cause the angle BCD of the triangle DCB is greater than its Book I. angle BDC, and that the greater c 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 together; therefore BA and AC together are greater than BC. In the fame manner it may be demonftrated, that the fides AB, BC are greater than CA, and BC, CA greater than AB. Therefore any two fides, &c. Q. E. D.

F from the ends of one fide of a triangle, there be N. drawn two ftraight lines to a point within the triangle, these two lines fhall be lefs than the other two fides of the triangle, but fhall contain a greater triangle.

I a

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 lefs than the other two fides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC.

D

Produce BD to E; and because two fides of a triangle a are a 20. 1. 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 thefe; therefore the fides CE, EB are greater than CD, DB; but it has been fhewn that BA, AC are greater than BE, EC; much more then are BA, AC greater than BD, DC.

B

Again, because the exterior angle of a triangle b is greater b 16. I. 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

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,

a 20. I.

a 3. I.

T

O conftruct a triangle of which the fides fhall 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 ftraight line DE terminated at the point D, but unlimited towards E,

and make DF equal

to A, FG to B, and
GH equal to C; and
from the centre F, at
the diftance FD, de-

bz. Poft. fcribeb the circle DKL; D
and from the centre G,
at the distance GH,
defcribeb another cir-
cle HLK; and join
KF, KG; the triangle
KFG has its fides equal
to the three ftraight
lines, A, B, C.

Because the point F is the centre of the circle DKL, FD is c 11. Def. 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 ftraight 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.

Book I.

AT

Ta given point in a given ftraight 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.

AA

Take in CD, CE any points D, E, and join DE; and make a the tri- D angle AFG, the fides of which fhall be equal to

the three ftraight lines

a 22. I.

CD, DE, CE, fo 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 bafe FG; the angle DCE is equal b to the angle FAG. Therefore, at the given b. 8. 1. 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.

PROP. XXIV. THEOR.

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 the one greater than the angle contained by the two fides of the other; the bafe of that which has the greater angle fhall be greater than the base of the other.

Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two DE, DF, each to each, viz. AB

equal

Book I. equal to DE, and AC to DF; but the angle BAC greater than the angle EDF; the bafe BC is alfo greater than the base EF.

Of the two fides DE, DF, let DE be the fide which is not greater than the other, and at the point D, in the straight line DE, make a the angle EDG equal to the angle BAČ: and make DG equal b to AC or DF, and join EG, GF.

Because AB is equal to DE, and AC to DG, the two fides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is e

D

DF, the angle

e 19. I.

but the
the angle

DGF is greater

F

that the angle EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF; and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater fide is oppofite to the greater angle; the fide EG is therefore greater than the fide EF; but EG is equal to BC; and therefore alfo BC is greater than EF. Therefore, if two triangles, &c. Q. E. D.

IF

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

Let ABC, DEF be two triangles which have the two fides AB, AC equal to the two fides DE, DF, each to each, viz.

AB

AB equal to DE, and AC to DF; but the base CB is greater than the bafe EF; the angle BAC is likewife greater than the angle EDF.

a

D

For, if it be not greater, it must either be equal to it, or less; but the angle BAC is not equal to the angle EDF, because then the base BC would be equal a to EF; but it is not; therefore the angle BAC is not equal to the angle EDF; neither is it less; because then the base BC would be lefs b than the base EF; but it is not; therefore the angle BAC

Book I.

a 4. I.

AA

B

E

b 24. I.

F

is not less than the angle EDF; and it was fhewn that it is not equal to it: therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D.

F two triangles have two angles of the one equal to two angles of the other, each to each; and one fide equal to onè fide, viz. either the fides adjacent to the equal angles, or the fides oppofite to the equal angles in each; then fhall the other fides be equal, each to each; and alfo the third angle of the one to the third angle of the other.

Let ABC, DEF be two triangles which have the angles ABC BCA equal to the G angles DEF, EFD, viz. ABC to DEF, and BCA to EFD;

alfo one fide equal to one fide; and first let thofe fides be equal

which are adjacent to

B

A

CE

D

the angles that are equal in the two triangles, viz. BC to EF;

the