A Refilineal figure (ABCD) is said to be inscribed in another reitilineal figure (EFGH), when all the angles (A, B, C, D) of the inscribed figure, are upon the sides of the figure in which it is inscribed (fig. 1). II. In like manner a rectilineal figure (EFGH) is said to be described about another rectilineal figure (ABCD); when all the sides (EF, FG, GH, HE) of the circumscribed figure pass thro’ the angular points (A, B, C, D) of the figure about which it is described, each thro' each (Fig. 1). III, A rectilineal figure (ABCD) is said to be inscribed in a circle, when all the angles (A, B, C, D) of the inscribed figure are upon the circumference of the cricle (ABCDE) in which it is inscribed (Fig. 2). IV. A redilineal figure (ABCDE) is said to be described about a circle, when each of the sides AB, BC, CD, DE, EA) touches the circumference of the circle (Fig. 3). V. A Circle (ABCD) is said to be inscribed in a refilineal figure (EFGH, when the circumference of the circle touches each of the sides (EF, FG, GH, HE) of the figure in which it is inscribed (Fig. 1). VI. A circle (ABCD) is described about a reelilineal figure (ABC), when the circumference of the circle passes thro' all the angular points (A, B, C) of the figure about which it is described (Fig. 2). VU. A straight line (AB) is said to be placed in a circle (ADBE), when the extremities of it (A & B) are in the circumference of the circle (fig. 3). IN PROPOSITION I. PROBLEM I. N a given circle (ABD), to place a straight line (AB) equal to a given straight line (N), not greater than the diameter of the circle (ABD). Given. Sought. A O ABD together with the firaight The straight line AB placed in the line N, not > the diameter of O ABD & = to the given straight this O. line N. Resolution. Draw the diameter AD of the O ABD. Pof. I. CASE I. If AD is = to N. Ther HER E has been placed in the given O ABD a straight line = to the given N. D. 7. B.4. CASEH. If AD is > N. Pof. 3. 1. Make AE = 10 N. P. 3. B. I. 2. From the center A at the distance AE describe the O EBF, and draw AB. DEMONSTRATION. Because AB is = 80 AE (D. 15. B. 1), and the straight line N is = to AE (Ref. 1.) 1. The straight line AB, placed in the O ABD, will be also = S Ax. 1. B. 1. 1 D. 7. B. I. Which was to be done. to N. PROPOSITION II. PROBLEM II. IN N a given circle (ABHC), to inscribe a triangle (ABC) equiangular to a given triangle (DFE). Given. Sought. AO ABHC together with the A The ABC inscribed in the ABHC, DFE. equiangular to tbe ADFE. Resolution. 1. From the point M, to the O ABHC draw the tangent MN. P. 17. B. 3. 2. At the point of contact A in the line MN make the V BAM = to the V FED, and the V CAN= to the V FDE. P. 23. B.1. 3. Draw BC. Pof. I. DEMONSTRATION. Because ECAUSE the BCA is = to the V BAM (P. 32. B. 3), and the V FED is = to the same V BAM (Ref. 2); also the V CBA is = to the V CAN (P. 32. B. 3.) and V FDE is = to V CAN (Ref. 2. 1. It follows that v BCA is = to V FED, and V CBA = to Ý FDE. Ax. I. B. 1. 2. Consequently, the third V BAC, of the ABC, is also to the third V DFE of the ADFE, and this A ABC is inscribed in the S P. 32. B. I. O ABHC. D. 3. B. 4 Which was to be done. P. 11. B. i. PROPOSITION III. PROBLEM III. ABOUT BOUT a given circle (EFG) to describe a triangle (ABD), equiangular a , to a given triangle (HKL). Given. Sought. The O EFG, together with the A The AABD described about the o HKL. EFG, equiangular to the A HKL. Resolution. ways. Pof. 2. 2. Find the center C of the O EFG, and draw the ray CE. P. 1. B. 3. 3. At the point C in CE, make the V ECF = to the V KHM, and y ECG = to v KLN. P. 23. B. 1. 4. Upon CE, CF, CG, erect the I AD, AB, DB produced. Preparation. Pos. 1. Ax. 8.B.1. 1. V FEA + EFA are < 2 L, & AD, AB meet some where in A. Ax. 11.B.1. It may be demonstrated after the same manner, that. 2. AD, DB & AB, DB meet somewhere in D & B. And since AD, AB, DB are I at the extremities E, F, G of the rays Er, CP, CG (Ref. 4.) 3. These straight lines touch the O EFG; and the A ABD formed P. 16. B. 3 by these straight lines is described about the O EFG. Cor.1.4.3.4 Moreover, the 4 V CEA + CFA + ECF + FAE of the quadrilateral figure AFCE being = to 4 L (P. 32. B. 1), and the V CEA +CFA = to 2 L (Ref. 4). 4. The V ECF + FAE are allo = to 2 L Ax. 3. B. 1. 5. Or = to V KHM + KHL as being allo = to 2 L. Ax. 1.B. 1. But ECF being = to V KHM (Řes: 3). P. 13.B.1. 6. The V FAE is = to v KHL, and Ý GDE = 10 V KLH. Ax. 3. B. I. 7. Hence the third V FBG of the A ABD is = to the third y HKL of A HKL. P. 32. B. I. 8. Therefore the A ABD described about the O EFG is equiangular to the given 4 HKL. Which was to be done, { |