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these add the angle ACB; therefore the angles ACD, ACB are Book I. greater than the angles ABC, ACB; but ACD, ACB are to- or-. gether equal b to two right angles; therefore the angles ABC, b 13. 1. BCA are less than two right angles. In like manner, it may be demonstrated, 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 side # every triangle is opposite to the greater angle. * * *

Let ABC be a triangle, of which A. the side AC is greater than the side AB; the angle ABC is also greater than the angle BCA.

Because AC is greater than AB, make - AD equal to AB, and join BD; and because ADB is the ex- terior angle of the trian-le BDC; B C it is greater b than the in-rior or opposite angle DCB, but b 16.1. ADB is equale to ABD, because the side AB is equal to the side c 5. 1. . AD; therefore the angle ABD is likewise greater than the angle ACB; wherefore much more is the angle ABC greater than ACB. Therefore the greater side, &c. Q. E. D.

PROP. XIX. THEOR.

THE greater angle of every triangle is subtended o by the greater side, or has the greater side opposite to it. *

Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB. * *

For, if it be not greater, AC must A either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal * to the a 5.1. angle ACB; but it is not; therefore AC is not equal to AB; neither is it less; because then the angle ABC would be less b than the angle ACB; B Cb 18.1.

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Book I.

but it is not; therefore the side AC is not less than AB; and it

wer- has been shown that it is not equal to AB; therefore AC is

See Note.

a 3.1.

b 5.1.

c 19. 1.

See Note.

greater than AB. Wherefore the greater angle, &c. Q. E. D.

PROP. XX. THEOR.

ANY two sides of a triangle are together greater than the third side.

Let ABC be a triangle; any two sides of it together are greater than the third side, viz. the sides BA, AC greater than the side BC; and AB, BC greater than AC; and BC, CA greater than AB.

Produce BA to the point D, and l) make a AD equal to AC; and join DC. A.

Because DA is equal to AC, the

angle ADC is likewise equal 5 to
ACD; but the angle BCD is great-
er than the angle ACD; therefore

the angle BCD is greater than the B C angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater • side is opposite to the greater angle; therefore the side 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.

PROP. XXI. THEOR.

IF, from the ends of the side 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 triangle, but shall contain a greater angle.

Let the two straight lines BD, CD be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC f the triangle, but contain an angle BDC greater than the an1e BAC. Produce BD to E; and because two sides of a triangle are greater than the third side, the two sides BA, AE of the tri

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angle ABE are greater than BE. To each of these add Ec; Book I. therefore the sides BA, AC are A * greater than BE, EC: again, because the two sidesCE, ED of the triangle CED are greater than CD,add DB to each of these; therefore thesides CE, EBare greater than CD, DB; but it has been shown that BA, AC are greater than BE, EC; much more then are BA, AC B • C 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 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.

PROP. XXII. PROB.

TO make a triangle of which the sides shall be see Note. equal to three given straight lines, but any two whatever of these must be greater than the third.” a 20. 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 than A. It is required to make a triangle of which the sides shallshe equal to

A, D, C, each to each. o
Take a straight line DE terminated at the point D, but un-
limited towards E, and ź.
make a DF equal to A,
FG to B, and GH equal
to C; and from the centre
F, at the distance FD, de-D
scribe b the circle DKL ;
and from the centre G, at
the distance GH, describe
* and ther circle H).K; and
§ KF, KG; the triangle
FG has its sides equal to Q — a
the three straight lines A, B, C. : _*
Because the point F is the centre of the circle DKL, FD is

a 3.1.

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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; 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.

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Which was to be done. -
PROP. XXIII. PROB.

AT 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 C A. 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 D E F G join DE; and make * the triangle AFG the sides of which shall B be equal to the three straight lines CD, DE, CE, so 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 b 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.

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 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 sides Book I. AB, Ac equal to the two DE, DF, each to each, viz; AB 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 DE, make a the angle EDG equal to the angle BAC; and make a 23.1. DG equal b to AC or DF, and join EG, GF. ... b3. 1.

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 equal A D to the angle EDG ; , 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 4 - d 5.1. to the angle DGF ; N ro but the angle DGF is F. ~g greater than the angle B C 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" side is opposite to the greater angle; the side EG e 19.1. is therefore greater than the side E.E; but EG is equal to BC; and therefore also BC is greater than SEF. Therefore, if two triangles, &c. Q. E. D.

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Paop. xxv. TheoR. s

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, shall be greater than the angle contained by the sides equal 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 wo DE, and AC to DF; but the base CB is greater than #5 * EF; the angle BAC is likewise greater than the angle

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