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To each of thefe add EC;

A

E

ingle ABE are greater than BE.
therefore the Gides BA, AC are
greater than BE, EC: Again,
because the two fides CE, ED
of the triangle CED are great-
er 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

D

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 demonftrated 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. PROB.

Book I.

O make a triangle of which the fides fhall be equal See N. to three given straight lines, but any two whatever

of these must be greater than the third".

Let A, B, C be the three given ftraight 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 fhall 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 a DF equal to A, FG
to B, and GH equal to C;
and from the centre F, at
the diftance FD, defcribe
the circle DKL; and D
from the centre G, at the
distance GH, describe b
another circle HLK, and
join KF, KG; the tri-
angle KFG has its fides.
equal to the three straight

lines, A, B, C.

a 20. I

a 3. I.

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

equal

Book I. equal to FK; but FD is equal to the ftraight line A; therefore FK is equal to A: Again, becaufe 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 fides KF, FG, GK equal to the three given ftraight lines A, B, C. Which was to be done.

2. 22. I.

b. 8. 1.

See N.

A

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Ta given point in a given ftraight line, to make a rectilineal angle equal to a given rectilineal angle.

Let AB be the given ftraight 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

ДД

A in the given straight
line AB, that fhall be
equal to the given rec-
tilineal angle DCE.
'Take in CD, CE, any
points D, E, and join
DE; and make a the
triangle AFG the fides D
of which fhall be equal
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 base DE to the bafe FG; the angle DCE is equal b to the angle FAG. Therefore, at the given 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.

IF

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F 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 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 fhall be greater than the base of the other.

Let ABC, DEF be two triangles which have the two fides

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 bafe BC is also greater than the bafe 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 ftraight line
DE, make the angle EDG equal to the angle BAC; and make a. 23. I.
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 equal

to the angle EDGA
therefore the base BC
is equal to the base
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

EGF; therefore the

D

E

F

b. 3. I.

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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. 19.1.
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.

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IF 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 alfo contained by the fides of that which has the greater base, fhall be greater than the angle contained by the fides equal to them, 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 equal to DE, and AC to DF; but the bafe CB is greater than the bafe EF; the angle BAC is likewife greater than the angle EDF.

For

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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 EDF, because then the bafe

2

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 lefs; because then the base BC would be less than the base EF; but it is not; therefore the

B

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angle BAC is not lefs 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.

IF

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F two triangles have two angles of one equal to two angles of the other, each to each; and one fide equal to one fide, viz. either the fides adjacent to the equal angles, or the fides oppofite to equal angles in each; then fhall the other fides 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; alfo one fide equal to one fide; and first let thofe fides be equal which are adjacent to the angles that are equal in the two triangles, viz.

BC to EF; the other A
fides fhall be equal,,
each to each, viz. ABG
to DE, and AC to DF;
and the third angle
BAC to the third
angle EDF.

For, if AB be not

D

بر

equal to DE, one of B

them must be the

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greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is

equal

Book 1.

equal to DE, and BC to EF, the two fides GB, BC, are equal to the two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the bafe 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 fides are oppofite; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothefis, equal to the angle BCA; wherefore alfo the angle BCG is equal to the angle BCA, the lefs to the greater, which is impoffible; 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 to the base DF, and the third angle BAC to the third angle EDF.

Next, let the fides which are oppofite to equal angles in each triangle be equal to one another, viz AB to DE; likewife in this cafe, the other fides fhall be equal, AC to DF, and BC to EF; and alfo the

D

AA

B

third angle BAC to the third EDF.

HC

E

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 bafe AH is equal to the bafe DF, and the triangle ABH to the triangle DEF, and the other angles fhall be equal, each to each, to which the equal fides are oppofite; therefore the angle BHA is equal to the angle EFD; but EFD is equal to the angle BCA; therefore alfo the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle. AHC is equal to its interior and oppofite angle ECA; which is impoffible 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, each to each; and they contain equal angles; wherefore the bafe AC is equal to the bafe DF, and the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q. E. D.

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