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. a 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 D to ABC, but placed subcontra N rily, nainely, so that the angle P AFD be equal to the angle G ABC; the section DEFG is L circle, of which the intersec M tion DF of the plains ABC T and DEF is a diameter. c Let the right line H K be the intersection of the plain DEF 11 with the plain of the base ; from any point E in the section DEFG, draw EKG 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 ER 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 Eų.), therefore the triangles FRM and DRL are equiangular (32. 1 Eu.), and FR is to RM, as LR to RD (4. 6 Eu.), 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 Di 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 cirr:'oference of a circle, whereof DF is a diameter. Scholium. A section of this kind is called, a subcontrary section 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 plam, 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 Š; the right lines EG and Oare bisected in R and S (3. 3 bu.), and LN 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 to 12. 2 Sup.; and the diameters LM and NP pass by the axis of the cone (Cor. 21. 2 Sup.); therefore the plain ANLBCMP passes by the axis of the cone ; let the right line DF be the intersection of this plain with the plain of the circle DEFG; since therefore DF passes through the points R and S, and therefore bisects the parallels EG and OQ, it is a diameter of the circle DEFG, for if any other tight line, bisecting one of them not passing through the centre, were a diameter, it would bisect it perpendicularly (3. 3 Eu.), and would therefore, because of the parallels, be perpendicular to the other (29. 1 Eu.), and therefore would bisect that other (3. 3 Eu.), which would of course be bisected in two points, which is absurd ; and that DF is a diameter appears also from Cor. 2. 32. 1 Sup., and it is perpendicular to the right lines EG and OQ (3. 3 Eu.); since therefore EG is perpendicular to the right lines LRM and DRF, it is perpendicular to the plain ABC passing by them (2. 2 Sup.), and therefore the plains DEFG and LEMG and of course the plain of the base, are perpendicular to the plain ABC passing by the axis (9. 2 Sup.) ; but because of the circles LEM and DEF, each of the rectangles LRM and DRF are equal to the square of ER (3 and 35. 3 Eu.), and therefore to each other, therefore DR s to LR, as RM is to RF (16.6 Eu.), and the angles DRL and FRM are equal (15. 1 Eu.), therefore the triangles DRL and MRF are equiangular (6. 6 Eu.), and the angle MFR is equal to DLR, or its equal (29. 1 Eu.), ABC; therefore the section DEFG is a subcontrary section (Schol. 22. 2 Sup). Cor. Hence the section made by a plain with a conical surface, which is neither a subcontrary section, nor made by a plain, which is parallel to the base, is not the circumference of a circle. PROP. XXIV. THEOR. If a cone be cut by a plain neither passing through its vertex, nor parallel to its base, nor placed subcontrarily; the section is an ellipse, hyperbola or parabola ; the intersection of that plain with a plain passing by the axis of the cone, and a right line drawn through the centre of its base, perpendicular to the common section of that plain with the base of the cône, being a diameter of the section ; the section being an ellipse ; when this diameter meets the conical surface of the cone twice ; a hyperbola, when it meets the opposite conical surfaces; and a parabola, when being parallel to a right line drawn from the vertex of the cone to the circumference of its base, it meets only one of the conical surfaces, and but once. Let a cone ABC, see fig. 1, 2 and 3, be cut by a plain PEF, neither passing through its vertex, nor parallel to its base, nor placed subcontrarily ; let HK be the common section of this plain with the plain of the base, D the centre of the base, DG a rigt line, drawn through D perpendicular to HK, ABC the triangular section of the cone with the plain passing by the axis and DG, and EG the intersection of the plains PEF and ABC; the section PEF is an ellipse, hyperbola or parabola, of which EG is a diameter; the section being an ellipse, when the diameter EG meets the conical surface of the cone ABC twice, as in E and L in fig. 1 ; a hyperbola, when it meets the opposite conical surfaces, as in E and L in fig. 2; and a parabola, when the diameter EG, being parallel to a right line AC, drawn from the vertex A of the cone, to the circumference of its base, meets only one of the conical surfaces, as in E in fig. 3, and but once. Fig. 1. A Let first EG meet the conical surface of the cone ABC in E and L, and let there be taken in the section any point P, and draw PO parallel to HK meeting EL in 0, and thro' 0 draw MN parallel to BC; the E plain passing by MN and PO is R parallel to that passing by BC and s HK (14. 2 Sup.); therefore the sec- M I N P к Cor. 21. 2 Sup.), and because the D angle BCH is a right angle (Flyp.), B G MOP is a right angle (12. 2 Sup.), H therefore the square of PO is equal to the rectangle MON (3 and 35. 3 Eu). In like manner, if any other point T be taken in the section PEL, and TQ be drawn to EL parallel to HK or PO, and through Q, RS be drawn parallel to BC, it may be shewn, that the square of TQ is equal to the rectangle RQS; therefore the square of PO is to the square of TQ, as the rectangle EOL is to the rectangle EQL (Cor. 1.7. 5 Eu.) ; but, because of the equiangular triangles EMO and ERQ, LON and LQS, MO is to RQ, as EO is to EQ, and ON is to Qs, as OL to QL (4. 6 and 16. 5 Eu.), therefore the ratios of the rectangle MON to the rectangle RQS, and of the rectangle EOL to the rectangle EQL, being compounded of these equal ratios (23. 6 Eu.), are equal (22. 5 Eu.); therefore the square of PO is to the square of TQ, as the rectangle EOL is to the rectangle EQL (11.5 Eu). Let an ellipse be described with the diameter EL, its vertices being E and L, and the ordinate PO (85. 1 Sup.), and because the square of TQ is to the square of PO, as the rectangle EQL is the rectangle EOL, the joint T is in the ellipse (Cor. 3. 40. 1 Sup.): in like manner it may be shewn that any other point in the section PEL is in the ellipse ; and since the section, being neither a subcontrary one, nor parallel to the base of the cone (Hyp.), is not a circle (Cor. 23. 2 Sup.), it is an ellipse, of which EL is a diameter. Secondly. Let EG produced meet the opposite conical surface in L, let Fig. 2. L H and K be the points in which HK meets the circuniference of the base of the cone, from any point P in the section HEK, draw PO parallel to HK, meeting EG in 0, and through o draw MN parallel lo BC ; as be A fore, the plain by PO and MO is parallel to that by HG and BC, or to the base of the cone, and the angle POM equal to HGB, and therefore a right angle, therefore the square of PO is equal to the rectangle MON (3 and 35. 3 Ew.); in like manner it may be jo М. Ni proved, that the square of HG is equal P1 to the rectangle BGC; therefore the square of PO is to the square of HG, as the rectangle MON is to the rect IK angle BGC Cor. 1. 7. 5 Eu.) ; but, BE because of the cquiangular triangles |