Circles are to each other as the fquares of their diameters. Let ABCD, EFGH be two circles, and BD, FH their diameters: then will the square of BD be to the square of FH as the circle ABCD is to the circle EFGH. For, if they have not this ratio, the fquare of BD will be to the fquare of FH, as the circle ABCD is to some fpace either lefs or greater than the circle EFGH. First, let it be to a space ST lefs than the circle EFGH; and infcribe the two fimilar polygons AROPQ, EKLMN fo that the circle EFGH may exceed the latter by less than it exceeds the space ST (VIII. S.) Then, fince the circle EFGH exceeds the polygon EKLMN by lefs than it exceeds the space ST, the polygon EKLMN will be greater than the space ST. And, because fimilar polygons, infcribed in circles, are to each other as the squares of their diameters (VIII. 2.), the fquare of BD will be to the fquare of FH as the polygon AROPQ is to the polygon EKLMN, But the fquare of BD is also to the fquare of FH as the circle ABCD is to the space ST (by Conft.); whence the circle ABCD will be to the space ST, as the polygon AROPQ is to the polygon EKLMN. 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, lefs and greater at the fame time, which is impoffible; confequently the fquare of BD is not to the fquare of FH as the circle ABCD is to any space lefs than the circle EFGH. And, in the same manner, it may be demonstrated, that the fquare of FH is not the fquare of BD as the circle EFGH is to any space lefs than the circle ABCD. Nor, is the fquare of BD to the square of FH as the circle ABCD is to a space greater than the circle EFGH. For, if it be poffible, let it be so to the space sx, which is greater than the circle EFGH. Then, fince the fquare of BD is to the fquare of FH as the circle ABCD is to the fpace sx, therefore, alfo, inverfely, the fquare of FH is to the square of BD as the fpace 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 fquare of FH is, therefore, to the fquare of BD as the circle EFGH is to a space lefs than the circle ABCD (V..11.), which has been fhewn to be impoffible. Since, therefore, the fquare of BD is not to the fquare of FH as the circle ABCD is to any space either lefs or greater than the circle EFGH, the fquare of BD must be to the fquare of FH as the circle ABCD is to the circle Q. E.D. COR. 1. Circles are to each other as the fquares of their radii; these being half the diameters. EFGH. COR. 2. If the radii or diameters of three circles be refpectively equal to the three fides of a right angled triangle, that whofe radius or diameter is the hypothenufe will be equal to the other two taken together (II. 14.) Every circle is equal to the rectangle of its radius, and a right line equal to half its circumference. H Let k mps 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 k m p s be equal to the rectangle ov. For if it be not, it must be either greater or lefs. Let it be greater; and let the rectangle oz be equal to the circle kmps; and infcribe a polygon In r t in the circle kmps that fhall differ from it by less than the magnitude wz (VIII. 3.) Then fince 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 infcribed in the circle k mps (by (by Hyp.), and ok is greater than ox (I. 17.), the rect angle ov will also be greater than the polygon / nrt. 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; confequently the polygon is greater than the rectangle ov. It is, therefore, both greater and lefs at the fame time, which is abfurd; whence the circle k mp s is not greater than the rectangle ov. Again, let it be less than ov, by the rectangle wy and let BDF H be a polygon circumfcribed about the circle, that shall differ from it by lefs than the magnitude wy (VIII. 4.) Then fince 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 lefs than half the perimeter of any polygon that can be circumfcribed about the circle (by Hyp.), and ok is common, the rectangle ov will also be lefs than the polygon BDFH. But the polygon differs from the circle, or from oy, by lefs than the magnitude wy (by Hyp.), and ov differs from oy by wy; consequently the rectangle ov will be greater than the polygon BDFH, which is abfurd. Since, therefore, the rectangle ov is neither greater nor lefs than the circle k mps, it must be equal to it, as was to be shewn. 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 femicircumferences DAB, HEF, and on the radii OA, SE make the squares OK SL (II. 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 fquares of their radii (VIII. 5. Cor.) the rectangle ON will also be to the fquare OK as the rectangle SR is to the fquare SL (V. 9.) But ON is to OK as OM to OD (VI. 1.), and SR 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 SP as twice OD is to twice SH. But |