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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 by the greater side, or has the greater side opposite to it.

et A B C be a triangle, of which the angle A BC is greater than the angle BCA; the side A c is likewise greater than the side A B.

For, if it be not greater, A c must either be equal to A B, or less than it; it is not equal, because then the angle ABC would be equal (1. 5.) to the angle A CB; but it is not ; therefore A c is not equal to AB; neither is it less ; because then the angle A B C would be less (1. 18.) than the angle ACB; but it is not ; therefore the side Ac is not less than AB; and it has been shown that it is not equal to AB; therefore A c is greater than A B. Wherefore the angle, &c.

Q. E. D. PROP. XX. THEOR. Any two sides of a triangle are together greater than the third side.

Let A BC 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 A B, BC greater than A C; and B C, C A greater than a B.

Produce bA to the point D, and make (1. 3.) AD equal to AC; and join DC.

Because DA is equal to AC, the angle adc is likewise equal (1. 5.) to ACD; but the angle B

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BCD is greater than the angle A CD; therefore the angle BCD is greater than the angle ADC; and because the angle B C D of the triangle Dce is greater than its angle B DC, and that the greater (1. 19.) side is opposite to the greater angle; therefore the side DB is greater than the side BC; but D B is equal to B A and AC; therefore the sides B A, AC are greater than Bc. In the same manner it may

be demonstrated, that the sides A B, B C are greater than C A, and B C, CA greater than A B.

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 B D, C D be drawn from B, C, the ends of the side B C, of the triangle A B C, to the point D within it; BD and DC are less than the other two sides B A, A c 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 side, the two sides BA, A E of the triangle A B E are greater than BE, to each of these add EC; therefore the sides B A, AC are greater than BE,

E EC: Again, because the two sides CE, ED of the triangle CED are greater than cd, add D B to each of these; therefore the sides CE, E B are greater than CD, DB; but it has been shown that BA, A C are greater than B E, EC; much more then are BA, A C greater than BD,

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Again, because the exterior angle of a triangle is greater than the interior and opposite angle (1. 16.), the exterior angle BDC of the triangle CDE is greater than CED; for the same reason, the exterior angle C E B of the triangle A B E is greater than BAC; and it has been demonstrated that the angle BDC is greater than the angle CE B; 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 equal to three given straight lines, but any two whatever of these must be greater than the third. (1. 20.)

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 shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but unlimited towards E, and make (1. 3.) DF equal to A, F G to B, and gh equal to c; and from the centre F, at a distance

describe (Post. 3.) the circle DKL; and from the centre G, at the distance GH, describe (Post. 3.) another circle HLK: and join K F, KG; the triangle K F G has its sides equal to the three straight lines, A, B, C.

Because the point F is the centre of the circle DKL; FD is equal (Def. 15.) to FK; but if FD is equal to the straight line A ; therefore Fk is equal

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to A: Again, because G is the centre of the circle LKH, GH is equal (Def. 15.) 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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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 point A in the given straight line A B, th:t shall be equal to the qiven rectilineal angle DCE.

Take in C1, CE any points D, E, and join DE; and make (1. 22.) the triangle AFG the sides of which shall 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, C E are equal to FA, A G, each to each, and the base DE to the base FG; the angle DCE is equal (1. 8.) to the angle FAG. Therefore, at the given point A in the given straight line A B, the angle Fag is made equal to the given rectilineal angle DC E. Which was to be done.

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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 A B C, D E F be two triangles which have the two sides A B, AC equal to the two DE, DF, each to each, viz. A B equal to D Е, and AC to D F; but the angle BAC greater than the angle EDF; the base BC is also greater than the base E F.

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 (1. 23.) the angle E D G equal to the angle BAC; and make D G equal (1. 3.) to a C or DF, and join E G, G F.

Because A B is equal to DE, and AC to D G, the two sides B A, AC are equal to the two E D, D G, each to each, and the angle BAC is equal to the angle EDG; therefore the base Bc is equal (I. 4.) to the

and because DG is equal to B

CE DF, the angle DFG is equal (1. 5.) to the angle DGF; but the angle DGF is greater than the angle EGF, therefore the angle DFG is greater than E GF; and much more is the angle EFG greater than the angle EGF; and because the angle EFG of the triangle E F G is greater than its angle EGF, and that the greater (I. 19.) side is opposite to the greater angle; the side EG is therefore greater

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