the Cirele equiangular to the Triangle DEF. Q. E. F. PROP. III. Probl. 3. To describe the Triangle LNM about a given Circle ABC, equiangular to a given Triangle DEF. D N DF, DG. arough touch the A L B M way of rea defcribed ,F,G; night ones Continue out the Side EF both ways. At the Centre I, make the Ang. AIB DEG, and the Ang. BIC=DFH; then let three right Lines LN, LM, MN touch the Circle in the Points A, B, C, and I say the thing is done. d For that the right Lines LN, LM, MN, will meet, will thus appear; because the Angles LAI, LBI, are right ones; and so a right Line AB being drawn, shall make the Angles LAB, LBA, less than two right ones. There32. fore fince the Ang. AIB + L=2. right Ang. = * DEG + DEF; and AIB & = DEG; the Ang. L shall be h = DEF. After the fame way of reasoning the Ang. M = DFE, therefore likewise the Ang. N* = D, and so the Triangle LNM is circumscribed about the Cire cle, Equiangular to the given Triangle EDF. the Triang Hence if we can fin Contact of Let Al AB+B = AE BF. Th or CG = B F To inscribe a Circle EFG in a given Tri angle ABC. a Bifect the two An- * 9. 1. gles B and C by the b C Perpendiculars DE, DF, DG. a Circle described from the Centre D through E shall pass through Gand F, and touch the three Sides of the Triangle. For the Angle DBE = DBF; and the Ang. DEB d = DFB; and the Side DB is common. Therefore DE DF. After the fame way of reasoning DG = DF, therefore a Circle described from the Centre D, shall pass through E, F, G; and fince the Angles at E, F, G, are right ones, it shall touch all the three Sides of the Triangle. Q. E. F. SCHOLIUM. Hence if the three Sides of a Triangle be known we can find their Segments made by the Points of Contact of the infcrib'd Circle. Thus, Let AB be 12, AC 18, BC 16. Then shall AB + BC = 28, from whence take 18 = AC = AE + FC, and there remains 10 = BE + BF. Therefore BE, or BF = 5. whence FC or CG = 11. and fo GA, or AE = 7. confir. d 12. ax. 26.1. PROP. PROP. V. Probl. 5. To describe a Circle ABC, about a given Triangle. * 10.11. 1. b conft. d 4. 1. a Bifect any two Sides (BA, AC) by the Perpendiculars DF, EF meeting in F, which will be the Centre of the Circle. For draw the right Lines FA, FB, FC. Now because AD=DB; and the Side DF is common; and the Ang. FDA=FDB; therefore shall FB = FA. In like manner FC = FA. Whence a Circle having the Point F for a Centre, shall pass thro' the three Angles B, A, C, of the given Triangle. COROL. * Hence, if the Triangle be acute-angled, the Centre will fall within the Triangle; if right-angled, on the Side opposite to the right Angle; and if obtufe-angled, without the Triangle. *31.3. SCHOL. SCHOL. By the fame way you may describe a Circle that shall pass through three given Points, not being in one streight Line, PROP. VI. Probl. 6. D : by the right Lines, AB, B BC, CD, DA; I say the thing is done. For because the four Angles at E are right ones the Arcs, and the Subtenses AB, BC, CD, DA, are equal. Therefore ABCD is an equalfided Figure; but all its Angles are in Semicircles, and so are right ones, whence ABCD is a a Square inscribed in the given Circle. PROP. VII. Probl. 7. H C 26.3. 29.3. 31. 3. 29 Def 1. > Draw the Diameters D AC, BD, mutually cutting one another at right Angles. And thro' their Extremes draw Tan- a 17.3. a I gents meeting in F,H,I,G; For because the Angles at A and Care right 18. 3. parall. to HI; in like man- 28. 1 ones, FG shall be H ner, C = CA=FH = GI. Whence FHIG is a Square describ'd in the given Circle. Q. E. F. will be inscribed in the Square. For because AH, BF are * equal and parallel, the Side AB shall be parall. to HF parall. to DC; therefore IA, ID, IB, IC are Parallelograms. Whence AHd = AE = HI = EI = IF= IG. And consequently a Circle describ'd from the Centre I through H, shall pass ro' H, E, F, G, and shall touch the Sides ter A which |