PROPOSITION 54 (PROBLEM)
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1 ook I roposition RSITI 54 (R) iven two bounded straight lines perpendicular to each other, one of them being produced on the side of the right angle, to find on the straight line produced the section of a cone called hyperbola in the same plane with the straight lines, so that the straight line produced is a diameter of the section the point at the angle is the vertex, where whatever straight line is dropped from the section to the diameter, making an angle equal to a given angle, will equal in square the rectangle applied to the other straight line having as breadth the straight line cut off by the dropped straight line beginning with the vertex projecting beyond by a figure similar similarly situated to that contained by the original straight lines. et there be the two bounded straight lines perpendicular to each other, let be produced to ; it is required then to find in the plane through the lines, an hyperbola whose diameter will be the straight line vertex, upright side the straight line, where the straight lines dropped from the section to at the given angle will equal in square the rectangles applied to having as breadths the straight lines cut off by them from projecting beyond by a figure similar similarly situated to the rectangle,. irst let the given angle be a right angle, on let a plane be erected at right angles to the plane of reference, let the circle R be described in it about, so that the segment of the circle s diameter within the
2 100 ook I roposition 54 sector has to the segment of the diameter within the sector a ratio not greater than that of to, * let be bisected at, let the straight line be drawn perpendicular from to the straight line * utocius, commenting, adds: et there be two straight lines, let it be required to describe a circle on so that its diameter is cut by in such a way that the part of it on the side of has to the remainder a ratio not greater than that of to. ow let it be supposed that they have the same ratio, let be bisected at, through it let the straight line be drawn perpendicular to, let it be contrived that : :: :, let be bisected; then it is clear that if = =, the point will be the midpoint of, if > the midpoint will be below, if >, <, it will be above. nd now [assuming > ] let it be below as, with center radius let a circle be described; then it will have to pass either within or without the points. nd if it should pass through the points, what was enjoined would be done; but let it fall beyond the points, let the straight line, produced both ways, meet the circumference at, let,, be joined, let be drawn through parallel to, parallel to, let be joined; then these will also be parallel to because = = is perpendicular to. nd since the angle at is a right angle, are parallel to, the angle at is a right angle; then for the same reasons also the angle at. nd so the circle described on will pass through the points (ucl. III. 31). et the circle be described. nd since is parallel to, : :: :. Then likewise also : :: :. nd : :: :. nd alternately, : :: : :: :. nd likewise if the circle described on cuts, the same thing could be shown. (Tr.)
3 ook I roposition let it be produced to ; the straight line is a diameter (ucl. III. 1). If then : :: : we use point, but if not, let it be contrived (ucl. VI. 12) that : :: : with <, through let be drawn parallel to, let,, be joined, through let be drawn parallel to. Since then angle = angle, but angle = angle, angle = angle, also angle = angle ; also =. R et a cone be conceived whose vertex is the point whose base is the circle about diameter at right angles to triangle. Then the cone will be a right cone; for =. Then let the straight lines,, be produced, let the cone be cut by a plane parallel to the circle ; then the section will be a circle (I. 4). et it be the circle R; so will be the diameter of the circle (I. 4, end). nd let the straight line R be the common section of circle of the plane of reference; then R will be perpendicular to both of the straight lines ; for both of the circles are perpendicular to triangle, the plane of reference is perpendicular to triangle ; their common section, the straight line R, is perpendicular to triangle ; it makes right angles also with all
4 102 ook I roposition 54 R the straight lines touching it in the same plane. nd since a cone whose base is circle vertex, has been cut by a plane perpendicular to triangle, has also been cut by another plane, the plane of reference, in the straight line R perpendicular to the straight line, the common section of the plane of reference of triangle, that is the straight line, produced in the direction of, meets the straight line at, by things already shown before (I. 12) the section R will be an hyperbola whose vertex is the point, where the straight lines dropped ordinatewise to will be dropped at a right angle; for they are parallel to straight line R. nd since : :: :, : :: : :: rect., : sq., : :: rect., : sq.. nd rect., = rect., (ucl. III. 35); : :: rect., : sq.. ut rect., : sq. :: : comp. : ;
5 ook I roposition but : :: : :: :, : :: : ; : :: : comp. :, that is sq. : rect.,. Therefore : :: sq. : rect.,. nd the straight line is parallel to the straight line ; the straight line is the transverse side, the upright side; for these things have been shown in the twelfth theorem (I. 12). RSITI 55 (R) Then let the given angle not be a right angle, let there be the two given straight lines, let the given angle be equal to angle ; then it is required to describe an hyperbola whose diameter will be the straight line, upright side, where the ordinates will be dropped at angle. et the straight line be bisected at, let the semicircle be described on, let some straight line, parallel to, be drawn to the
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