ELEMENTS. A BOOK IV. DEFINITION S. I. A Right-lined Figure is faid to be inscribed in a Right-lined Figure, when every one of the Angles of the infcribed Figure touches every one of the Sides of the Figure wherein it is infcribed. II. In like Manner a Figure is faid to be defcribed about a Figure, when every one of the Sides of the Figure, circumfcribed, touches every one of the Angles of the Figure, about which it is circumfcribed. III. A Right lined Figure is faid to be infcribed in a Circle, when every one of the Angles of that Figure, which is infcribed, touches the Circumference of the Circle. IV. A Right-lined Figure is faid to be defcribed about a Circle, when every one of the Sides of the circumfcribed Figure touches the Circumference of the Circle. V. So likewife a Circle is faid to be inscribed in a Right-lined Figure, when the Circumference of H 3 the 3. I. the Circle touches all the Sides of the Figure, in which it is infcribed. VI. A Circle is faid to be defcribed about a Figure, when the Circumference of the Circle touches all the Angles of the Figure which it circumfcribes. VII. A Right Line is faid to be applied in a Circle, when its Extremes are in the Circumference of the Circle. PROPOSITION I. PROBLEM. To apply a Right Line in a given Circle, equal to a given Right Line, whofe Length does not exceed the Diameter of the Circle. L ET the Circle given be ABC, and the given Right Line not greater than the Diameter, be D.. It is required to apply a Right Line in the Circle A B C, equal to the Right Line D. Draw B C the Diameter of the Circle; then, if BC be equal to D, what was required is done: Fer in the Circle A B C there is applied the Right Line BC, equal to the Right Line D: But if not, the Diameter B C is greater than D, and put * C E equal to D; and about the Centre C, with the Dittance CE, ler the Circle A E F be defcribed; and join C A. Then, because the Point C is the Centre of the Circle AEF, CA will be equal to CE; but D is equal to CE. Wherefore C A is equal to D. And fo, in the Circle AB C, there is applied a Right Line CA, equal to the given Right Line D, not greater than the Diameter ; which was to be done. PRO PROPOSITION II. PROBLEM.' In a given Circle to defcribe a Triangle, equian- LET ABC be a Circle given, and DEF a given in the Circle A B C, equiangular to the Triangle DEF. Draw the Right Line GAH touching the* 17. 3. Cicle ABC in the Point A, and with the Right Line A H, at the Point A, make † an Angle H A ̊C,† 23. 1. equal to the Angle DEF. Likewife, at the fame Point A, with the Line A G, make the Angle G A B equal to the Angle DFE; and join BC. Then, becaule the Right Line HAG touches the Circle A B C, and AC is drawn from the Point of Contact in the Circle, the Angle HAC fhall be ‡‡ 32. 3. equal to A B ̊C, the Angle in the alternate Segment of the Circle. But the Angle H A C is equal to the Angle D E F; therefore allo the Angle A B C is equal to the Angle DEF. For the fame Reason, the Angle ACB is likewife equal to the Angie DFE. Wherefore the other Angie B A C fhall be † equal to the + Cor. 2. other Angle EDF. And, confequently, the Triangle 32. 1. ABC is equiangular to the Triangle DEF, and is defcribed in the Circle A B C; which was to be done. PROPOSITION III. About a given Circle to defcribe a Triangle, equi LE ET ABC be the given Circle, and DEF the given Triangle. It is required to defcribe a Triangle about the Circle A BC, equiangular to the Triangle D E F. Produce the Side E F, both Ways, to the Points G and H, and find the Centre of the Circle K, and any how draw the Line K B. Then at the Point K, H 4 with 23. I. $17.3. 18.3. I Cor. 2. 32. I. 9. 1. + 12. 1. with KB make *the Angle BK A equal to the Angle DEG; and the Angle B K C, at the fame Point K on the other Side the Line K B, equal to the Angle DFH; and thro' the Points A, B, C, I let the Right Lines LAM, MBN, NCL, be drawn, touching the Circle A B C. Then, because the Lines L M, MN, NL, touch the Circle A B C in the Points A, B, C, and the Lines KA, K B, KC, are drawn from the Centre K to the Points A, B, C; the Angles at the Points A, B, C, will be + Right Angles. And becaufe the four Angles of the quadrilateral Figure A M B K are equal to four Right Angles (for it may be divided into two Triangles), and the Angles KAM, KBM, are each Right Angles; therefore the other Angles AK B, AM B, are equal to two Right Angles. But DEG, DEF, are equal to two Right Angles; therefore the Angles A K B, AM B, are equal to the Angles DEG, DEF, whereof AK B is equal to DEG. Wherefore the other Angle A M B is equal to the other Angle DEF. In like manner we demonftrate, that the Angle L N B is equal to the Angle D F E. Therefore the other Angle L M N is t equal to the other Angle EDF. Wherefore, the Triangle L N M is equiangular to the Triangle DEF, and is defcribed about the Circle ABC, which was to be done. PROPOSITION IV. PROBLEM. To infcribe a Circle in a given Triangle. ET ABC be a Triangle given. It is required to infcribe a Circle in the fame. LE Cut the Angles A B C, B C A, into two equal Parts by the Right Lines BD, DC, meeting each other in the Point D); and from this Point draw DE, DF, DG + perpendicular to the Sides AB, BC, AC. Now, because the Angle E B D is equal to the Angle F B D, and the Right Angle BED is equal to the Right Angle BFD; then the two Triangles E BD, DBF, have two Angles of the one, equal to two Angles of the other, and one Side D B common to both, * viz. that which fubtends the equal Angles; therefore the other Sides of the one Triangle fhall be equal to * 26. 1. the other Sides of the other; and fo DE fhall be equal to DF. And, for the fame Reafon, D G is equal to DF; therefore DE is alfo equal to DG: And fo the thee Right Lines DE, D F, DG, are equalbetween themselves. Wherefore a Circle defcribed about the Centre D, with either of the Distances DE, DF, DG, will alfo pafs thro' the other Points. And the Sides A B, BC, A C, will touch it; because the Angles at E, F, and G, are Right Angles. For if it fhould cut them, a Right Line drawn on the Extremity of the Diameter of a Circle at Right Angles, will fall within the Circle; which is * abfurd. * 16. x. Therefore a Circle defer bed about the Centre D, with either of the Distances DE, D F, DG; will not cut the Sides A B, BC, CA; wherefore it will touch them, and will be a Circle defcribed in the Triangle ABC. Therefore, the Circle E F G is defcribed in the given Triangle ABC; which was to be done. PROPOSITION V. To defcribe a Circle about a given Triangle. to defcribe a Circle about the fame. B fect the Sides A B, A C, in the Points D, E; 10. I. from which Points let D F, E F, be drawn + at Right +11. 1. Angles to A B, A C, which will meet either within the Triangle ABC, or in the Side B C, or without the Triangle. Firft, Let them meet in the Point F within the Triangle; and join B F, F C, F A. Then, because AD is equal to DB, and D F is common, and at Right Angles to A B; the Bafe A F will be equal ‡4. 1. to the Bafe F B. And after the fame manner we prove, that the Bafe CF is equal to the Bafe F A. Therefore alfo is B F equal to CF: And fo the three Right Lines F A, F B, B C, are equal to each other. Wherefore, a Circle defcribed about the Centre F, with either of the Distancese, FA, FB, FC, will pass alfo thre |