KB make the Angle BKA equal to the Angle* 23. 1. Then because the Lines LM, MN, NL, touch the Circle ABC in the Points A, B, C, and the Lines KA, KB, KC, are drawn from the Center K to the Points A, B, C; the Angles at the Points A, B, C, will be + Right Angles. And because the four An-† 18. 3 gles of the quadrilateral Figure AMBK 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 AKB, AMB are equal to two Right Angles. But DEG, DEF, are equal to two Right Angles; therefore the Angles AKB, AMB, are equal to the Angles DEG, DEF, whereof AKB 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 LNB is equal to the Angle DFE. Therefore the other Angle MLN is equal to the other Anglet Cor. EDF. Wherefore the Triangle LNM is equian-32. I gular to the Triangle DEF, and is described about the Circle ABC; which was to be done. PROPOSITION IV. PROBLEM. To infcribe a Circle in a given Triangle. LET. ABC be a Triangle given. It is required * Cut the Angles ABC, BCA, into two equal⭑9. L 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, † 12 BC, AC. Now because the Angle EBD is equal to the Angle FBD, and the Right Angle BED is equal to the Right Angle BFD; then the two Triangles EBD, PBF, have two Angles of the one, equal to two H 3 Angles Angles of the other, and one Side DB common to both, viz. that which fubtends the equal Angles; therefore the other Sides of the one Triangle fhall be equal to the other Sides of the other; and fo DE fhall be equal to DF. And for the fame Reason, DG is equal to DF: Therefore DE is alfo equal to DG. And fo the three Right Lines DE, DF, DG, are equal between themselves. Wherefore a Circle defcribed about the Center D, with either of the Distances DE, DF, DG, will alfo pafs thro' the other Points. And the Sides AB, BC, AC, 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. Therefore a Circle described about the Center D, with either of the Diftances DE, DF, 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. There fore the Circle EFG is defcribed in the given Triangle ABC; which was to be done, * PROPOSITION V. PROBLEM. To defcribe a Circle about a given Triangle. ET ABC be a given Triangle. It is required to defcribe a Circle about the fame. Bifect the Sides A B, AC, in the Points D, E; from which Points let DF, EF, be drawn † at Right Angles to AB, AC, 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 BF, FC, FA. Then because AD is equal to DB, and DF is common, and at Right Angles to AB; the Bafe AF will be equal to the Base FB. And after the fame Manner we prove, that the Bafe CF is equal to the Bafe F A. Therefore alfo is BF equal to CF: And fo the three Right Lines FA, FB, FC, are equal to each other. Wherefore a Circle described about the Center F, with either of the Distances Distances FA, FB, FC, will pafs alfo thro' the other Points, and will be a Circle described about the Triangle ABC. Therefore defcribe the Circle ABC. Secondly, let DF, EF, meet each other in the Point F, in the Side BC, as in the second Figure, and join AF. Then we prove, as before, that the Point F is the Center of a Circle defcribed about the Triangle ABC. Laftly, let the Right Lines DF, EF, meet one another again in the Point F, without the Triangle, as in the third Figure; and join AF, FB, FC. And because AD is equal to DB, and DF is common, and at Right Angles, the Bafe AF fhall be equal to the Bafe BF. So likewife we prove, that CF is alfo equal to AF. Wherefore BF is equal to CF. And fo again, if a Circle be defcribed on the Center F, with either of the Distances FA, FB, FC, it will pass through the other Points, and will be described about the Triangle ABC; which was to be done. Coroll. If a Triangle be Right-angled, the Center of the Circle falls in the Side oppofite to the Right Angle; if acute-angled, it falls within the Triangle; and if obtufe-angled, it falls without the Triangle. L PROPOSITION VI. PROBLEM. To infcribe a Square in a given Circle. ET ABCD be a Circle given. It is required Draw AC, BD, two Diameters of the Circle cutting one another at Right Angles, and join A B, BC, CD, DA. * Then because BE is equal to ED, (for E is the Center) and EA is common, and at Right Angles to BD, the Bafe B A fhall be equal to the Bafe AD; 4. r and for the fame Reason BC, CD, as alfo BA, AD, are all equal to each other. Therefore the quadrilateral Figure ABCD, is equilateral. I fay it is alfo rectangular, For because the Right Line DB is a Diameter of the Circle ABCD, BAD, will be a Semicircle. H 4 31. 3. 17.3. + 18.3. $28. 1. $34. I. micircle. Wherefore the Angle BAD is * a Right Angle. And for the fame Reason every one of the Angles ABC, BCD, CDA, is a Right Angle. Therefore ABCD is a rectangular quadrilateral Figure: But it has also been proved to be equilateral. Wherefore it fhall neceffarily be a Square, and is defcribed in the Circle ABCD; which was to be done. L PROPOSITION VII, PROBLEM. To defcribe a Square about a given Circle. ET ABCD be a Circle given. It is required to defcribe a Square about the fame. Draw AC, BD, two Diameters of the Circle cutting each other at Right Angles, and through the Points A, B, C, D, draw FG, GH, HK, KF, Tangents to the Circle ABCD. * Then because F G touches the Circle A B CD, and EA is drawn from the Center E to the Point of Contact A, the Angles at A will be † Right Angles. For the fame Reason, the Angles at the Points B, C, D, are Right Angles. And fince the Angle AEB is a Right Angle, as alfo EBG, GH fhall bet parallel to AC; and for the fame Reafon, AC to KF. In this Manner we prove likewise, that GF and HK are parallel to BED; and fo GF is parallel to HK. Therefore GK, GC, AK, F B, BK, are Parallelograms; and fo GF is equal to HK, and GH to FK. And fince AC is equal to BD, and AC* equal to either GH, or FK, and BD equal to either GF, or HK, GH, or FK, is equal to GF, or HK. There fore FGHK is an equilateral quadrilateral Figure: I fay it is alfo equiangular. For because GBEA is * Parallelogram, and AEB is a Right Angle, then AGB fhall be alfo a Right Angle. In like manner we demonstrate, that the Angles at the Points H, K, F, are Right Angles. Therefore the quadrilateral Figure FGHK is rectangular; but it has been proved to be equilateral likewife. Wherefore it must neceffarily be a Square, and is defcribed about the Circle ABCD; which was to be done. PRO PROPOSITION VIII. PROBLEM. To defcribe a Circle in a given Square. ET the given Square be ABCD. It is required LET t I. Bifect the Sides AB, AD, in the Points F, E;* 10. 1. and draw + EH thro' E, parallel to AB, or DC; and † 31. I FK thro' F, parallel † to BC, or AD. Then AK, KB, AH, HD, AG, GC, BG, GD, are all Parallelograms, and their oppofite Sides are equal. And be- 34 1, caufe DA is equal to AB, and AE is half of AD, and AF half of AB, AE fhall be equal to AF; but the oppofite Sides are alfo equal. Therefore F G is equal to GE. In like manner we demonftrate, that GH, or GK, is equal to either F G, or GE. Therefore GE, GF, GH, GK, are equal to each other: And fo a Circle being defcribed about the Center G, with either of the Distances GE, GF, GH, GK, will also pafs thro' the other Points, and hall touch the Sides A B, BC, CD, DA, because the Angles at E, F, H, K, are Right Angles. For if the Circle fhould cut the Sides of the Square, a Right Line, drawn from the End of the Diameter of a Circle at Right Angles, will fall within the Circle; which is abfurd. Wherefore a Circle defcribed about the* 16. Center G, with either of the Distances GE, GF, GH, GK, will not cut AB, BC, CD, DA, the Sides of the Square. Wherefore it fhall neceffarily touch them, and will be described in the Square ABCD; which was to be done. PROPOSITION IX. PROBLEM. To defcribe a Circle about a Square given. LET ABCD be a Square given. It is required to circumfcribe a Circle about the fame.. Join AC, BD, mutually cutting one another in the Point E. And |