For (E. 34. 1.) the opposite of a□ are equal to one another; and (E. 22. 3.) if a circle could be described about it, the two opposite would, together, be equal to two right ; .., since these are equal, they would be each of them a right <; but (E. 32. def. 1.) the angles of a rhombus, which (E. 32. def. 1. and S. 18. 1.) is a □, are not right; a circle cannot be described about a rhombus, nor about any other, which has not its opposite right, that is (S. 19. 1.) which is not rectangular. PROP. XXX. 40. THEOREM. If from any point, in the circumference of a given circle, straight lines be drawn to the three angles of an inscribed equilateral triangle, the greatest of them shall be equal to the aggregate of the two less. Let the equilateral A ABC be inscribed in the B D E circle ADBC, and from any point D in the circumference, let there be drawn to the three angular points A, B, C, the straight lines DA, DB, DC, of which DC is the greatest: Then DC=DA+ DB. From the centre A, at the distance AD, describe a circle cutting DC in E, and join A, E; .. (E. 15. def. 1.) AD AE; .. (E. 5. 1.) the 2 ADE AED; also (E. 21. 3.) the 4 ADC or ADE = ABC; and (hyp. and E. 5. 1.) the ABC= ACB; .. (S. 26. 1.) the DAE 2 BAC; ..the A ADE is equiangular; .. (E. 6. 1.) AD =DE. L Again, since (E. 22. 3.) the ▲ ACB+ 2 ADB= two right, and (E. 13. 1.) the 2 AED+ AEC two right, and that the AED has been shewn to be equal to the ▲ ACB, .. the ▲ AEC= LADB; also (E. 21. 3.) the ACD or ACE= Z ABD; and (hyp.) the side AC, of the A AEC, is equal to the side AB of the A ADB, .. (E. 26. 1.) EC=DB: And DA has been proved to be equal to DE; ... DE + EC=DA+ DB; that is, DC DA+DB. PROP. XXXI. 41. THEOREM. The first, third, fifth, &c. angles of any polygon, of an even number of sides, which is inscribed in a given circle, are together equal to the remaining angles of the figure; any angle whatever being assumed as the first. L Let ABCDEF be any polygon, of an even num ber of sides, inscribed in the given circle ACE: Then A being assumed as the first 4, the A+ <C+E+, &c. = 2B+4D+2 F+, &c. First, let the inscribed figure have six sides, and join B, E. Then, since BAFE is a quadrilateral figure inscribed in a circle, .. (E. 22. 3.) the BAF +2 FEB= 2 EFA+ 2 EBA: Also, the BCD+ 4 BED LEDC+2 EBC. Wherefore, equals being added to equals, it will be manifest, that the BAF+ 2 BCD + ≤ FED =2CBA+2 EDC+ 2 AFE: i. e. the A+<C+<E=<B+<D+< F. And, in a similar manner, the proposition may be demonstrated, when the figure inscribed in the given circle has eight, ten, twelve, or any other even number of sides. PROP. XXXII. 42. PROBLEM. To make a trapezium, about which a circle may be described, having its four sides respectively equal to four given straight lines, two of which are equal to each other, and any three together greater than the fourth; the two equal sides of the trapezium, also, being opposite to each other. Let AB, CD, DE be three given straight lines: It is required to make a trapezium having two of its opposite sides each of them equal to AB, and its two other sides equal to CD and CE, each to each, about which a circle may be described. Take GH=CD; and CD and CE being placed in the same straight line, bisect (E. 10. 1.) DE in F; produce GH, both ways, and make GI and HK each of them equal to DF or FE; .. IK= CE: From the points G, H draw (E. 11. 1.) GX and HY to IK; from I and K, as centres, at distances equal to AB, describe two circles, cutting GX and HY in L and M, respectively; and join I, L and K, M; .. (E. 15. def. 1.) ILAB and KM AB; join L, M. And, because (constr.) LI=MK, and IG= KH, and that the IGL, KHM, are right 4, (S. 74. 1.) GL HM; and since, the at G and H are right , GL is (E. 28. 1.) parallel to HM; .. (E. 33. 1.) LM is parallel and to EH; but (constr.) GHCD,... LM = CD. L = ; Again, since GLMH is a □, the GLH=L GHM (E. 34. 1.) which (constr.) is a right ; also, since the two sides IL, LG, of the ▲ LGI, are equal to the two sides KM, MH of the A MHK, and the base IG is equal to the base KH, .. (E. 8. 1.) the ILG= KMH; but (constr. and E. 32. 1.) the HKM+ 2 KMH a right .. the HKM, or IKM, + ILG= a right ; to each of these add the right GLM; .. the IKM + ILG + GLM = two right ; that is the IKM + 2 ILM = two right ; .. (S. 28. 3.) a circle may be described about the trapezium ILMK, which, as hath been shewn, has two equal sides LI, MK, each of them equal to AB, has its side LM equal to CD, and its remaining side IK equal to CG. L |