or AD or BC. therefore each of the figures AK, KB, AH, HD, AG, Book IV. А E D G K B H H C To describe a circle about a given square. . a. 8. I, Let ABCD be the given square; it is required to describe a Join AC, BD cutting one another in E. and because DA is equal E B Book iv. lines EC, ED are each of them equal to EA or EB, therefore the four straight lines EA, EB, EC, ED are equal to one another ; and the circle described from the center E, at the distance of one of them, shall pass thro' the extremities of the other three, and bę described about the square ABCD. Which was to be done, PROP. X. PROB. To describe an ifosceles triangle , having each of the angles at the base double of the third angle. Take any straight line AB, and divide • it in the point C, fo that the rectangle AB, BC be equal to the square of CA; and from the center A, at the distance AB describe the circle BDE, in which place b the straight line BD equal to AC, which is not greater than the diameter of the circle BDE ; join DA, DC, and about the triangle ADC describe the circle ACD. the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the angle BAD. Because the rectangle AB,BC is equal to the square of AC, and E A d. 37. 3. e. 32. 3. f. 32. . equal 6 to the angle CBD, because the side AD is equal to the Book IV. fide AB ; therefore CBD, or DBA is equal to BCD; and consequently the three angles BDA, DBA, BCD are equal to one ano- 8. 5. 1. ther. and because the angle DBC is equal to the angle BCD, the fide BD is equal b to the side DC. but BD was made equal to CA, h. 6. 1. therefore alfo CA is equal to CD, and the angle CDA equal 6 to 8. 5. 1. the angle DAC. therefore the angles CDA, DAC together, are double of the angle DAC. but BCD is equal to the angies CDA, DAC; therefore also BCD is double of DAC. and BCD is equal to each of the angles BDA, DBA; each therefore of the angles BDA, DBA is double of the angle DAB. wherefore an isosceles triangle ABD is described having each of the angles at the base double of the third angle. Which was to be done. PRO P. XI. PROB. gon in a given circle. Let ABCDE be the given circle ; it is required to inscribe an equilateral and equiangular pentagon in the circle ABCDE. Describe * an Isosceles triangle FGH having each of the angles a. 10. 4. at G, H double of the angle at F; and in the circle ABCDE infcribe'o the triangle ACD equiangular to the triangle FGH, fo b. 2. 4. that the angle CAD be equal A to the angle at F, and each of the angles ACD, CDA equal F to the angle at Gor H; where B E fore each of the angles ACD, CDA is double of the angle CAD. Bisect - the angles ACD, CDA by the straight lines CE, DB, and join AB, H BC, DE, EA. ABCDE is the pentagon required. Because each of the angles ACD, CDA is double of CAD, and are bifected by the straight lines CE, DB, the five angles DAC, ACE,ECD, CDB, BDA are equal to one another. but equal angles stand upon equal circumferences; therefore the five circumferen- d. 26. 3. ces AB, BC, CD, DE, EA are equal to one another. and equal C. 9. 1. Book IV. circumferences are subtended by equal o straight lines; therefore the five straight lines AB, BC, CD, DE, EA are equal to one A F В. G H ABC, BCD, CDE is equal to the angle BAE or AED. therefore the pentagon ABCDE is equiangular ; and it has been sewn that it is equilateral. Wherefore in the given circle an equilateral and equiangular pentagon has been inscribed. Which was to be done." PRO P. XII. PROB. gon about a given circle. 2. 11. 4. Let ABCDE be the given circle ; it is required to describe an equilateral and equiangular pentagon about the circle ABCDE. Let the angles of a pentagon inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumfe rences AB, BC, CD, DE, EA are equal"; and thro' the points A, Þ. 17. 3. B, C, D, E draw GH, HK, KL, LM, MG touching the circle; take the center F, and join FB, FK, FC, FL, FD. and because the straight line KL touches the circle ABCDE in the point C, to c. 18. 3. which FC is drawn from the center F, FC is perpendicular to KL; therefore each of the angles at C is a right angle. for the same rea son, the angles at the points B, D are right angles. and because FCK d. 47, I. is a right angle, the square of FK is equal to the squares of FC, CK. for the same reason the square of FK is equal to the squares of FB, BK. therefore the squares of FC, CK are equal to the squares of FB, BK, of which the square of FC is equal to the square of FB; the remaining square of CKis therefore equal to the remaining square f. 27. 3. of BK, and the straight line CK equal to BK. and because FB is Book IV. G А E M F K may be shewn that HK is double of BK. and because BK is equal to KC, as was demonstrated, and that KL is double of KC, and HK double of BK, HK shall be equal to KL. in like manner it may be shewn that GH, GM, ML are each of them equal to HK or KL. therefore the pentagon GHKLM is equilateral. It is also equiangular ; for since the angle FKC is equal to the angle FLC, and that the angle HKL is double of the angle FRC, and KLM double of FLC, as was before demonstrated; the angle HKL is equal to KLM. and in like manner it may be shewn, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM. therefore the five angles GHK, HKI, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular. and it is equilateral, as was demonstrated ; and it is described about the circle ABCDE. Which was to be done. ner, it ! |