angles CBA, ABD; these are either two right angles, or are together equal to two right angles.

For if the angle CBA be equal to ABD, each of them is a

right (Def. 10.) angle; but; if not, from the point B draw BE at right angles (11. 1.) to CD; therefore the angles CBE, EBD are two right angles; and because CBE is equal to the two angles CBA; ABE together, add the angle EBD to each of these equals; therefore, the angles CBE, EBD are equal (2 Az.) to the three angles CBA, ABE, EBD.

Again, because, the angle DBA is equal to the two angles DBE, EBA, add to these equals the angle ABC; therefore the angles DBA ABC are equal to the three angles DBE, EBA, ABC; but the angles CBE, EBD have been demonstrated to be equal to the same three angles; and things that are equal to the same are equal (1 Ax.) to one another; therefore the angle CBE, EBD are equal to the angles DBA, ABC; but CBE, EBD are two right angles; therefore DBA, ABC are together equal to two right angles. Wherefore, when a straight line, &c. Q. E. D.

If, at a point in a straight line, two other straight lines, upon the opposite sides of it, make the adjacent angles, together equal to two right angles, these two straight lines shall be in one and the same straight line.

At the point B in the straight line AB, let the two straight lines BC, BD upon the opposite sides of AB, make the adjacent angles ABC, ABD equal together to two right angles, BD is in the same straight line with CB.

For, if BD be not in the same straight line with CB, let BE be in the same straight line with it; therefore, because the straight line AB makes angles with the straight line CBE, upon one side of it, the angles ABC, ABE are together equal (13. 1.) to two right angles; but the angles ABC, ABD are likewise together equal to two right angles; therefore the angle CBA, ABE are equal to the angles CBA, ABD. Take away the common angle ABC, the remaining angle ABE is equal (3. Ax.) 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. point, &c. Q. E. D.

Wherefore, if at a

If two straight lines cut one another, the vertical, or opposite, angles shall be equal.

Let the two straight lines AB, CD, cut one another in the point E; the angle AEC shall be equal to the angle DEB, and CEB to AED.

Because the straight line AE makes with CD the angles CEA, AED, these angles are togethe equal (13. 1.) to two right angles. Again, because the straight line DE makes with AB the angles AED, DEB, these also are together equal (13. 1.) to two right angles; and CEA, AED have been demonstrated to be equal to two right angles; wherefore the angles CEA, AED, are equal to the angles AED, DEB.

Take away the common angle AED, and the remaining angle CEA is equal (3. Ar.) to the remaining angle DEB. In the same manner it can be demonstrated, that the angles CEB, AEB are equal. Therefore, if two straight lines, &c. Q. E. D.

COR. 1. From this it is manifest that, if two straight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles.

COR. 2. And consequently that all the angles made by any number of lines meeting in one point, are together equal to four right angles.

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

Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAC.

Bisect (10.1.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G.

Because AE is equal to EC, and BE to EF; AE, EB are equal to CE, EF, each to each; the angle AEB is equal (15.1.) to the angle CEF, because they are opposite vertical angles; therefore the base AB is equal (4. 1.) to the base CF, and the triangle AEB to the triangle CEF,

B

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 the angle ACD, is greater than BAE.

In the same manner, if the side BC be bisected, it may be demon

strated that the angle BCG, that is, (15. 1.) the angle ACD, is greater than the angle ABC. Therefore, if one side, &c. Q. E. D.

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

Let ABC be any triangle; any two of its angles together are less than two right angles.

B

C

D

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

Proposition XVIII. Theorem.

The greater side of every triangle is opposite to the greater angle. Let ABC be a triangle of which 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, B make (3 1.) AD equal to AB, and

A.

D

join BD; and because ADB is the exterior angle of the triangle BDC it is greater (16. 1.) than the interior and opposite angle DCB; bu ADB is equal (5. 1.) to ABD, because the side AB is equal to the side 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.

The greater angle of every triangle is subtended 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 either

be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal (5. 1.) to the angle ACB; but it is not; there fore AC is not equal to AB; neither is it less; because then the angle ABC would be less

(18. 1,) than the angle ACB; 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. Wherefore the greater angle, &c. Q. E. D.

Proposition XX. Theorem.

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 make (3. 1.)

AD equal to AC and join DC.

Because DA is equal to AC, the angle ADC is likewise equal (5. 1.) to ACD; but the angle BCD is greater than the angle ACD; therefore 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 (19. 1.) 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.

B

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.

Proposition. XXI. Theorem.

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 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, AE of the triangle ABE are greater than BE.

To each of these add EC; therefore the sides BA, AC are greater than BE, EC.

Again, because the two sides CE, ED of the triangle CED are greater than CD, add DB to each of these; therefore the sides CE, EB are greater than CD, DB; but it has been shewn that BA, AC are greater than BE, EC, much more then are BA. AC greater than BD, DC.

B

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

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 must be greater than the third. (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 side 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 (3. 1.) DF equal to A, FG to B, and GH equal to C; and from the centre F, at the distance FD, describe (3. Post.) the circle DK L; and from the centre G, at the distance GH, describe (3. Post.) another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines A. B. C.

D

Б

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

Again because G is the centre of the circle LKH, GH is equal (15. Def.) 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

Proposition XXIII. Problem.

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 AB, that shall be equal to the given rectilineal angle DCE.

Take in CD, CE any points D, E, and join DE; and make (22. 1.) the triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC, so that CD be equal to AF, CE to AG, and DE to FG; and because DC, CE are equal to FA, AG,

E

each to each, and the base DE to the base FG; the angle DCE is equal (8.1.) to the angle FAG.