Proclus has given a direct demonstration of this theorem, and is translated by Stone into his edition of the Elements, page 15. It is the converse of the fourth, although Euclid has not added so much as in that theorem, viz., that the triangles and remaining angles are equal; the reason is manifest, for the equality of the vertical angles being demonstrated, it follows, that all are equal to all by the fourth. Whence it was only necessary to demonstrate this, and assume the rest as consequents. PROPOSITION IX. PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal parts. Let BAC be the given rectilineal angle; it is required to bisect it. Take any point D in a the right line AB, and from the to the base EF: therefore the angle DAF is equal to the angle EAF. Wherefore the given rectilineal angle BAC is bisected by the right line AF. Deduction. Q. E. F. Divide a given rectilineal angle into any even number of equal parts.. PROPOSITION X. PROBLEM.* To bisect a given finite right line, that is, to divide it into two equal parts. Let AB be the given finite right line; it is required to bisect it. Upon ita describe the equilateral triangle ABC; and bisect the angle ACB by the right line CD. The right line AB is bisected in the point D. For because Acis equal to CB, and CD common; the two AC, CD, are equal to the two BC, A D B * A given finite right line may also be bisected by means of the construction to the first proposition of this book, and joining the commou sections of the circles. с CD, each to each; and the angle ACD is equal to the angle BCD: therefore the base AD is equal to the base © 4. 1. BD. And consequently the finite right line AB is bisected in the point D. Q. E. F. Deduction. From the vertex of a given scalene triangle, to draw, to the base, a straight line which shall exceed the less of the two sides, as much as it is itself exceeded by the greater. PROPOSITION XI. PROBLEM. To a given right line, from a given point in it; to draw a right line at right angles to the former. a A D F a 3. 1. b 1.1. с E B Let AB be the given right line, and c a given point in it, it is required to draw from the point c a right line at right angles to AB. Take any point D in ac, and make CE equal to CD,a and upon DE describe the equilateral triangle FDE, and join Fc. The right line Fc is drawn at right angles to the given right line AB from the point c given in it. For because DC is equal to CE, and Fc common; the two DC, CF, will be equal to the two EC, CF, each to each; and the base DF is equal to the base EF; wherefore the angle DCF is equal to the angle ECF, and they are adjacent angles. But when a right line standing upon a right line makes the adjacent angles equal to one another, each of them is a right angle: therefore each of the angles DCF, ECF, is a right angle. Wherefore the right line Fc is drawn at right angles to the given right line AB, from the point c given in it. Q.E. F. Deductions. 1. Describe a circle which shall pass through three given points which are not in the same right line. 2. In a right line given in position, but indefinite in a 10. 1. b 8.1. length, to find a point, which shall be equidistant from each of two given points, either on contrary sides, or both on the same side of the given line, and in the same plane with it; but not situated in a perpendicular to it. PROPOSITION XII.* PROBLEM. Upon a given infinite right line from a given point which is without it; to draw a perpendicular right line. Let AB be the given infinite right line, and c a given point which is without it, it is required to draw upon the given infinite right line AB a perpendicular from the given point c, which is without it. Take any point D upon the other side of AB, and A from the centre c at the distance CD describe the circle EDG; and bisect GE in H; and join CG, CH, CE. The perpendicular CH is drawn upon the given infinite right line AB from the point c, which is without it. b H D B For because GH is equal to HE, and HC common, the two GH, HC, are equal to the two EH, HC, each to each; and the base CG is equal to the base CE. Therefore the angle CHG is equal to the angle CHE, and they are adjacent angles. But when a right line standing upon another right line makes the adjacent angles equal to one another, each of them is a right angle, and the right line standing upon the other is called a perpendi• 10. Def. 1. cular; wherefore upon a given infinite right line AB, from the given point c, which is without it, CH has been drawn perpendicular. Q. E. F. C Deduction. Given the vertex of a triangle, the perpendicular from the vertex to the base and also the base, to construct the triangle. * Euclid did well in proposing an infinite right line, for otherwise the given point might be situated in a direct position with the given line, and consequently the problem would not succeed. PROPOSITION XIII. THEOREM. When a right line standing upon a right line makes angles, these are either two right angles, or are equal to two right angles. a A For let a certain right line AB standing upon the right line CD make the angles CBA, ABD. The angles CBA, ABD, are either two right angles, or are equal to two right angles. For if CBA be equal to ABD, they are right angles; but if less, draw from the point B, BB at right angles to DC; the angles CBE, DBE, are therefore two right angles; and because CBE is equal to the two CBA, ABE, add • Def. 10. b 11. 1. D B EBD, which is common; therefore the two angles CBE, EBD, are equal to the three angles CBA, ABE, EBD.C‹ Ax. 2. Again, because the angle DBA is equal to the two DBE, EBA, and ABC, which is common; therefore the two angles DBA, ABC, are equal to the three DBE, EBA, ABC. But it was shown that the angles CBE, EBD, are equal to the same three, and things which are equal to the same are equal to one another; therefore the angles CBE, EBD, are equal to DBA, ABC; but CBE, EBD, are two right angles; therefore the angles DBA, ABC, are equal to two right angles. Therefore when a right line standing upon a right line, &c. Q. E. D. PROPOSITION XIV. THEOREM. If to a certain right line, and to a point in it, two right lines not placed towards the same parts, make the adjacent angles equal to two right angles; the right lines will be in one and the same straight line. For to a certain right line AB, and to a point in it B, let there be two right lines BC, BD, not placed toward the same parts, make the adjacent angles ABC, ABD, equal to two right angles. Then the line BD is in the same straight line with CB; for if BD is not in the same straight line A B a 13. 1. b Ax. 3. a 13. 1. b Ax. 3. with CB, let BE be in the same straight line with it. Therefore, because the right line AB stands upon the right line CBE, the angles ABC, ABE, are equal to two right angles. But also the angles ABC, ABD, are equal to two right angles. Therefore the angles CBA, ABE, will be equal to the angles CBA, ABD. Take away ABC, common to both. Therefore the remaining angle ABE is equal to the remaining angle ABD, the less to the greater, which is impossible. Therefore EB will not be in the same straight line with BC. In like manner we may show, that not any other can be except BD. Therefore BD will be in a right line with BC. If therefore to a certain right line, &c. Q. E. D. b PROPOSITION XV. THEOREM.* If two right lines cut one another, they will make the vertical angles equal to one another. For let the two right lines AB, CD, cut one another in the point E. Then the angle AEC is equal to the angle DEB; and the angle CEB equal to the angle AED. For because the right line AE standing upon the right line CD makes the angles CEA, AED; these will be equal to two right angles. Again, because the right C B line DE standing upon the right line AB makes the angles AED, DEB; the angles AED, DEB, will be equal to two right angles. But it was shown that the angles CEA, AED, are equal to two right angles. Therefore the angles CEA, AED, are equal to the angles AED, DEB. Take away the common angle AED. Therefore the remaining angle CEA is equal to the remaining angle BED. In like manner it may be shown that the angles CEB, AED, are equal. If therefore two right lines cut one another, &c. Q. E. D. *This proposition was discovered by Thales, according to the account given by Eudemus; and the corollaries ought, I think, to be rather placed after the thirteenth, as they are the natural consequences of that theorem, upon which the demonstration of the fifteenth entirely depends. |