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CBE, upon one side of it, therefore the angles ABC and ABE are together equal (I. 13) to two right angles; but the angles ABC and ABD are likewise together equal to two right angles ;

therefore the angles CBA and ABE are equal to the angles CBA and ABD: take away the common angle CBA, and the remaining angle ABE is equal (Ax. 3) to the remaining angle ABD, the less to the greater, which is impossible; therefore BE is not in the same straight line with BC. And, in like manner, it may be demonstrated that no other can be in the same straight line with it but BD, which, therefore, is in the same straight line with CB.

PROPOSITION XV. THEOREM. If two straight lines cut one another, the vertical or opposite angles shall be equal. . Given the two straight lines AB and CD, cutting one another in the point E; to prove that the angle AEC shall be equal to the angle DEB, and CEB to AED.

(Dem.) For the angles CEA and AED, which the straight line AE makes with the straight line CD, are together equal (I. 13) to two right angles; and the angles AED and DEB, which the straight line 0 DE makes with the straight line AB, are also together equal to two right angles; A

therefore the two angles CEA and AED are equal to the two AED and DEB. Take away the common angle AED, and the remaining angle CEA is equal (As. 3) to the remaining angle DEB. In the same manner it can be demonstrated that the angles CEB and AED are equal. COR. 1.-From this it is manifest that, if two straight lines cut

one another, the angles which they make at the point of

their intersection, are together equal to four right angles. Cor. 2.-And hence, all the consecutive angles made by any

number of lines meeting in one point, are together equal to four right angles.

PROPOSITION XVI. THEOREM. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Given a triangle ABC, and its side BC produced to D, to prove that the exterior angle ACD is greater than either of the interior opposite angles CBA or BAC.

(Const.) Bisect (I. 10) AC in E, join BE, and produce BE to F, making EF equal to BE; join also FC, and produce AC to G.

(Dem.) Because AE is equal to EC, and BE to EF; AE and EB are equal to CE and EF, each to each; and the angle AEB is equal to the angle CEF (I. 15), because they are opposite vertical angles;

therefore (I. 4) the base AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite; wherefore the angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore (Xx. 14) the angle ECD, that is, ACD, is greater than BAE. In the same manner, if the side BC be bisected, and a similar construction made below the base, it may be demonstrated that the angle BCG, that is (I. 15), the angle ACD, is greater than the angle ABC; therefore the exterior angle ACD is greater than either the angle CAB or ABC.

PROPOSITION XVII. THEOREM. Any two angles of a triangle are together less than two right angles.

Given any triangle ABC; to prove that any two of its angles together are less than two right angles.

(Const.) Produce BC to D; (Dem.) and because ACD is the exterior angle of the triangle ABC, ACD is greater than the interior and opposite angle ABC (1. 16); to each of these add the angle ACB; therefore the angles ACD and ACB (Ax. 4) are greater than the angles ABC. and ACB; but ACD and ACB are together equal to two right angles (I. 13); <

therefore the angles ABC and BCA are less than two right angles. In like manner it may be demonstrated, that BAC and ACB, as also CAB and ABC, are less than two right angles. COR.--Hence there cannot be two perpendiculars drawn from a

point to a line.

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PROPOSITION XVIII. THEOREM. The greater side of every triangle has the greater angle opposite to it.

Given a triangle ABC, of which the side AC is greater than the side AB ; to prove that the angle ABC is also greater than the angle BCA.

: (Const.) Because AC is greater than AB, make (I. 3) AD equal to AB, and join BD; (Dem.) and because ADB is the exterior angle of the triangle BDC, it is greater than the interior and opposite angle DCB (I. 16);

but ADB is equal to ABD (I. 5), because the side AB is equal to the side come AD; therefore (Ax. 15) the angle ABD is likewise greater than the angle ACB; wherefore still more is the angle ABC greater than ACB. COR.—The perpendicular from a point to a straight is the

least line that can be drawn from that point to it. For if the angle B were a right angle, the angle C would be less than a right angle (I. 17), therefore the perpendicular AB would be less than AC.

PROPOSITION XIX. THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.

Given a triangle ABC, of which the angle ABC is greater than the angle BCA ; to prove that the side AC is likewise greater than the side AB.

(Dem.) For, if AC be not greater than AB, AC must either be equal to AB, or less than it; AC is not equal to AB, because then the angle ABC would be equal to the angle ACB (1.5); but it is not; therefore AC is not equal to AB; neither is AC less than AB; because then the angle ABC would be less than the angle ACB (I. 18); but it is not; therefore the side AC is not less than AB; - and it has been shewn that it is not equal to AB; therefore AC is greater than AB. “

PROPOSITION XX. THEOREM. Any two sides of a triangle are together greater than the third side.

Given a triangle ABC; to prove that any two sides of it together are greater than the third side;

namely, the sides BA and AC greater than the side BC; AB and BC greater than AC; and BC and CA greater than AB. (Const.) Produce BA to the point D, B

C . and make (I. 3) AD equal to AC; and join DC. Dory (Dem.) Because DA is equal to AC, the angle ADC is equal to ACD (I. 5); but the angle BCD is greater than the angle ACD; therefore (Ax. 14) the angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater angle has the greater side opposite to it (I. 19); therefore the side DB is greater than the side BC; but DB is equal to BA and AC; therefore the sides BA and AC are greater than BC. In the same manner it may be demonstrated, that the sides AB and BC are greater than CA, and BC and CA greater than AB.

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PROPOSITION XXI. THEOREM. * If from the ends of one side of a triangle, there be drawn two straight lines to a point within the triangle, these two lines shall be less than the other two sides of the triangle, but shall contain a greater angle.

Given the two straight lines BD and CD, drawn from B and C, the ends of the side BC of the triangle ABC, to the point D within it; to prove that BD and DC are less than the other two sides BA and AC of the triangle, but contain an angle BDC greater than the angle BAC. **(Const.) Produce BD to E; (Dem.) and because two sides of a triangle are greater than the third side, the two sides BA and AE of the triangle ABE are greater than BE.

To each of these add EC; therefore the sides BA and AC are greater than BE and EC. Again, because the two sides CE 6 and ED of the triangle CED are greater than CD, add DB to each of these; therefore the sides CE and EB are greater than CD and DB; but it has been shewn that BA and AC are greater than BE and EC; still more (Ax. 16), then, are BA and AC greater than BD and 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; still more (Ax. 16), then, is the angle BDC greater than the angle BAC.

PROPOSITION XXII. PROBLEM. To make 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 third (I. 20).

Given A, B, and C, three straight lines, of which any two

whatever are greater than the third ; namely, 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, and C.

(Const.) Take a straight line DE terminated at the point D, but unlimited towards E, and make (1.3) DF equal to A, FG to B, and GH equal to C; and from the centre F, with the radius FD, describe the circle DKL (Post. 3); and from the centre G, with the radius GH, describe another circle HLK, and join KF and KG; the triangle KFG has its sides equal to the three straight lines A, B, and C.

(Dem.) Because the point F is the centre of the circle DKL, FD is equal to FK (Def. 17); but FD is equal to the straight line A; therefore FK (Ax. 1) 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, and GK, are equal to the three A, B, and C. And therefore the triangle KFG has its three sides KF, FG, and GK, equal to the three given straight lines, A, B, and C.

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PROPOSITION XXIII. PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle.

Given the straight line AB, and the point A in it, and the rectilineal angle DCE; it is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE.

(Const.) Take in CD and CE, any points D'and 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, and CE, so that AF be equal to CD, AG to CE, and FG to DE; (Dem.) and because DC and CE are equal to FA and AG, each to each, · and the base DE to the base FG; the angle DCE is equal to the angle FAG (I. 8). Therefore, at the given point A in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE.

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