Book I. these the angle BGH, therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal to two right angles; therefore also BGH, GHD are equal to two right angles. wherefore if a straight line, &c. Q. E. D. C. 13. 1. a. 29. 1. b. 27. 1. S PROP. XXX. THEOR. TRAIGHT 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. Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal a to G B H F D K nate angles; therefore AB is parallel to CD. wherefore straight lines, &c. Q. E. D. Lct A be the given point, and BC the given straight line; it is required to draw a straight line thro' the point A, parallel to the straight E line BC. AF 2. 23. I. b. 27. 1. In BC take any point D, and join AD; and at the point A in the straight line AD make the angle DAE e-B qual 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 b to BC. therefore the straight line EAF is D C drawn thro' the given point A parallel to the given straight line Book I. BC. Which was to be done. I F a fide 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 fides BC be produced to D. the exterior angle ACD is equal to the two interior and oppofite angles CAB, ABC; and the three interior angles of the triangle, viz. ABC, BCA, CAB are together equal to two right angles. Thro' the point C draw CE parallel to the straight line AB. a. gr. 1. and because AB is parallel to 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 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; c. 13. 1. therefore also the angles CBA, BAC, ACB are equal to two right angles. wherefore if a fide of a triangle, &c. Q. E. D. COR. 1. All the interior angles of any rectilineal figure, together D with four right angles, are equal F to twice as many right angles as C the figure has fides. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has fides, by drawing straight lines from a point F within the figure to each of its F A B Book I. 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 fides of the figure. and the fame angles are equal to the angles of the figure, together with the angles at a. 2. Cor. the point F which is the common Vertex of the triangles; that is a, 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 fides. 15. 1. COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. Because every interior angle b. 13. 1. ABC with its adjacent exterior A a. 29. 1. b. 4. 1. gether with all the exterior angles foregoing Corollary, they are equal D to all the interior angles of the figure, together with four right angles. therefore all the exterior angles are equal to four right angles. T PROP. XΧΧΙΙΙ. ΤHEOR. HE straight lines which join the extremities of two fame parts, are also themselves equal and parallel. wards the fame parts by the straight A Join BC, and because AB is pa rallel to CD, and BC meets them; the alternate angles ABC, BCD C D are equal ; and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two fides 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 to the bafe BD, and the triangle ABC to the triangle BCD, and the other angles to the other an : b. 4. 1. gles b, each to each, to which the equal fides are opposite. there- Book I. fore 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 to be equal to it. therefore c. 27. 1. straight lines, &c. Q. E. D. THE HE opposite sides and angles of parallelograms are them, that is, divides them into two equal parts. N. B. A Parallelogram is a four fided figure of which the opposite fides are parallel. and the diameter is the traight line joining two of its opposite angles. Let ABCD 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. Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another, and because AC is A B 2 29. 1. parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal a to one another. wherefore the two triangles ABC, CBD have two angles ABC, BCA C D in one, equal to two angles BCD, CBD in the other, each to each, and one fide BC common to the two triangles, which is adjacent to their equal angles; therefore their other fides shall be equal, each to each, and the third angle of the one to the third angle of the other, viz. the fide AB to the fide CD, and AC to BD, and b. 26. 1. 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 shewn to be equal to the angle BDC; therefore the opposite fides 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 Book I. angle BCD; therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. C. 4. I. See N. See the 2d and 3d Figures. P E. D. PROP. XXXV. THEOR. ARALLELOGRAMS upon the fame base and between the fame parallels, are equal to one another. Let the parallelograms ABCD, EBCF be upon the fame base BC and between the same parallels AF, BC. the parallelogram ABCD shall be equal to the parallelogram EBCF. If the fides AD, DF of the parallelograms ABCD, DBCF oppo But if the fides AD, EF opposite to the base BC of the parallelograms ABCD, EBCF be not terminated in the same B point; then because ABCD is a parallelogram, AD is equal to BC; for the fame reason, EF is equal to BC; wherefore AD is h. 1, Ax. equal to EF; and DE is common; therefore the whole, or the c. 2. or 3. remainder, AE is equal to the whole, or the remainder DF; AB also is equal to DC; and the two EA, AB are therefore equal Ax. A DE FAEDF d. 29. 1. e. 4. 1. to the two FD, DC, each to each; and the exterior angle FDC is equal d to the interior EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal to the triangle FDC. take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle E AB; the remainders therefore are f. 3. Ax. equalf, that is, the parallelogram ABCDis equal to the parallelogram EBCF. therefore parallelograms upon the same base, &c. Q. E. D. |