PROPOSITION XI. Similar polygons inscribed in circles are to one another as the squares on their diameters, also the perimeters of the polygons are as the diameters of the circles. Let ACG, ZXV be similar polygons inscribed in the Os AG, ZV. Through A, Z draw the diameters AP, ZQ. Then polygon ACG polygon ZXV as square on AP square on ZQ. Also the perimeters of the polygons are as the diameters, ZY as AP : ZQ; for AB (VI. 4) .. perimeter of ACG perimeter of ZXV in same ratio. (V. II) PROPOSITION XII. Circles are to one another as the squares on their diameters, and their circumferences are as their diameters. K ӨӨС Let ABG, CFK be two circles; AB, CF their dia meters. Then shall Os ABG, CFK be to one another as the squares on AB, CF. For if not, if possible, suppose that square on AB : square on CF as ○ ABG : some space X less than o CFK. Inscribe in the ◇ CFK a polygon which differs from the CFK by a space less than the excess of CFK above the space X; (IV. 14, Cor. 2) so that the polygon CFK is > the space X. Now let a similar polygon to this be inscribed in ABG. (IV. 10) but polygon in © ABG is < © ABG, .. polygon in © CFK is < X; but it is also greater, which is impossible; (v. 9) .'. square on AB is not to square on CF as ABG : any space less than © CFK. Again, if possible, suppose that square on AB : square on CF as ABG some space T, greater than CFK; A B K F T .. square on CF: square on AB as T: and .. as CFK: some space less than which is impossible. Hence the ratio of the square on AB to the square on and is... the same as that of the ABG to the CFK. Also, The circumferences of circles are to one another as their diameters. For the perimeters of similar polygons inscribed in circles are as their diameters. (VI. II) Also a polygon may be inscribed in a whose perimeter shall differ from the Oce of the O by less than able magnitude. any assign(IV. 15, Cor. 2) Hence it may be shewn by a demonstration precisely similar to the foregoing, that the circumferences of circles are as their diameters. DEFINITION. A sector is the figure contained by an arc of a circle and the radii drawn to its extremities. PROPOSITION XIII. In equal circles, angles at the centre have the same ratio as the arcs on which they stand; Let AKG, BLH be angles at the centres K, L of equal circles. Then shall AKG have to BLH the same ratio as arc AG to arc BH. For if AKG be divided into any number whatever of equals AKM, MKN, &c., and is BLP, PLQ, &c. equal to them be successively cut off from BLH: then the arc AG will contain the same number of arcs each equal to AM as AKG does angles each equal to LAKM; also the arc BH will contain the same integral number of arcs each equal to AM as ▲ BLH does angles each equal to AKM; .. LAKG: L BLH as arc AG arc BH. (Def.) So also sector AKG : sector BLH as arc AG : arc BH. For by applying the sector KAM to the sectors KMN, LBP, &c. it may be shewn that they are equal to one another. The remainder of the proof is similar to the foregoing. PROBLEM (a). To divide a straight line in a given ratio. Let AB be a given straight line. It is required to divide AB in the ratio of P to Q. PQ A X B H Through A draw a straight line AGH, making any 4 with AB; cut off AG= P and GH = Q. Join HB, and through G draw GX || to HB. Then shall AB be divided in X in the ratio of P to Q. •. GX is || to HB; ;. AX : XB as AG: GH; (vI. 1) i. e. as P: Q. Hence AB has been divided in the required ratio. PROBLEM (6). To produce a given straight line to a point such that the whole line thus produced shall be to the part produced in a given ratio. Through A draw any straight line AHG making an with AB; Join HB, and through G draw GX || to HB meeting AB produced in X. Then AX: XB as P: Q. GX is to HB; .. AX: XB as AG: GH; (VI. I) |