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Book I. ABC would be less than the angle ACB; but it is not; there, m fore the fide AC is not less than AB; and it has been shewn € 18, 1.

that it is not equal to AB; therefore AC is greater than AB. Wherefore the greater angle, &c. Q. E. D.

See N.

AN

3.1.

PROP. XX. THEOR.
NY two sides of a triangle are together greater than

the third fide.
Let ABC be a triangle ; any two sides of it together are
greater than the third fide, viz. the ides 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 • 1D equal to AC; and join DC.

because DA is equal to AC, the angle ADC is likewise equal © to ACD; but the angle BCD is greater than the angle ACD; therefore the angle BCD is great

с er than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater dide is opposite to the greater angle; therefore the Gde DB is greater than the side BC; but DB is equal to BA and AC ; therefore the sides BA, AC are greater than BC. In the same manner it may be demonstrated, that the sides AB, BC are greater than CA, and BC, CA greater than AB. Therefore any two sides, &c. Q. E. D.

C 19.1.

PROP. XXI. THEOR.

See N.

F, 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 sides of the tri angle, 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 sides BA, AC of the triangle, but contain an angle BDC greater than the angle BAC.

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

angle

angle ABE are greater than BE. To each of these add EC; Book I. therefore the sides BA, AC are greater than BE, EC: Again,

E
because the two fides CE, ED
of the triangle CED are great-
er than CD, add DB so each
of these ; therefore the fides
CE, EB are greater than CD,
DB; but it has been shewn that

B
BA, AC are greater than BE,
EC; much more then are BA,
AC greater than BD, DC.

Again, because the exterior angle of a triangle is greater than the anterior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED; for the same 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.

a 20. 1.

å 3. I.

PRO P. XXII. PRO B. To make a triangle of which the sides shall be equal Sce Ni

to three given ftraight 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 un limited towards E, and makeo DF equal to A, FG to B, and GH equal to C; and from the centre F, at the distance FD, defcribe

the circle DKL; and D from the centre G, at the

G HE distance GH, debeb another circle HLK, and

ΤΑ join KF, KG; the triangle KFG has its fades equal to the three straight lines, A, B, C. Because the point F is the centre of the circle DKL, FD is

equal

b 3. Poft, Book I. equal to FK; but FD is equal to the straight line A: there.

fore FK is equal to A : Again, because G is the centre of the c 15. Def. circle LKH, GH is equal to GK; but GH is equal to C;

therefore allo GK is equal to C; and FG is equal to B ; there-
fore the three straight lines KF, FG, GK are equal to the three
A, B, C: And therefore the triangle KFG has its three sides
KF, FG, GK equal to the three given straight lines A, B, C:
Which was to be done.

PRO P. XXIII. P R O B.
A

T 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 angie; it is required to make at angle at the given point A in the given straight

C

A
line AB, that Thall be
cqual to the given rec-
tilineal angle DCE.

Takein CD, CE, any
points D, E, and join
DE ; and make a the
triangle AFG the fides D

E
of which shall be equal

F
to the three straight
lines CD, DE, CE, fo

B
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 bafe DE | 8. 1. to the base FG; the angle DCE is equalb 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.

SAA

2 22. I,

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See N.

IF
F two triangles haye two sides of the one equal to two

sides of the other, each to each, but the angle con-
tained by the two sides of one of them greater than the
angle contained by the two sides 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,

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AB, AC equal to the two DE, DF, each to each, viz. AB equal Book I. to DE, and AC to DF; but the angle BAC greater than the angle EDF; the base BC is also greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater than the other, and at the point D, in the straight line DE, make the angle EDG equal to the angle BAC; and make a 23. r. DG equal o to AC or DF, and join EG, GF.

b 3. I. 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 equal to the angle EDG;A

D therefore the base BC is equal to the base

C 4. 1 EG; and because DG is equal to DF, the angle DFG is equal d to the angle DGF; but the angle DGF is

E greater than the angle B

G EGF; therefore the

F 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 lide is opposite to the greater angle; the side EG e 19.1. is therefore greater than the side EF; but EG is equal to BC; and therefore also BC is greater than EF. Therefore, if two triangles, &c. Q. E. D.

d s.l.

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IF two triangles have two sides of the one equal to two

sides of the other, each to each, but the base of the one greater than the base of the other; the angle also contained by the sides of that which has the greater base, Thall be greater than the angle contained by the sides ea qual to them, of the other.

Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DĚ, and AC to DF; but the base CB is greater than the base EF; the angle BAC is likewise greater than the angle EDF.

For,

Book 1.

* 4. I.

For, if it be not greater, it must either be equal to it, or
lefs; but the angle BAC is not equal to the angle EPF, be-
cause then the base
BC would be equal · A

D
to EF; but it is not ;
therefore the angle
BAC is not equal to
the angle EDF; nei-
ther is it lefs; becaufe
then the base BC
would be less than
the base EF; but it B

E

F is not; therefore the angle BAC 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.

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PRO P. XXVI. THEOR.
IF two triangles have two angles of one equal to two

angles of the other, each to each ; and one fide equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each ; and also the third angle of the one to the third angle of the other.

G

Let ABC, DEF be two triangles which have the angles ABC,
BCA equal to the angles DEF, EFD, viz. ABC to DEF, and
BCA to EFD ; also one side equal to one side ; and first let those
fides be equal which are adjacent to the angles that are equal in
the two triangles, viz.
BC to EF; the other
А

D
Gdes shall be equal,
each to each,viz.AB
to DE,and AC to DF;
and the third angle
BAC to the third
angle EDF.

For, if AB be not
equal to DE, one of B
them must be the
greater. Let AB be the greater of the two, and make
BG equal to DE, and join GC; therefore, because BG is

equal

E

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