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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.

PROP. XXII. PROB.

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TO construct a triangle of which the sides shall be equal to three given straight lines; but any two whatever of these lines must be greater than the thirda.

a 20. 1.

1

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 greater than A. It is 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

unlimited towards E, a 3. 1. and makea DF equal

to A, FG to B, and
GH equal to Cand
from the centre F, at
the distance FD, des-

G H 6 3. Post. cribeb the circle DKL;

and from the centre G,
at the distance GH,

u
describeb another cir-
cle HLK; and join

A
KF, KG; the triangle

B-
KFG has its sides equal
to the three straight

C-
lines A, B, C.

Because the point F is the centre of the circle DKL, FD is e 11. Def. 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 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; therefore 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.

Book 1.

PROP. XXIII. PROB.

AT a given point in a given straight line to make a rectilineal angle equal to a given rectilineal angle.

F

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

C
given point A, in the
given straight line AB,
that shall be equal to the
given rectilineal angle
DCE.
Take in CD, CE any

E
points D, E, and join
DE ; and makea the tri-
D

a 22. 1, angle AFG, the sides of which shall be equal to the three straight lines

B CD, DE, CE, so that CD be equal to AP, 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 equalb to the angle FAG. Therefore, at the given b 8. 1: 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.

PROP. XXIV. THEOR.

IF two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides 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 sides AB, AC equal to the two DE, DF, each to each, viz. AB

D

Book I. equal 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 a 23. 1. DE, makea the angle EDG equal to the angle BAC, and b 3. 1. make DG equalb to AC or DF, and join EG, GF.

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

D
qual to the angle
EDG; therefore

the base BC is eI c 4. 1.

qual to the base
EG; and because
DG is equal to
DF, the angle
DFG is equal to
the angle DGF; B В

c
but

the angle DGF is greater

I than 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 e 19. 1.

EFG is greater than its angle EGF, the side EG is greatere
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.

p 5. 1.

E

PROP. XXV. THEOR.

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 contained by the sides of that which has the greater, base shall be greater than the angle contained by the sides 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 DE, and AC to DF; but the base CB is greater Book I. than the base EF; the angle BAC is likewise greater than the angle EDF.

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

D would be equala to А.

a 4. 1. 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 lessb

b 24. 1. than the base EF;

E but it is not; there B

F fore the angle BAC is not less than the angle EDF; and it was shown 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.

PROP. XXVI. THEOR.

IF two triangles have two angles of the 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 the 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

A

D two triangles which have the angles ABC, BCA equal to the G angles DEF, EFD, viz. ABC to DEF, and BCA to EFD; also one side equal to one side; and first let those sides be equal B.

CE

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

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Book 1. the other sides shall be equal, each tò 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 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 to DE, and BC to EF, the two sides GB, BC are equal to

the two DE, EF, each to each; and the angle GBC is equal a 4. 1.

to the angle DEF: therefore the base GČ is equala to the
base DF, and the triangle GBC to the triangle DEF, and
the other angles to the other angles, each to each, to which
the equal sides 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 im-
possible; 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 equals to the base DF, and the third angle BAC to the
third angle EDF.
· Next, let the sides
which are opposite to

A

D
equal angles in each
triangle be equal to
one another, viz. AB
to DE; likewise in
this case, the other
sides shall be equal,
AC to DF, and BC
to EF; and also the
'third angle BAC to

B
HC E

F 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 base DF, and the triangle ABH to the triangle DEF, and the other angles are equal, each to each, to which the equal sides are opposite ; therefore the angle BHA is equal to the angle

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