and it was proved that the angle AGK is equal to the angle GHF; therefore AGK is also equal to GKD; and they are alternate angles; therefore AB is parallel * to CD. Wherefore straight lines, &c. Q. E. D. * 27. 1. PROP. XXXI. PROB. A F с 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. In BC take any point D, and B join AD; and at the point A, in the straight line AD make the angle DAE equal to the angle ADC; and * 23. 1. produce the straight line EA to F. Because the straight line AD meets the two straight lines BC, EF, and 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. PROP. XXXII. THEOR. is equal to the two interior and opposite angles; and 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 be together equal to two right angles. B D *1 Ax. D Through the point C draw * 31. 1. CE parallel * to the straight line BA; and because CE meets them, the alternate an* 29. 1. gles BAC, ACE are equal. * Again, because CE is pa rallel to AB, and BD falls upon them, the exterior angle * 29. 1. ECD is equal * to the interior and opposite angle ABC; but the angle ACE was proved be equal the angle * 2 Ax. 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 * 2 Ax. ACD, ACB are equal* to the three angles CBA, BAC, * 13. 1. ACB; but the angles ACD, ACB are equal * to two right angles; therefore the angles CBA, BAC, ACB, are also equal* to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D. 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 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 * 2 Cor. 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. 15. 1. COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. Because every interior angle ABC, with its adjacent exterior angle ABD, is equal * to two right angles ; * 13. 1. 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; D that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles. PROP. XXXIII. THEOR. с 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. 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 com- * 29. 1. mon to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC was proved to be equal to the angle BCD; therefore the base AC is equal * to the base BD, and * 4. 1. the triangle ABC to the triangle BCD, and the other angles to the other angles,* each to each, to which the * 4. 1. 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 B the alternate angles ACB, CBD equal to one another, * 27.1. AC is parallel * to BD; and it was proved to be equal to it. Therefore, straight lines, &c. Q. E. D. PROP. XXXIV. THEOR. equal to one another, and the diameter bisects them, 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. Because AB is parallel to CD, A and BC meets them, the alter nate angles ABC, BCD are * 29. 1. equal * to one another; and be cause AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are * 29. 1. 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, which is adjacent to their equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle * 26. 1. 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 CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: and the angle BAC has been proved to be equal to the angle BDC; 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, с D 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 * 4. 1. BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D. * 6 Ax. PROP. XXXV. THEOR. Parallelograms upon the same base, and between the same parallels, are equal to one another. Let the parallelograms ABCD, DBCF be upon the See the 2d and same base BC, and between the same parallels AF, BC; 3d the parallelogram ABCD shall be equal to the paral- figures. lelogram DBCF. If the sides AD, DF of the parallelograms ABCD, DBCF 0 opposite to the base BC be terminated in the same point D; it is plain that each of the parallelo- B grams is double* of the triangle BDC; and they are * 34. 1. therefore equal * to one another. But, if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD is equal * to BC; for the same reason EF * 34. 1. is equal to BC; wherefore AD is equal * to EF; and * 1 Ax. DE is common; therefore the whole, or the remainder, AE is equal * to the whole, or the remainder DF; AB * 2 or 3 is also equal to DC; and therefore the two EA, AB Ax. are equal to the two A FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB, therefore the base EB is equal to the base FC, and the triangle EAB equal* to the triangle * 4. 1. FDC; take the triangle FDC from the trapezium * 29. 1. E |