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AB equal to A, and at the end erect a per. pendicular Line equal to your fhortest Line BC, and so proceed, as you were taught in the last Problem.

PROBLEM XIII.

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Fig. 3i. 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 Rhombus is made by 2 equilateral or Isosceles Triangles.

PROBLEM XIV.

Fig. 32. To make a Trapeziam, ABCD, 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 h i.

Fig. 32. Fird, Make the Line AB equal to one of the given Lines, as f. 2dly. Upon the point B, (by Prob. 10.) make the Angle A 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 feta ting one Foot upon C, with the other describe the Arch b b. Lastly, Take the fourth line i in your Compasses, letting one foot in A

with

with the other describe the Arch oi, and draw the lines DA and DC, which will conAitute a Trapeziam, the sides 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 exceedirg Ten. First, D:scribe a Circle, and cross it with 2 Diameters, AC and BD, passing through the point or Conter E, and make Ao and A Q equal to BE, and join OQ; so is OQ the third part of the Circle: then join AB together, lo will AB be the fourth part; upon L, and the distance LB, describe the Arch Bm, and join Bm, which line is the sth part; AE or BE is the sixth part, and LO or LQ are the seventh pirt, and k A, will be the eighth part. Divide the ArchQAO into 3 equal parts at S, and join SQ, which will be the ninth part, EM is the tenth part.

So you may inake the Figures called Pentagon, Hexagon, Heptagon, Octogon, Nona

gon, &c.

PROBLEM XVI.

Fig

. 34. Any three points A. B, and C, which are not in a streight Line, being given ; how

to find the Center o of the Circle BAC, which shall pass through those three given Points. Fir], Set one foot of the Compasses 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 Compasses not altered, set one foot in B, and with the other cross the form mer Arch with two final] Arches in the points D and E, and draw the right Line DE,- Thirdly, Set one foot of the Compasses in the 3d point C; they still keeping the same distance, and with the other food cross the first drawn Arch GFD in the points F and G, and draw the right Line FG, crossing the former right line DE, in the point Gs ro is O the Center fought for ; upon which if you describe a Circle at the distance GA, it shall pass through all the 3 given points AB, and C, as was required.

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PROBLEM XVII.

How to make an Oval several ways.

Fig 35. Make three Circles whose Diameters may be in a streight Line, as B: Cross the line with another Perpendicular to it at the center of the middleCircle; as cd; draw the lines ce, ch, dg, df; set one foot of the Compasses in D, and extend the other to g, describe the Arch gf, with the fame extent, setting one foot in c, describe the other part he, then from the Center with the distance BO, describe the Arch f BE. Again, with the fame distance on the Center @ describe the Archa EAH, and it is done.

then

THEOREM I.

Fig36. If any Triangle QR Shath two sides, QR, ant Qs, equel io two others or, and as, in any other Triangle, and if alro the <

Q included by those lides = to <q, inc!!!. ded by the other fides; I say, each part in one Triangle is = to its corresponding part in the other and therefore the whole Triangle QRS is = the Triangle grs.

Demon. For suppose the Triangle are be placed upon the A QRS, the side or will fall exactly on QR ( by the seventh Maxim) and the side qs will fall on its equal Qs; because <Q=< 9 fo the point S will fall on s, and Ron t,and therefore the wilole Triangles, q rs and QRS do murally agree, and consequently each part in one is equal to its corresponding one in the other.

Scholium to the first Prop.

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By the fame Reasoning we may Demonstrate the following Theorem.

If the sides R S, and rs of the two Triangles QRS, qrs were equal, and the <s

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