quired to inscribe a circle in the trapezium ABCD. Bisect (E. 9. 1.) each of the BAD, ADC, by AK and DK, which meet in K; from K draw (E. 12. 1.) KE1 to AD, KF 1 to AB, KL 1 to BC, and KN to CD: Then (demonstr. of S. 3. 4.) KE, KF, and KN are equal to one another; as are, also, AF and AE, and DE and DN; and KL is equal to KF or KN: For if KL be not equal to KF or KN, it is either greater or less; if it be possible, let KL > KF or KN; and join K, A, and K, D, and K, C, and K, B: Then (constr. and E. 47. 1.) KB*=KF + BF; and, likewise, KB*= KL*+ BL'; ... KF2+ BF2=KL*+BL*; but KL' >KF', ... BL*< BF, and BL< BF: In the same manner it may be shewn that LC< CN; .. BL+ LC, or BC, < BF+CN; add AD to BC, and AF +DN, which=AD, to BF+CN; ... AD+BC< 2 AB+DC, which is contrary to the supposition; .. KL is not > KF; and in a similar manner it may be shewn that KL is not <KF; .. KL=KF, or KN, or KE: From K, .., as a centre, at the distance KF, describe a circle EFLN, and it will pass through the points L, G, and E, and (E. 16. 3. cor.) will touch AB, BC, CD and DA, respectively, in the points F, L, N, and E. 21. COR. If two opposite sides of a trapezium be together equal to the other two sides, taken together, the four straight lines, which bisect the four of the figure, all of them meet in the same point. PROP. XIV. 22. PROBLEM. Upon a given finite straight line, to describe an equilateral and equiangular decagon. Let AB be the given straight line: It is re quired to describe upon it an equilateral and equi angular decagon. Describe (E. 10. 4.) the ▲ PQR, having each of the PQR, PRQ, double of the P; at the points A and B, in AB, make (E. 23. 1.) the BAK, ABK, each of them equal to the PQR, or PRQ;.. (S. 26. 1.) the 4 AKB=4QPR, and the KAB, KBA, are each of them the double of the Z AKB, which is, .., the fifth part of two right ; from the centre K, at the distance KA or KB, describe the circle ABCD, cutting AK and BK, produced, in C and D; bisect (E. 9. 1.) the 4 DKA by EKF, which (E. 15. 1.) also bisects the CKB; again bisect the DKE, EKA, by GKH, LKM, which also bisect the FKB, CKF; lastly, draw BH, HF, FM, MC, CD, DG, GE, EL and LA: The ten-sided figure ABHFMCDGEL is an equilateral and equiangular decagon. For the AKB has been shewn to be the fifth part of two right. ; ..(E. 13. 1.) it is the fifth part of the AKB, AKD; .. the ▲ AKD=44 AKB; .. (constr.) the BKA, AKL, LKË, EKG, GKD, and (E. 15. 1.) their vertical are equal to one another; .. (E. 26. 3. and E. 29. 3.) the ten-sided figure is equilateral; and since (constr, and E. 32. 1.) the isosceles, into which it is divided by the straight lines drawn from K to . its angular points, have the at their bases all equal, the figure is also equiangular. 23. COR. 1. It is manifest from this proposition, and from E. 10. 4., that if the semi-diameter of a circle be divided into two parts, so that the rectangle contained by the whole and the lesserpart may be equal to the square of the greater part, the greater segment shall be equal to the side of an equilateral and equiangular decagon inscribed in the circle; and thus may such a decagon be inscribed in a given circle. 24. COR. 2. In the solution of the proposition, is shewn the method of describing, upon a given finite straight line as a base, an isosceles A, having each of the at the base double of the third angle. 25. Cor. 3. The figure ABHFMCDGEL being an equilateral and equiangular decagon, if the points A, H, and H, M, and M, D, and D, E, and E, A, be joined, it may be shewn, from E. 4. 1., that the figure AHMDE is an equilateral and equiangular pentagon. In the same manner, if an equilateral and equiangular rectilineal figure of any even number of sides be given, a similar figure, having half that number of sides, may be constructed: Also, if a circle be described about the given figure, which can always be done by the method used in E. 14. 4., and, each of the equal of the figure having (E. 9. 1.) been bisected, if the points in which the circumference is met by the bisecting lines, and the angular points of the given figure be joined, a figure of twice as many sides as the given figure will have been constructed, which (E. 26. 3., and E. 29. 3.), is equilateral, and, .., equiangular : For in the same manner, that an equilateral pentagon, inscribed in a circle, is shewn (E. 11. 2.) to be equiangular, may any other equilateral rectilineal figure, inscribed in a circle, be shewn to be equiangular. Thus, by the help of E. 9. 1, E. 2. 4, E. 6. 4, E. 11. 4, and E. 16. 4., equilateral and equiangular figures may be inscribed in a given circle, of three, six, twelve, &c., equal sides; of four, eight, sixteen, &c. equal sides; of five, ten, twenty, &c. equal sides; and of fifteen, thirty, sixty, &c. equal sides. PROP. XV. 26. PROBLEM. Upon a given finite straight line, to describe an equilateral and equiangular pentagon. If upon the given finite straight line an isosceles ▲ be described (S. 14. 4. cor. 2.) having each of the at the base double of the third 4, and if, also, a circle be described (E. 5. 4.) about that A, it will be manifest, that the equilateral and equiangular pentagon inscribed in the circle, according to the method used in E. 11. 4., is the figure which was to be constructed. |