The circle ABCD, therefore, being greater than the polygon AROPQ, which is contained in it, the space st will also be greater than the polygon EKLMN. It is, therefore, less and greater at the same time, which is impossible; consequently the square of BD is not to the square of Fh as the circle ABCD is to any space less than the circle EFGH. And, in the same manner, it may be demonstrated, that the square of th is not to the square of BD as the circle EFGH is to any space less than the circle ABCD. Nor, is the square of BD to the square of Fh as the circle ABCD is to a space greater than the circle EFGH. For, if it be posible, let it be fo to the space sx, which is greater than the circle EFGH. Then, since the square of bp is to the square of Fh as the circle ABCD is to the space sx, therefore, also, inversely, the square of FH is to the square of BD as the space sx is to the circle ABCD (V.7.) But the space sx is greater than the circle EFGH (by Hyp.); whence the space sx is to the circle ABCD as the circle EFGH is to some space less than the circle ABCD (V. 14.) 'The square of FH is, therefore, to the square of BD as the circle EFGH is to a space less than the circle ABCD (V. 11.), which has been shewn to be impoffible. Since, therefore, the square of BD is not to the square of FH as the circle Abcd is to any space either less or greater than the circle EFGH, the square of bd must be to the square of Fh as the circle ABCD is to the circle Q. E. D. Cor. 1. Circles are to each other as the squares of their radii ; these being half the diameters, EFGH. COR. 2. If the radii or diameters of three circles be respectively equal to the three sides of a right angled triangle, that whose radius or diameter is the hypothenuse will be equal to the other two taken together (II. 14.) PROP. VI. THEOREM. Every circle is equal to the rectangle of its radius, and a right line equal to half its circumference. Let kmps be a circle, and ov a rectangle contained under the radius ok and a right line ow equal to half the circumference; then will the circle kmps be equal to the rectangle ov. For if it be not, it must be either greater or less. Let it be greater ; and let the rectangle oz be equal to the circle kmps; and inscribe a polygon Inrt in the circle kmps that shall differ from it by less than the magnitude wz (VIII. 3.) Then since the triangle kot is equal to half a rectangle under the base kt and the perpendicular ox (I. 32.), the whole polygon will be equal to half a rectangle under its perimeter and the perpendicular ox. And because ow is greater than half the perimeter of any polygon that can be inscribed in the circle kmps (by (by Hyp.), and ok is greater than ox (I. 17.), the rectangle ov will also be greater than the polygon Inrt. But the polygon differs from the circle, or from the rectangle oz, by less than the magnitude wz (by Conft.), and ov differs from oz by wz; consequently the polygon is greater than the rectangle ov. It is, therefore, both greater and less at the same time, which is abfurd; whence the circle kmps is not greater than the rectangle ov. Again, let it be less than ov, by the rectangle wy; and let BDFH be a polygon circumscribed about the circle, that shall differ from it by less than the magnitude wy (VIII. 4.) Then since the triangle boa is equal to half a rectangle under the base BA and the perpendicular ok (I. 32.), the whole polygon will be equal to half a rectangle under its perimeter and perpendicular ok. And because ow is less than half the perimeter of any polygon that can be circumscribed about the circle (by Hyp.), and ok is common, the rectangle or will also be less than the polygon BDFH. But the polygon differs from the circle, or from oy, by less than the magnitude wy (by Hyp.), and ov differs from Oy by wy; consequently the rectangle oy will be greater than the polygon Bokh, which is absurd. Since, therefore, the rectangle ov is neither greater nor less than the circle k mps, it must be equal to it, as was to be thewn. The circumferences of circles are in proportion to each other as their diameters. 'Let ABCD, EFGH be any two circles, whose diameters are BD, FH; then will the circumference ABCD be to the circumference EFGH as the diameter BD is to the diameter FH. For let om, sp be two right lines equal to the semicircumferences DAB, HEF, and on the radii o A, se make the squares ok, SL (11. 1.), and complete the rectangles ON, SR: Then fince the rectangles ON, sr are equal to the circles ABCD, EFGH (VIII. 6.), and the circles are to each other as the squares of their radii (VIII. 5. Cor.) the rectangle on will also be to the fquare ok as the reétangle sr is to the square sl (V. 9.) But on is to ok as om to od (VI. 1.), and so to SL as sp to sh (VI. 1.); therefore, by equality, OM will be to od as sp is to sh (V. 11.) And because any equimultiples of four proportional quantities, are also proportional (V. 13.), twice om will be to twice od as twice sp is to twice sh; or, by alternation, twice om is to twice so as twice od is to twice SH. Q But But twice om and twice sp are equal to the circum. ferences ABCD, EFGH (by Const.); and twice oD and twice SH are equal to the diameters BD, FH; whence the circumference ABCD is to the circumference EFGH as the diameter bp is to the diameter FH. Q. E. D. PROP. VIII. THEOREM. If a prism be cut by a plane parallel to its base, the section will be equal and like the base. 1 Let AG be a prism, and KLMN a plane parallel to the base ABCD ;, then will KLMN be equal and like ABCD. For join the points NL, and DB: Then since KM, ac are parallel planes (by Hyp.), and the plane an cuts them, the section KN will be parallel to the section AD (VII. 12.) And since AK is also parallel to DN (VIII. Def. 3.), the figure An is a parallelogram; and consequently KN is equal to AD (I. 30.) In like manner it may also be shewn, that Kl is equal to AB, LM TO BC, and mn to cd. And since KN, KL in the plane KM, are parallel to AD, as in the plane Ac, the angle NKL will be equal to the angle DAB (VII. 7.) |