* Let the straight line GHK cut AB, EF, CD: and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal to the angle GHF. Again, because the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD: and it was shewn that the angle AGK is equal to the angle GHF; * * 29. 1. A -B H -F K -D 29. 1. E C † 1 Ax. therefore also AGK is equal to GKD: and they are alternate angles; therefore AB is parallel to CD. 27.1. Wherefore, straight lines, &c. Q. E. D. To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line; it is required to draw a straight line through the point A, parallel to the straight line BC. B D 23. 1. In BC take any point D, and join E AD; and at the point A, in the straight line AD, make* the angle DAE equal to the angle ADC; and produce the straight line EA to F: EF shall be parallel to BC. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel to BC. 27.1. Therefore the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Which was to be done. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D: the exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC: and the three interior angles of the triangle, viz. ABC, BCA, CAB, shall together be equal to two right angles. * B A E Through the point C draw CE parallel to the straight line AB: and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal *. Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC: but the angle ACE was shewn to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC to each of these equals add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal to two right angles; therefore also the angles CBA, BAC, ACB, are equal to two right angles. Wherefore, if a side of a triangle, &c. * COR. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. For any rectilineal figure ABCDE, can be divided into as many triangles as the figure has sides, by drawing straight Q. E. D. E A lines from a point F, within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure: and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. * COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. * to two Because every interior angle ABC, with its adjacent exterior ABD, is equal to right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the D B figure, together with four right angles; therefore* all *3 Ax. the exterior angles are equal to four right angles. PROP. XXXIII. THEOR. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD: AC, BD shall be equal and parallel. B N * 29. 1. Join BC; and because AB is parallel to CD, and BC meets them, the alternate angles* ABC, BCD are equal: and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides. AB, BC, are equal to the two DC, CB, each to each: and the angle ABC was proved to be equal to the angle BCD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles*, each to each, to * 4.1. which the equal sides are opposite: therefore the angle ACB is equal to the angle CBD: and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD: and it was shewn 27.1. to be equal to it. Therefore, straight lines, &c. Q. E. D. PROP. XXXIV. THEOR. The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. N. B. A parallelogram is a four-sided figure, of which the opposite sides are parallel: and the diameter is the straight line joining two of its opposite angles. Let ACDB be a parallelogram, of which BC is a diameter: the opposite sides and angles of the figure shall be equal to one another; and the diameter BC shall bisect it. * *29.1. * 29. 1. * 26. 1. +2 Ax. * 4. 1. See N. See the 2d and 3d figures. * 34. 1. + 6 Ax. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD, are equal to one another: and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal to one another: wherefore the two triangles ABC, CBD have two angles ABC, BCA in the one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other*, viz. the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC: and because the angle ABC is equal to the angle BCD, and the angle ČBD to the angle ACB,, the whole angle ABD is equal to the whole angle ACD: and the angle BAC has been shewn to be equal to the angle BDČ; therefore the opposite sides and angles of parallelograms are equal to one another. Also, their diameter bisects them: for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC has been proved equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts: Q. E. D. PROP. XXXV. THEOR. Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC: the parallelogram ABCD shall be equal to the parallelogram EBCF. A If the sides AD, DF of the parallel- But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not termi * 1 Ax. 2 or 3 Ax. nated in the same point; then, because ABCD is a parallelogram, AD is equal to BC; for the same 34. 1. reason EF is equal to BC; wherefore AD is equal to EF; and DE is common; therefore the whole, or the remainder, AE is equal to the whole, or the remainder DF: AB also is equal to DC; therefore the two EA, AB are equal to the two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB: therefore the base EB is DE FAE D +34. 1. 29. 1. 4. 1. the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, and the remainders are equal; that is, the parallelogram 3 Ax. ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E. D. PROP. XXXVI. THEOR. Parallelograms upon equal bases, and between the same parallels, are equal to one another. Let ABC, DEFGH, be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG: the parallelogram ABCD shall be equal to EFGH. * A DE Η 34. 1. † 1 Ax. Join BE, CH; and because BC, is equal to FG, and + Hyp. FG to* EH, BC is equal to+ EH; and they are+ parallels, and joined towards the same parts by the straightHyp. lines BE, CH: but straight lines which join the extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel; * 33. 1. therefore EB, HC, are both equal and parallel; and therefore EBCH is at parallelogram; and it is equal* to ABCD, because they are upon the same base BC, and between the same parallels BC, AH: for the like reason, the parallelogram EFGH is equal to the same EBCH: therefore the parallelogram ABCD is equal+ + 1 Ax. to EFGH. Wherefore parallelograms, &c. PROP. XXXVII. THEOR. Q. E. D. Triangles upon the same base, and between the same parallels, are equal to one another. + Def. 34. 1. * 35. 1. |