: VI. A circle is faid to be described about a rec- VII. A ftraight line is said to be placed in a circle, when the extre- Book IV. PROP. I. PROB. IN a given circle to place a straight line, equal to a gi- circle. Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle. Draw BC the diameter of the circle ABC; then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed e qual to D: But, if it is not, BC A is greater than D; make CE equal to D, and from the cen a 3. 1. ter C, at the distance CE, de scribe the circle AEF, and join B F the center of the circle AEF, CA is equal to CE; but Dis D equal to CE; therefore D is equal to CA: Wherefore in the PROP. II. PROB. IN a given circle to inscribe a triangle equiangular to a Book IV. Let ABC be the given circle, and DEF the given triangle; ~ it is required to inscribe in the circle ABC a triangle equiangu 2 17 3. b 23. 1. lar to the triangle DEF. in the straight line G AG, make the angle A GAB equal to the H AC is drawn from F C the point of contact, the angle HAC is C 32. 3. equal to the angle d 32. I. ABC in the alternate segment of the circle: But HAC is equal to the angle DEF; therefore also the angle ABC is equal to DEF: For the same reason, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is equald to the remaining angle EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Let ABC be the given circle, and DEF the given triangle; it is required to describe a triangle about the circle ABC equiangular to the triangle DEF. Produce EF both ways to the points G, H, and find the center K of the circle ABC, and from it draw any straight line KB, at the point K in the straight line KB, make the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the ftraight lines LAM, MBN, NCL touching the circle ABC: Therefore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the center are drawn KA, KB, KC, the angles at the points A, B, C, are right angles: And because the four angles of the quadrilateral fi a 13. I. bry. 3. c. 18. 3. gure fore gure AMBK are equal to four right angles, for it can be divided in- Book IV. to two triangles; and that two of them KAM, KBM are right an gles, the other two AKB, AMB are L angles DEG, B N DEF, of which AKB is equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DEF: In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal to the e 32, 1. remaining angle EDF: Wherefore the triangle LMN is equiangular to the triangle DEF: And it is described about the circle ABC. Which was to be done. Let the given triangle be ABC; it is required to inscribe a circle in ABC. See N. Bifect the angles ABC, BCA by the straight lines BD, CD a. 9. 1. meeting one another in the point D, from which draw DE, DF, b 12. 1. DG perpendiculars to AB, BC, CA: And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and that the right angle BED is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other, and the fide BD, which is C с 26. 8. Book IV. fore DE is equal to DF: For the fame reason, DG is equal to DF; therefore the three ftraight lines DE, DF, DG are equal to one another, and the circle described from the center D, at the distance of any of them, shall pass through the extremities of the other two, and touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches d the circle: Therefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done. ₫ 16.3. PROP. V. PROB. Sce N. a 10. I. bII. 1. TO describe a circle about a given triangle. Let the given triangle be ABC; it is required to describe a circle about ABC. Bisect AB, AC in the points D, E, and from these points draw DF, EF at right angles to AB, AC; DF, EF produced 4.1. meet one another: For, if they do not meet, they are parallel, wherefore AB, AC, which are at right angles to them, are parallel; which is absurd: Let them meet in F, and join FA; also, if the point F be not in BC, join BF, CF: Then, because AD is equal to DB, and DF common, and at right angles to AB, the bafe AF is equal to the base FB: In like manner, it may be shewn that CF is equal to FA; and therefore BF is equal to FC: and FA, FB, FC are equal to one another, other; wherefore the circle described from the center F, at the Book IV. distance of one of them, shall pass through the extremities of the other two; and be described about the triangle ABC, which was to be done. Cor. And it is manifest, that, when the center of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a femicircle; but, when the center is in one of the fides of the triangle, the angle oppofite to this fide, being in a femicircle, is a right angle; and, if the center falls without the triangle, the angle opposite to the fide beyond which it is, being in a segment less than a femicircle, is greater than a right angle: Wherefore, if the given triangle be acute angled, the center of the circle falls within it; if it be a right angled triangle, the center is in the fide oppofite to the right angle; and, if it be an obtufe angled triangle, the center falls without the triangle, beyond the fide opposite to the obtuse angle. Let ABCD be the given circle; it is required to inscribe a square in ABCD. Draw the diameters AC, BD at right angles to one another; and join AB, BC, CD, DA; because BE is equal to ED, for E is the center, and that EA is common, and at right angles to A BD; the base BA is equal to the bangle; for the fame reason each of the angles ABC, BCD, b 31. 3. CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shewn to be equilateral; therefore it is a square; and it is infcribed in the circle ABCD. Which was to be done. PROP. |