will mutually agree with each other, and therefore be equal; for if we consider these two triangles to be made of the same three given "lines, they are manifestly one and the same triangle. Otherwise, Upon the centres A and B, at any distance, let two arcs, DE, FG, be described ; make the arc FG=DE, and through B and G draw the line BG, and it is done. For since the chords ED, GF, are equal, the angles A and B are also equal, as before (by def.17.) PROB. III. PL. 2. fig. 9. To bisect or divide into two equal parts, any given right lined angle, BAC. In the lines AB and AC, from the point A set off equal distances AE,= AD, then, with any distance more than the half of DE, describe two arcs to cut each other in some point F; and the rightline AF, joining the points A and F, will bisect the given angle BAČ. For if DF and FE be drawn, the triangles ADF, AEF, are equilateral to each other, viz. AD=AE, DF=FE, and AF common, where fore DAF=EAF, as before. PROB. IV. PL. 2. fig. 10. To bisect a right-line AB. With any distance; more than half the line, from K A and B, describe two circles CFD, CGD, cutting each other in the points C and D; draw CD intersecting AB in E, then AE=EB. For, if AC, AD, BC, BD, be drawn, the triangles ACD, BCD, will be mutually equilateral, and consequently the angle ACE=BCE: therefore the triangle ACE, BCE, having AC=BC, CE common, and the angle ACE=BCE; (by theo. 6.) the base AE= the base BE. Cor. Hence it is manifest, that CD not only bisects AB, but is perpendicular to it, (by def. 11.) PROB. V. PL. 2. fig. 11. Ona given point 1, in a right line EF, to erect a perfien dicular. From the point A lay off on each side, the equal distances, AC, AD; and from C and D, as centres, with any interval greater than AC or AD, describe two arcs intersecting each other in B ; from A to B draw the line AB, and it will be the perpendicular required. For, let CB, and BD be drawn; then the triangles CAB, DAB, will be mutually equilateral and equiangular, so CAB=DAB, a right angle, (by def. 10.) PROB. VI. PL. 2. fig. 12. To raise a perpendicular on the end B of a right line AB, From any point D not in the line AB, with the distance from D to B, let a circle be described cut ting AB in E ; draw from E through D the right line EDC, cutting the periphery in C, and join CB; and that is the perpendicular required. EBC being a semicircle, the angle EBC will be a right angle (by cor. 5. theo. 7.) PROB. VII. PL. 2. fig. 13. From a given point A, to let fall a perpendicular upon a given right line BC. From any point D, in the given line, take the distance to the given point A, and with it describe a circle AGE, make GE=AG, join the points A and E, by the line AFE, and AF will be the perpendicular required. Let DA,DE,be drawn;the angle ADF=FDE, DA=DE, being radii of the same circle, and DF common; therefore (by theo. 6.) the angle DFA =DFE, and FA a perpendicular. (By def. 10.) PROB. VIII. PL. 2. fig. 14. Through a given point A, to draw a right line AB, parallel to a given right line CD. From the point A, to any point F, in the line CD, draw the line AF; with the interval FA, and one foot of the compasses in F, describe the arc AE, and with the like interval and one foot in A, describe the arc BF, making BF= AE; through A and B draw the line AB, and it will be parallel to CD By prob. 2. The angle BAF=AFE, and by theo. 11. BA and CD are parallel. PROB. IX. PL. 1. fig. 17. ABCD. 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 equilateral 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 DÅB 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. PROB. X. PL. 2. fig. 15. To divide a given right line AB, into any proposed number of equal parts. Draw the indefinite 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 mn being equal and parallel, FN will be parallel to EM; 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. 10.) that AĖ, EF, FG, &c. will likewise be equal PROB. XI. PL. 2. fig. 16. To find a third proportional to two given right lines, A and B. Draw two indefinite blank lines CE, CD, anywise to make any angle. Lay the line A, from C to P; and the line B, from C, to G; and draw the line FG ; lay again the line A, from C to H; and through H, draw HI parallel to FG (by prob. 8.) so is ČI the third proportional required. For by cor. 1. theo. 20, CG : CH::CF: CI. Or, B; A::A: CI. PROB. XII. PL. 2. fig. 17. Three right lines A, B, C, given to find a fourth proportional. Having made an angle DEF anywise, by two indefinite blank right lines, ED, EF, as before; lay the line A; from E to G; the line B, from E to l; and draw the line IG; lay the line C, from E to |