AB equal to A, and at the end erect a per. pendicular Line equal to your shortest Line BC, and so proceed, as you were taught ina the last Problem. PROBLEM XIII. Fig. 31. To make a Rhombus ABCD. Make an Angle, as ABC, and make the fides AB, BC equal, then taking the length of one of them and setting your Compasses in A, describe the Arch mm; also put one Foot in Cand strike the Arch a a.Laflly draw the Lines DC and D and it's finished. Note, A Rhornbus is made by 2 equilateral or Isosceles Triangles. PROBLEM XIV. Fig. 32. To make a Trapeziam, A B CD, which shall have one Angle at C equal to a given Angle E, and the four fides equal to four given Lines, viz. the Lines fg hi. Fig. 32. Firf, Make the Line AB equal to one of ihe given Lines, as f. 2dly. Upon the point B, (by Prob. 10.) make the Angle Å BC equal to the Angle E, makirg the fide CB equal to the given Line g. 3dly, Take another of your given lines, as the lineh, and fettirg one Foot upon C, with the other describe the arch b b. Lastly, Take the fourth line i in your Compasses, setting one foot in A witha with the other defcribe the Arch oo, and draw the lines DA and DC, which will conAtitute a Trapeziam, the fides whereof are equal to the four lines given, and it has an Angle equal to the Angle given,, which was to be done. PROBLEM XV. Fig 33.To divide a Circle ABCD,into any Number of equal parts, not exceeding Ten. Firft, Defcribe a Circle, and crofs it with 2 Diameters, AC and BD, paffing through the point or Center E, and make Ao and A Qequal to BE, and join OQ; fo is OQ the third part of the Circle: then join AB together, fo will AB be the fourth part; upon L, and the diftance LB, defcribe the Arch Bin, and join Bm, which line is the 5th part; AE or BE is the fixth part, and LO or LQ are the feventh part, and k A, will be the eighth part. Divide the Arch QAO into 3 equal parts at S, and join SQ, which will be the ninth part, EM is the tenth part. So you may make the Figures called Pentagon, Hexagon, Heptagon, Octogon, Nonagon, &c. PROBLEM XVI. Fig 34. Any three points A. B, and C, which are not in a ftreight Line, being given; how to find the Center O of the Circle BAC, which fhall pass through thofe three given Points. Fir, Set one foot of the Compaffes in one of the given points, as in A, and extend the other foot to B,another of the points, and draw the Arch of a circle GFD.Secondly, The Compaffes not altered, fet one foot in B, and with the other crofs the former Arch with two finall Arches in the points D and E, and draw the right Line DE,- Thirdly, Set one foot of the Compaffes in the 3d point C; they ftill keeping the fame diftance, and with the other foot crofs the first drawn Arch GFD, in the points F and G, and draw the right Line FG, croffing the former right line DE, in the point G, fo is O the Center fought for; upon which if you defcribe a Circle at the diftance GA, it fhall pafs through all the 3 given points AB, and C, as was required. PROBLEM XVII. How to make an Oval feveral ways. Fig 35.Make three Circles whofe Diameters may be in a ftreight Line, as B: Crofs the line with another Perpendicular to it at the center of the middle Circle; as cd; draw the lines ce, ch, dg, df; fet one Foot of the Compaffes in D, and extend the other to g, defcribe the Arch gf, with the fame extent,fetting one foot in c, describe the other part he, then then from the Center O,with the distance BO, defcribe the Arch f BE. Again, with the fame diftance on the Center Q defcribe the Arch EAH, and it is done. THEOREM I Fig 36. If any Triangle QRShath two fides, QR, and QS, equal to two others gr, and qs, in any other Thangle, and if alfo the < Qincluded by thofe fides to < q, inclu ded by the other fides; I fay, each part in one Triangle is to its correfponding part in the other,and therefore the whole Triangle ORS is the Triangle grs. Demon. For fuppofe the Triangle q r s be placed upon the A QRS, the fide qr will fall exactly on QR (by the feventh Maxim) and the fide qs will fall on its equal QS; becaufe <Q=< q fo the point S will fall on s, and R on r,and therefore the whole Triangles, qrs and QRS do mutally agree, and confequently each part in one is equal to its correfponding one in the other. Scholium to the firft Prop. By the fame Reafoning we may Demonftrate the following Theorem. If the fides RS, and rs of the two Triangles QRS, qrs were equal, and the <s adjacent |