PROPOSITION XXIX.* THEOREM. If a right line falls upon two parallel right lines, it will make the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle towards the same parts, and the interior angles towards the same parts equal to two right angles. E Let the right line EF fall upon the two right lines AB, CD. The alternate angles AGH, GHD, are equal to one another, the exterior angle EGB towards the same parts, equal to the interior and opposite angle GHD. And the interior angles BGH, GHD, towards the same parts equal to two right angles. For if AGH be unequal to GHD, one of them is the greater. Let AGH be the greater. And A C G B H a because the angle AGH is greater than the angle GHD, add BGH, which is common. Therefore the angles AGH, BGH, are greater than the angles BGH, GHD. But the angles AGH, BGH, are equal to two right angles. There- 13. 1. fore the angles BGH, GHD, are less than two right angles. But right lines which with another right line falling upon them make the adjacent angles less than two right angles, do meet, if produced far enough.bb Ax. 12. Therefore the right lines AB, CD, produced far enough, will meet. But they do not meet, since they are parallel. Therefore the angle AGH is not unequal to the angle GHD; wherefore it is equal. But the angle AGH is equal to the angle EG B. Therefore EGB will be equal 15. 1. to GHD; add BGH, common to both. Therefore the angles EGB, BGH, are equal to the angles BGH, GHD; but EGB, BGH, are equal to two right angles. Therefore BGH, GHD, will be equal to two right angles. If, therefore, a right line, &c. Q. E. D. *This and the preceding 27th Proposition show the excellency of Euclid's definitions of parallels, and its superiority to many others given by the moderns; for he here employs the negative property of these lines with great success, and the addition of their being always at the same perpendicular distance from each other would have been useless, as it is not wanted in any part of the Elements. с a 29. 1. L 27.1. a 23. 1. b 27. 1. Deduction. Trisect a right angle; that is, to divide it into three equal parts, trisect any rectilineal angle which is an even aliquot part of a right angle, PROPOSITION XXX. THEOREM. Right lines which are parallel to the same right line, are parallel to each other. A Let AB, CD, be each of them parallel to EF; then AB, CD, are parallel to one another. Let GK, a right line, fall upon them. And because the right line GK falls upon the parallel right lines AB, EF, the angle AGH is equal to the angle c GHF. Again, because the right line GK falls upon the parallel right E K H lines EF, CD, the angle GHF is equal to the angle GK D. But it was shown the angle AGK is also equal to the angle GHF; therefore AGK will also be equal to GKD; and they are alternate angles. Therefore AB is parallel to CD. Wherefore the right lines, &c. PROPOSITION XXXI. PROBLEM. Q. E. D. Through a given point to draw a right line parallel to a given right line. Let a be a given point, and вc a given right line. It is required through the point A to draw a right line parallel to the right line BC. Take E A F B D For in BC any point D, and join AD; Deductions. 1. To draw to a right line from a given point without it another right line, which shall make an angle equal to a given rectilineal angle. 2. From a given isosceles triangle to cut off a trapezium, which shall have the same base as the triangle, and shall have its three remaining sides equal to each other. PROPOSITION XXXII.* THEOREM. One side of any triangle being produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of a triangle are equal to two right angles. E Let ABC be a triangle; and let one of its sides BC be produced to D. The exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB, are equal to two right angles. For through the point c draw CE parallel to the right line AB ; and because AB is parallel to CE, and AC falls upon them, the alternate angles BAC, ACE, are equal to one another. Again, because AB is parallel to CE, and the right line BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. But the angle ACE was shown to be equal to the angle BAC. Wherefore the whole exterior angle ACD is equal to the two interior and opposite angles BAC, ABC. Add ACB, which is common: therefore the angles ACD, ACB, B D D A E *The three angles of a triangle may be demonstrated to be equal to two right angles without the aid of the first part of the proposition, as Eudemus relates was done by the Pythagoreans, in the following manner. Let there be a triangle ABC, and let there be drawn through the point a, a line DE parallel to BC. Because therefore the right lines DE, BC, are parallel, the alternate angles are equal. Hence the angle DAB is equal to the angle ABC, and the angle EAC equal to the angle ACB. Let the common angle BAC be added. The angles, therefore, DAB, BAC, CAK; that is, the angles DAB, BAE, and that is two right angles, are equal to the three angles of the triangle. B C a 31. 1. b 29. 1. c 13. 1. a 29. 1. are equal to the three angles, ABC, BCA, CAB; but the angles ACD, ACB, are equal to two right angles.c Therefore the angles ACB, CBA, CAB, will also be equal to two right angles. Therefore if one side of every triangle, &c. Q. E. D. Deductions. 1. The angle at the base of an isosceles triangle is equal to half the difference between the vertical angle and two right angles. 2. The difference of the angles at the base of any triangle is double the angle contained by the line bisecting the vertical angle and another drawn from the vertex perpendicular to the base. 3. Given the difference of the angles at the base of a triangle, the perpendicular drawn from the vertex to the base, and one of the segments made by the perpendicular, to construct the triangle. 4. To construct a triangle, which shall have its three sides taken together equal to a given finite right line, and its three angles equal to three given angles, each to each; the three given angles being together equal to two right angles. PROPOSITION XXXIII.* THEOREM. Right lines, which join equal and parallel right lines towards the same parts, are themselves equal and parallel. Let AB, CD, be equal and parallel, and let the right lines AC, BD, join them towards the same parts. AC, BD, are equal and parallel to one another. Draw BC, and because AB is parallel to CD, and BC falls upon them, the alternate angles ABC, BCD, A are equal. Again, because AB is B * I think the following mode of enunciating this proposition is clearer than the one in the text. If the extremes of two equal and parallel right lines be joined by the extremes of two other right lines not cutting one another, these two right lines shall also be equal and parallel. AB, BC, are equal to the two BC, CD, and the angle ABC is equal to the angle BCD. Therefore the base AC is equal to the base BD, and the triangle ABC to the b 4. 1. triangle BCD; also the remaining angles will be equal to the remaining angles, each to each, to which the equal sides are opposite. Therefore the angle ACB is equal to the angle CBD. And because the right line BC falls upon the two right lines AC, BD, makes the alternate angles ACB, CBD, equal to one another, AC с is parallel to BD. But it was shown that it was equal © 27. 1. to it. Therefore right lines, &c. Q. E. D. PROPOSITION XXXIV. THEOREM. The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it. "A parallelogram is a four-sided figure, whose opposite sides are parallel." Let ABDC be a parallelogram, whose diameter is BC; the opposite sides and angles of the parallelogram ABDC are equal to one another, and the diameter BC bisects it. For because AB is parallel to CD, and the right line BC falls upon them, the alternate angles ABC, BCD, are equal to one another. Again, because AC is parallel to BD, and Bс falls upon them, the alternate angles ACB, CBD, are equal A B D a 29. 1. b to one another. Therefore the two triangles ABC, CBD, have the two angles ABC, BCA, equal to the two angles BCD, CBD, each to each, and one side equal to one side, viz. BC, which is common to both. Therefore they will have the remaining sides equal to the remaining sides, each to each, and the remaining angle equal to the remaining angle. Therefore the side AB is equal ↳ 26. 1. to the side CD, and the side AC to the side BD; also the angle BAC is equal to the angle BDC. And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD will be equal to the whole angle ACD. But it was shown that the angle BAC is equal to the angle BDC. Therefore |