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angle ABE are greater than BE. To each of these add EC ;

Book I.
therefore the Gides BA, AC are А.
greater than BE, EC: Again,

E
because the cwo sides CE, ED

D
of the triangle CED are great-
er than CD, add DB to each
of these; therefore the sides
CE, EB are greater than CD,
DB; but it has been shewn that

B
BA, AC ace 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 interior and opposite angle, the exterior angle BDC of the

triangle CDE is greater than CED; for the same reason, the exIterior 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.

PRO P. XXII. PRO B.

a
to three given straight lines, but any two whatever
of these must be greater than the third'.

a 20. I,
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

It is
ihan C; A and C greater than B; and B and C than A.
required to make a triangle of which the sides 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
makea DF equal to A, FG

a 3.1.
to B, and GH equal to C;
and from the centre F, at
the distance FD, describe
o the circle DKL ; and D

b 3. Poft,

G НЕ
from the centre G, at the

L
distance GH, describe
another circle HLK, and

A
join KF, KG; the tri-

В.
angle KFG has its fides

С
equal to the three straight
lines, A, B, C.
Because the point F is the centre of the circle DKL, FD is

equal

T

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 also 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 Gides KF, FG, GK equal to the three given straight lines A, B, C. Which was to be done.

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

C
line AB, that shall be

А
equal to the given rec-
tilineal angle DCE.

Take in CD,CE, any
points D, E, and join
DE; and make a the

E
triangle AFG the sides D
of which shall be equal

F.
to the three straight

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

2. 22. 1.

b. 8. 1.

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Sec N.

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

fides of the other, cach to each, but the angle con. tained 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 shal be greater than the base of the other. Let ABC, DEF be two triangles which have the two sides

AB,

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 Gdes 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 BAC; and make a. 23. I.
DG equal 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. I. EG; and because DG is equal to DF, the angle DFG is equal d to the angle DGF; but the angle DGF is 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 e fide is opposite to the greater angle; the side EG C 19.1. is therefore greater than the fide EF; but EG is equal to BC; and therefore also BC is greater than EF. Therefore, if two triangles, &c. Q. E. D.

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d. 5.1.

E

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

, , one greater than the base of the other; the angle also contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides co. qual to them, of the other.

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

For,

32

Book 1. For, if it be not greater, it must either be equal to it, or mless; but the angle BAC is not equal to the angle EDF, be

cause then the base
a 4. I. 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 less; because

then the base BC
b 24. I. would be less b than
the base EF; but it B

C
E

F
is not; therefore the
angle BAC is not lets than the angle EDF; and it was shewn
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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IF

two triangles have two angles of one equal to two

angles of the other, each to each ; and one side 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.

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 fide 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 A

D
fides 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

с E 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

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equal to DE, and BC to EF, the two Gdes GB, BC, are equal Book 1. to the two DE, EF, each to each ; and the angle GBC is equal to the angle DEF; therefore the base GC is equal a to the base a 4. I. DF, and the triangle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal lides are opposite ; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothesis, equal to the angle BCA ; wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is impossible; therefore AB is not unequal to DE, that is, it is equal to it; and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each ; and the angle ABC is equal to the angle DEF; the base therefore AC is equal a to the base DF, and the third angle BAC to the third angle EDF. Next, let the Gdes A

D which are opposite to equal angles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the other fides shall be equal, AC to DF, and BC

B HC E to EF; and also the third angle BAC to the third EDF.

For, if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH; and because BH is equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal angles ; therefore the base AH is equal to the bale DF, and the triangle ABH to the triangle DEF, and the other angles shall be equal, each to each, to which the equal Gides are opposite ; therefore the angle BHA is equal to the angle EFD; but EFD is equal to the angle BCA; therefore also the angle BHA is equal to the angle BCA, that is, the exterior angle BH A of the triangle AHC is equal to its interior and opposite angle ECA ; which is impossible b; wherefore BC is not unequal to EF, that is, b 16. 1. it is equal to it; and AB is equal to DE; therefore the two AB, BC are equal to the two DE, EF, cach to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q. E. D.

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