necessary) the angle DFE equal to ABC, and draw any line GH parallel to FD, cutting ED in I; the rectangle EI, ID is equal to the rectangle GI, IH. For the angle AGH=AFD=ABC =ADE, and the vertical angles at I are equal, .. the triangles GEI, HID are equiangular; and HI: ID :: IE : IG, .. the rectangle EI, ID is equal to the rectangle HI, IG. (47.) If the sides, or sides produced, of a triangle be cut by any line; the solids formed by the segments which have not a common extremity are equal. Let ABC be a triangle having the sides (produced if necessary) cut by the line DEF; then AF × CD × BE = AE × DB× CF. F D GE EG Draw BG parallel to AC; the triangles AEF, BEG will be similar, as also CDF, BDG; :. AF: AE :: BG : BE, and CD CF :: BD: BG, :. AF× CD : AE× CF :: BD : BE, (48.) If through any point within a triangle, three lines be drawn parallel to the sides; the solids formed by the alternate segments of these lines are equal. Through any point D within the triangle ABC, let HG, EF, IK, be drawn parallel to the sides; then ID x DGx DF-ED× DK× DH. Since the lines are drawn parallel to the sides, the triangles IED, GDK, HDF are similar to ABC, and to one another; .. ID: DE :: AC: CB GD: DK :: AB AC DF: DH:: BC: AB, whence ID x DG × DF: DE× DK x DH :: ACX BC× AC× AB, ABX BC i. e. in a ratio of equality. (49.) If through any point within a triangle lines be drawn from the angles to cut the opposite sides; the segments of any one side will be to each other in the ratio compounded of the ratios of the segments of the other sides. B H D Through any point D within the triangle ABC, let lines AE, BF, CG be drawn from the angles to the opposite sides; the segments of any one A of them as AC, will be in the ratio compounded of the ratios AG GB, and BE EC. F Draw IBH parallel to AC, meeting AE and CG produced in H and I. Then the triangles GCA, GBI, and EAC, EBH, as also ADF, BDH, and FDC, IDB, are respectively equiangular, .. BH BI: AG BE: GBX CE. But BH: BI :: AF: FC, .. AF FC: AG BE: GBX CE. (50.) If from each of the angles of any triangle, a line be drawn through any point within the triangle, to the opposite side; the solid contained by the segments thereof, intercepted between the angles and the point, will have to the solid contained by the three remaining segments, the same ratio that the solid contained by the three sides of the triangle, has to either of the (equal) solids contained by the alternate segments of the sides. Let ABC be the given triangle, and through any point D within it, let AE, BF, CG be drawn from the angles to the opposite sides; then will ADX DBx DC: ED× DFX DG :: G N B M D H F E K Let fall the perpendiculars AH, BI, CK; EL, GM, FN. Since EL is parallel to BI, CB CE: BI EL, and GM being parallel to AH, BA: BG:: AH: GM, also FN and CK being parallel, AC: AF:: CK: FN, .. ACX ABX BC: CEx BGx AF:: BIx AHx CK: [ELx GMx FN. Again, since EL is perpendicular to DC, and CK to DK, the triangles DEL, DCK are equiangular, and.. DC: DE: CK: EL. In the same manner, DB: DG :: BI : GM, and DA DF: AH: FN, .. DA× DBX DC: DEX DFX DG :: BI× AH x GMx FN:: ABX BCX CA: AFX CE × CK: EL BG. SECT. V. (1.) A straight line of given length being drawn from the centre at right angles to the plane of a circle; to determine that point in it, which is equally distant from the upper end of the line and the circumference of the circle. From the centre of the circle, let OA be drawn at right angles to its plane; draw OB perpendicular to OA; join AB, and make the angle ABC equal to BAC. C is the point required. Since the angle ABC=CAB, .:. AC= CB. B (2.) To determine a point in a line given in position, to which lines drawn from two given points may have the greatest difference possible. Let A and B be the given points, and CD the line given in position. Let fall the perpendicular BC, and U produce it, so that CE may be equal to CB; join AE, and produce it to meet CD in D. Join BD. Dis the point required. For DE=DB; and... AE is equal to the difference between AD and DB. If then any other point F be taken, BF E F = EF; and the difference between AF and BF is equal to the difference between AF and EF, which is less than AE (iii. 1.). The same may be proved for every other point in CD. (3.) A straight line being divided in two given points; to determine a third point such that its distances from the extremities may be proportional to its distances from the given points. Let AB be the given line, divided in C and D. On AD and CB let semicircles be described intersecting in C FD E. From E let fall the perpendicular EF; F is the point required. FE FE: FD, and FE FB :: FC : FE, .. AF: FB :: FC: FD. (4.) In a straight line given in position, to determine a point, at which two straight lines, drawn from given points on the same side, will contain the greatest angle. |