PROP. XIX. THEOR. The section of a plain with the surface of a sphere (ADBE), is circle, whose centre is in the diameter of the sphere which is perpendicular to the plain. Part 1. If the section pass through the centre C of the Part 2. But if the section do not pass through the centre of the sphere, as the section HLKM, let CO be a perpendicular drawn from the centre C the plain HLKM (6. 2 Sup.); through O draw, in the plain HLK, any two right lines whatever HK and LM, meeting the section HLKM in the points H and K, L and M, and draw CM and CK; and because the triangles COK and COM have the angles at O right (Constr. and Def. 3. 2 Sup.), the sides CK and CM equal to each other (Def. 12. 2 Sup.), and CO common to both, the sides OK and OM are equal (Cor. 7. 6 Eu.) ; in like manner, may all right lines drawn from 0 to the perimeter of the section HLKM be proved equal to each other, which section is therefore a circle (Def. 10. 1 Eu.), whose centre is in the diameter of the sphere DE, which is perpendicular to the plain HLKM. Cor. From this proposition it appears, how the centre of a given sphere may be found, namely, by finding the centre O of any circle HLK-formed by the intersection of a plain with its surface (1. 3 Eu), drawing through that centre a perpendicular to the plain (7. 2 Sup.), to meet the surface of the sphere both ways as in D and E, and bisecting DE in C; the point C is the centre of the sphere. If the centre of the sphere be in DE, 'tis plain, C must be that centre; if the centre were in any point without DE, right lines being drawn, in the plain passing by DE and that point from that point to O and the intersections of the same plain with the circle HLK, a like absurdity might be shewn to follow, as in 1, 3 Eu. PROP. XX. THEOR. Any plain, passing through the vertex of a cone, and cutting the circumference of its base, cuts the opposite surfaces in two right lines, and in them only. For if from the points in which this plain cuts the circumfer ence of the base, two right lines be drawn to the vertex, they are in the cutting plain (Def. 4 and 7. 1 Eu.), and in the conical surface (Def 17. 2 Sup.), and, being produced beyond the vertex, in the opposite surface (Def. 22. 2 Sup.); therefore the plain passing through the vertex cuts the opposite surfaces in these two right lines; and since the intersection of this plain with the base, which is a right line (1. 2 Sup.), cannot cut the circumference of the base in more than two points, that plain cannot cut the coni-· cal surface in more than these two right lines. Cor. The intersection of a plain, passing through the vertexof a cone, with the conical surface and base, is a triangle. 50 PROP. XXI. THEOR. If either of the opposite surfaces of a cone be cut by a plain parallel to the base, the section is the circumference of a circle whose centre is in the axis of the cone. Let ABD be a cone, whose vertex is A, and its base the circle BKDL, of which C is the centre; AC is its axis (Def. 20. 2 Sup.); let EGFH be the section of a plain parallel to the base, with one of the opposite surfaces; EGFH is a circle, whose centre is in the axis AOC. B F H Let the cone be cut by any two plains passing by the axis AC, making the triangles ABD and AKLor. 20. 2 Sup.), and let these trianges, produced if necessary, meet the pain EGFH in the right lines Er and GOH; because of the parallel plains, the right lines EF and BD, as also GH and KL, are parallel (15. 2 Sup.); therefore the triangles AOF and ACD, as also AOH and ACL, are equiangular, and therefore the ratios of CD to OF, and of CL to OH, are each of them equal to that of AC to AO (4. 6 and 16.5 Eu.), and therefore to each other (11. 5 Eu.) ; whence, CD and CL being, on account of the circular base of the cone, equal, OF and OH are equal (14. 5 Eu). In like manner it may be shewn, that any other two right lines, drawn from O to the section EGFH are equal; since therefore all right lines drawn from 0 to that section are equal, that section is the circumference of a circle whose centre is 0, and therefore in the axis AC of the cone. Cor. Hence it follows, that any diameter of such a section passes through the axis of the cone; and any right line, drawn in the plain of such a section, through the axis of the cone, is a diameter of the section. PROP. XXII. THEOR. If an oblique cone be cut by a plain passing by the axis and at right angles to the base, and be cut by another plain perpendicular to the former, and cutting off from the triangle, which is the section of the former with the cone, a triangle similar to that triangle, but placed subcontrarily, namely, so that the equal angles be at different right lines formed by the section of the first plain with the conical surface; the section of the conical surface by the latter plain, is the circumference of a circle, whereof the intersection of the two drawn plains is a diameter. Let ABC be a cone, A its vertex, BC its base, and the triangle ABC the section with the cone, of a plain passing by the axis and at right angles to the base (Cor. 20. 2 Sup.) ; let DEFG be the section with the cone, of a plain perpendicular to the plain ABC, and cutting off from the triangle ABC a triangle AFD similar to ABC, but placed subcontrarily, namely, so that the angle AFD be equal to the angle ABC; the section DEFG is a circle, of which the intersection DF of the plains ABC and DEF is a diameter. Let the right line HK be the intersection of the plain DEF with the plain of the base; from any point E in the sec tion DEFG, draw ERG parallel to HK, meeting DF in R, and through R, in the plain ABC, draw LM parallel to BC; the plain passing by ERG and LRM is parallel to the base BC (14.2 Sup.), and therefore its intersection with the surface of the cone is a circle whose diameter is LM (21. 2 and Cor. 21. 2 Sup.) and because both the base and the plain DEF are perpendicular to the plain ABC (Hyp.), their intersection HK is perpendicular to the same plain produced as necessary (10. 2 Sup.), and therefore ER, which is parallel to HK (Constr.), is perpendicular to the plain ABC (5. 2 Sup), and therefore to both the right lines LM and DF (Def. 3. 2 Sup.); whence, because of the circle LEM. the square of EK is equal to the rectangle LRM (3 and 35. 3 Eu.); but the angle MFR is equal to the angle ABC (Hup.), or its equal (29, 1 Eu.), DLR, and the angles FRM and DRL are equal (15, 1 Eu.), therefore the triangles FRM and DRL are equiangular (32. 1 Eu.), and FR is to RM, as LR to RD (4, 6 Fu.), and therefore the rectangle DRF is equal to the rectangle LRM (16. 6 Eu.), or to the square of ER. Therefore the point E is in the circumference of a circle described about the diameter DF in the plain DEF, for if RE met the circumference of a circle so described, in any other point on the part E of DF, the rectangle DRF would not be equal to the segment of RE between DF and the circle, contrary to 3 and 35. 3 Eu. In like manner any other point in the section DEFG may be proved to be in the circumference of a circle described about the diameter DF in the plain DEF, which section is therefore the circumference of a circle, whereof DF is a diameter. tion Scholium. A section of this kind is called, a subcontrary sec Cor. From this proposition, the preceding, and Def. 17. 2 Sup., it follows, that any circle formed by a plain parallel to the base of a cone, or by a subcontrary section, may be considered as the base of a cone having the same vertex, as the original one. PROP. XXIII. THEOR, If the intersection of a plain not parallel to the base of a cone, with the conical surface, be the circumference of a circle, the section is a subcontrary one, Let the intersection DEFG, see fig. to the preceding proposition, of a plain, not parallel to the base BC of a cone, with the conical surface, be the circumference of a circle; the section DEFG is a subcontrary one. Let the conical surface be cut by two plains parallel to the base, making the circles LEMG. and NOPQ, and intersecting the plain DEFG in the right lines EG and OQ, which are parallel (15. 2 Sup); draw diameters LM and NP, of the circles LEM and NOP, perpendicular to the right lines EG and OQ, and meeting them, in the points R and S; the right lines EG and O are bisected in R and S (3. 3 Eu.), and LM and NP are parallel, for if any other right line drawn through S in the plain NOP, except NP, were parallel to LM, the angle made by that parallel with SQ would not be equal to the angle LRG, contrary |