Book I.

a. 15. I.

b. 15. 1.

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 opposite angles CВА, ВАС.

Bifecta AC in E, join BE
and produce it to F, and make
EF equal to BE; join also
FC, and produce AC to G.
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 to the angle CEF, be-
cause they are opposite verti-
cal angles. therefore the base

A

F

E

C

D

G

C. 4. I.. 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 opposite. 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 bisected, it may be demonstrated that the angle BCG, that is d, the angle ACD, is greater than the angle ABC. therefore if one side, &c. Q. E. D.

d. 15. 1.

A

PROP. XVII. THEOR.

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

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

A

a. 16. 1.

Produce BC to D; and be-
cause ACD is the exterior angle
of the triangle ABC, ACD is
greater a than the interior and B
opposite angle ABC; to each of

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 to two right angles; therefore the angles ABC, BCA are b. 13. 1. less than two right angles. in like manner it may be demonftrated that BAC, ACB, as also CAB, ABC are less than two right angles. therefore any two angles, &c. Q. E. D.

PROP. XVIII. THEOR.

THE greater fide of

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

Let ABC be a triangle of

which the fide AC is greater than the fide AB; the angle ABC is also greater than the angle BCA.

A

Because AC is greater than

AB, make a AD equal to AB,
and join BD. and because ADB B

D

a. 3. 1.

1

is the exterior angle of the tri

angle BDC, it is greater than the interior and opposite angle b. 16. 1.

DCB. but ADB is equal to ABD, because the fide AB is c. 5. 1.

equal to the fide AD; therefore the angle ABD is likewife greater

than the angle ACB; wherefore much more is the angle ABC

greater than ACB. therefore the greater fide, &c.

E. D.

PROP. XIX. THEOR.

is fubtended by

THE greater angle of every triangle

the greater fide, or has the greater fide opposite 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.

b. 18. 1.

Book I. 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.

a. 3. 1.

b. 5. 1.

PROP. XX. THEOR.

NY two fides of a

than the third side.

triangle are together greater

Let ABC be a triangle; any two fides of it together are greater than the third side, 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,
and make a AD equal to AC,
and join DC.

Because DA is equal to AC,
the angle ADC is likewise equal
b to ACD. but the angle BCD
is greater than the angle ACD; B
therefore the angle BCD is great-

D

A

C

er than the' angle ADC. and because the angle BCD of the tric. 19. 1. angle DCB is greater than its angle BDC, and that the greater 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 demonftrated that the fides AB, BC are greater than CA; and BC, CA greater than AB. therefore any two fides, &c. Q. E. D.

See N.

PROP. XXI. THEOR.

IF from the ends of the fide of

a

triangle there be drawn two straight lines to a point within the triangle, these 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 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 great-er than the third side, the two fides BA, AE of the triangle ABE

are greater than BE. to each of these add EC, therefore the fides Book I.

BA, AC are greater than BE,

EC. 'again, because the two fides

A

E

CE, EB are greater than CD,

DB. but it has been shewn that
BA, AC are greater than BE, B

EC; much more 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 demonstrated 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...

10 make

T

PROP. XXII. PRO B.

a triangle of which the fides shall be equal See N. to three given straight lines; but any two whatever 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 shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unli

mited towards E, and make

2 DF equal to A, FG to B, and GH equal to C; and from the center F, at the distance FD describe the D circle DKL. and from the

a. 20. 1.

center G, at the distance GH describe another cir

cle HLK, and join KF, KG.

ABC

the triangle KFG has its

fides equal to the three straight lines A, B, C.

Because the point F is the center of the circle DKL, FD is

A

Book I. equal to FK; but FD is equal to the straight line A; therefore FK is equal to A. again, because G is the center of the circle LKH, c. 15. Def. 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 furaight 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.

a. 22. I.

1.8. 1.

ee N.

A

T

PROP. XXIII. PROB.

a 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 straight line AB that shall be equal to the given rectilineal angle DCE.

Take in CD, CE, any

points D, E, and join DE;
and make
a the triangle D

AFG the fides of which
shall be equal to the

B

G

three ftraight lines CD, DE, EC, 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 base FG; the angle DCE is equal to the angle FAG. therefore at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Which was to be done.

IF

PROP. XXIV. THEOR.

two triangles have two sides of the one equal to two fides of the other, each to each, but the angle contained by the two sides 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 ABC, DEF be two triangles which have the two fides AB,