PROB. VIII. Thro' a given point A, to draw a right line AB, parallel to a given right line CD. fig. 49. From the point A, to any point, F, in the line CD, draw the line AF; with the interval FA, and one foot in F, describe the arc AE, and with the like interval and one foot in A, describe the arc BF, making BF= AE; thro' A and B draw the line AB, and will be parallel to CD. By prob. 2. The angle BAF=AFE, and by theo. 41. BA and DC are parallel. PROB. IX. Upon a given line AB to describe a square ABCD. fig. 52. Make BC perpendicular and equal to AB; and from A and C, with the line AB, or BC, let two arcs be described, cutting each other in D; from whence to A and C, let the lines AD, DC be drawn ; so is ABCD the square required. For all the sides are equal by construction; therefore the triangles ADC and BAC, are mutually equi B 1 lateral and equiangular, and ABCD is an equilateral parallelogram, whose angles are right. For B being right, D is also right, and DAC, DCA, BAC, ACB, each half a right angle [by lemma preceding theo. 7. and cor. 2. theo. 5.] whence DAB and BCD will each be a right angle and [by def. 44.] ABCD is a square. SCHOLIUM. By the same method a rectangle or oblong, may be described, the sides thereof being given. PROR. X. To divide a given right line AB, into any proposed number of equal parts. fig. 50. Draw the infinite right line AP, making any angle with AB, also draw BQ parallel to AP, in each of which, let there be taken as many equal parts AM, MN, &c. Bo, on, &c. as you would have AB divided into; then draw Mm, Nn, &c. intersecting AB in E, F, &c. and it is done. For MN and m n, being equal and parallel, FN will be parallel to EN; and in the same manner, GO to FN [by theo. 12.] therefore AM, MN, NO, being all equal by construction, it is plain [from theo. 20.] that AE, EF, FG, &c. will likewise be equal. PROB. XI. To find a third proportional to two given right lines, A and B. fig. 51. Draw two infinite blank lines CE, CD, anywise to make an angle. Lay the line A, from C to F; and the line B, from C to G; and draw the line FG; lay again the line A, from C to Hand thro' H draw H1 parallel to FG [by prob. 8.] so is CI the third proportional required. For by cor. 1. theo, 20. CG: CH:: CF: CI. Three right lines, A, B, C, given to find a fourth proportional. fig. 52. Having made an angle DEF anywise, by two infi A, from E to G; the line B from E to I; and draw the line IG; lay the line C, from E to H and (by prob. 8.) draw HK parallel thereto, so will EK bẹ the fourth proportional required. For, by cor. 1. theo. 20. EG: EI:: EHEK. Or, A: B:: C: EK. PROB. XIII. Two right lines A and B, given to find a mean proportional. fig. 53. Draw an infinite blank, line as AF, on which lay the line A, from A to B, and the line B, from B to C, on the point B, which is the joining of the lines A and B; erect a perpendicular BD (by prob. 5.) bisect AC in E (by prob. 4.) and describe the semicircle ADC; and from the point D, where its periphery cuts the perpendicular BD, draw the line BD, and that will be the mean proportional required. For if the lines AD, DC, be drawn, the angle ADC is a right angle (by cor. 5. theo. 7.) being an angle in a semicircle. The angles ABD, DBC, áre right ones (by def. 11.) the line BD being a perpendicular; wherefore the triangles ABD, DBC, are similar, thus the angle ABD DBC, being both right, the angle DAC is the complement of BDA to a right angle (by cor. 2 theo. 5.) and is therefore equal to BDC, the angle ADC being a right angle as before; consequently (by cor. 1. theo. 5.) the angle ADB DCB, wherefore (by theo. 16.) To divide a right line AB, in the point E, so that AE shall have the same proportion to EB, as two given lines C and D have. fig. 54. Draw an infinite blank line, AF, to the extremity of the line AB, to make with it any angle: lay the line C, from A to C and D from C to D; and join the points B and D, by the line BD; thro' C, draw CE parallel to BD (by prob. 8.) so is E the point of division. For, by cor. 1. theo. 20. AC: AD :: AE: AB. |