by the hypothesis; therefore the angle AGH is not unequal to the angle GHD, that is, it is equal to it; but the angle AGH is equal (15. 1.) to the angle EGB; therefore likewise EGB is equal to GHD; add to each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal (13. 1.) to two right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D. PROP. XXX. THEOR. STRAIGHT lines which are parallel to the same straight line are parallel to one another. Let AB, CD, be each of them parallel to EF, AB is also parallel to CD. B K F K C D E Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal (29. 1.) to the angle GHF. Again, be- A cause the straight line GK cuts the parallel straight lines EF, CD, the angle GHF is equal to (29. 1.) the angle GKD; and it was shown that the angle AGK is equal to the angle GHF; therefore also AGK is equal to GKD; and they are alternate angles; therefore AB is parallel (27. 1.) to CD. Wherefore straight lines, &c. Q. E. D. PROP. XXXI. PROB. 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 join AD; and at the point A, in the straight line AD, make (23. 1.) the E B A F angle DAE equal to the angle ADC; and produce the straight line EA to F. 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 (27. 1.) to BC. 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. Ir 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. B A C D F 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 together equal to two right angles. Through the point C draw CE parallel (31. 1.) to the straight line AB; and because AB is parallel to CE and AC meets them, the alternate angles BAC, ACE are equal (29. 1.) Again; because A 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 shown 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 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 (13. 1.) 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. 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 E angles as the figure has sides. For any rectilineal figure ABCDE can be divided into ás 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 propo D F A B sition, 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, (2 Cor. 15. 1.) 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. Because every interior angle ABC, with its adjacent exterior ABD, is equal (13. 1.) to two 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 D are sides of the figure; that is, by the foregoing corollary, they are A C B 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. 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 are also equal and parallel. A C B D Join BC; and because AB is parallel to CD; and BC meets them, the alternate angles ABC, BCD are equal (29. 1.); 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; and the angle ABC is equal to the angle BCD; therefore the base AC is equal (4. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles, (4. 1.) each to each, to 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 (27. 1.) to BD; and it was shown 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 are equal to one another and the diameter BC bisects it. C B D Because AB is parallel to CD, and A BC meets them, the alternate angles ABC, BCD are equal (29. 1.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (29. 1.) to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in 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 shall be equal, each to each, and the third angle of the one to the third angle of the other, (26. 1.) 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 shown 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, BC are equal to the two DC, CB, each to each; and the angle ABC is equal to the angle BCD; there h fore the triangle ABC is equal (4. 1.) 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.f If the sides AD, DF of the parallelo- A grams ABCD, DBCF opposite to the base BC be terminated in the same point D; it is plain that each of the parallelograms is double (34. 1.) of the triangle BDC; and they are therefore equal to one another. B D C F 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 (34. 1.) to BC; for the same reason EF is equal to BC; wherefore AD is equal (1. Ax.) to EF; and DE is common; therefore the whole, or the remainder AE, is equal (2. or 3. Ax.) to the whole, or the remainder DF; AB also is equal to DC; and the two F A E D F A D E B C B C EA, AB are therefore equal to the two FD, DC, each to each; and the exterior angle FDC is equal (19. 1.) to the interior. EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal (4. 1.) to the triangle FDC; take the triangle FDC from the trapezium, ABCF, and from the same trapezium take the triangle EAB; the remainders therefore are equal, (3. Ax.) that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q. E. D. |