make the angle HAC equal to the angle DEF; (I. 23.) and at the point Å, in the straight line AG, Because HAG touches the circle ABC, and AC is drawn from the point of contact, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle: (111. 32.) but HAC is equal to the angle DEF; (constr.) therefore also the angle ABC is equal to DEF: (ax. 1.) for the same reason, the angle ACB is equal to the angle DFE: therefore the remaining angle BAC is equal to the remaining angle EDF: (1. 32. and ax. 3.). wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Q.E.F. PROPOSITION III. PROBLEM. About a given circle to describe a triangle equiangular to a given triangle. 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, II; at the point K in the straight line KB, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC. (111. 17.) Then LMN shall be the triangle required. Because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the center are drawn KA, KB, KC, therefore the angles at the points A, B, C are right angles: (111. 18.) and because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two triangles; and that two of them KAM, KBM are right angles, therefore the other two AKB, AMB are equal to two right angles : (ax. 3.) but the angles DEG, DEF are likewise equal to two right angles ; (1. 13.) therefore the angles AKB, AMB are equal to the angles DEG, DEF; (ax. 1.) of which AKB is equal to DEG; (constr.) wherefore the remaining angle ÀMB is equal to the remaining angle DEF. (ax. 3.) In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining angle EDF: (1. 32 and ax. 3.) therefore the triangle LMN is equiangular to the triangle DEF: and it is described about the circle ABC. Q.E.F. PROPOSITION IV. PROBLEM. Let the given triangle be ABC. D/ B F Bisect the angles ABC, BCA by the straight lines BD, CD meeting one another in the point D, (1. 9.) from which draw DE, DF, DG perpendiculars to AB, BC, CA. (1.12.) And because the angle EB) 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; (ax.11.) therefore the two triangles EBD, FBD have two angles of the one equal to two angles of the other, each to each; and the side BD, which is opposite to one of the equal angles in each, is common to both; therefore their other sides are equal; (1. 26.) wherefore DE is equal to DF: therefore DE is equal to DG: (ax. 1.) therefore the three straight lines DE, DF, DG are equal to one another; and the circle described from the center D, at the distance of any of them, will 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 the circle: (III. 16.) therefore the straight lines AB, BC, CA do each of them touch the circle, and therefore the circle EFG is inscribed in the triangle ABC. Q.E.F. PROPOSITION V. PROBLEM. Let the given triangle be ABC. Bisect AB, AC in the points D, E, (1. 10.) and from these points draw DF, EF at right angles to AB, AC; (1.11.) *DF, EF produced 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 B°C, join BF, CF. Then, because AD is equal to DB, and DF common, and at right angles to AB, therefore the base AF is equal to the base FB. (1. 4.) In like manner, it may be shewn that CF is equal to FA; and therefore BF is equal to FC; (ax. 1.) and FA, FB, FC are equal to one another: wherefore the circle described from the center F, at the distance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC. Q.E.F. 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, (I11. 31.) each of them being in a segment greater than a semicircle; but, when the center is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, (111. 31.) is a right angle; and, if the center falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, (111. 31.) is greater than a right angle: therefore, conversely, if the given triangle be acute-angled, the center of the circle falls within it; if it be a rightangled triangle, the center is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the center falls without the triangle, beyond the side opposite to the obtuse angle. PROPOSITION VI. PROBLEM. Let ABCD be the given circle. Draw the diameters, AC, BD, at right angles to one another, (111. 1. and 1. 11.) and join AB, BC, CD, DA. The figure ABCD shall be the square required. Because BE is equal to ED, for E is the center, and that EA is common, and at right angles to BD; the base BA is equal to the base AD: (1. 4.) and, for the same reason, BC, CD are each of them equal to BA, or AD; It is also rectangular; for the straight line BD being the diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right angle: (I11. 31.) for the same reason, each of the angles ABC, BCD, CDA is a right angle: and it has been shewn to be equilateral, therefore it is a square: (1. def. 30.) PROPOSITION VII. PROBLEM. Let ABCD be the given circle. G A F н с ќ Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D, draw FG, GH, HK, KF touching the circle. (111. 17.) The figure GIKF shall be the square required. Because FG touches the circle ABCD, and EA is drawn from the center E to the point of contact A, therefore the angles at A are right angles: (111. 18.) for the same reason, the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as likewise is EBG, therefore GH is parallel to AC: (1. 28.) for the same reason AC is parallel to FK: and in like manner GF, HK may each of them be demonstrated to be parallel to BED: therefore the figures GK, GC, AK, FB, BK are parallelograms; and therefore GF is equal to HK, and GH to FK: (1. 34.) and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK: It is also rectangular; therefore AGB is likewise a right angle: (1. 34.) and in the same manner it may be shewn that the angles at H, K, F are right angles : and it was demonstrated to be equilateral; therefore it is a square; (1. def. 30.) PROPOSITION VIII. PROBLEM. To inscribe a circle in a given square. Let ABCD be the given square. A E D B HC and through F draw FK parallel to AD or BC: therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a right-angled parallelogram; and their opposite sides are equal: (1. 34.) and because AD is equal to AB, (1. def. 30.) therefore AE is equal to AF; (ax. 7.) wherefore the sides opposite to these are equal, viz. FG to GE: in the same manner it may be demonstrated that GH, GK are each of them equal to FG or GE: therefore the four straight lines GE, GF, GH, GK are equal to one another; and the circle described from the center G at the distance of one of them, will pass through the extremities of the other three, and touch the straight lines AB, BC, CD, DA; because the angles at the points E, F, H, K, are right angles, (1. 29.) and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle: (II. 16. Cor.) therefore each of the straight lines AB, BC, CD, DA touches the circle, which therefore is inscribed in the square ABCD. Q.E.F. |