

drawn two ftraight lines to a point within the triangle, these two lines shall be less than the other two fides of the triangle, but shall contain a greater angle.

Let the two straight lines BD, CD be drawn from B, C, the ends of the fide BC of the triangle ARC, 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 BAС.

A

E

Produce BD to E; and because two fides of a triangle are greater than the third fide, the two fides BA, AE of the triangle ABE are greater than BE. To each of these add EC; therefore the fides BA, AC are greater than BE, EC: Again, - because the two fides CE, ED of the triangle CED are greater than CD, add DB to each of these; therefore the fides CE, EB are greater than CD, DB; but it has been shewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC.

B

D

C

Again, because the exterior angle of a triangle is greater than the interior and opposite 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

Book I. 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.

á 20. 1.

PROP. XXII. PROB.

To make a triangle of which the fides shall be equal to three given straight lines; but any two whatever of these lines 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 shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but

unlimited towards E,

a 3. 1.

and makea DF equal
to A, FG to B, and
GH equal to C; and
from the centre F, at
the distance FD, de- D

b 3. Poft. fcribeb the circle DKL; and from the centre G, at the distance GH, describe b another circle HLK; and join KF, KG; the triangle KFG has its fides equal to the three straight lines, A, B, C.

Because the point F is the centre of the circle DKL, FD is C II. Def. equal e to FK; but FD is equal to the straight 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.