.. BH: BI :: AG BE: GB× CE. But BH: BI :: .. AF: FC :: X AF: FC, AG × BE: GB× 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 B drawn from the angles to the opposite sides; then will AD× DB x DC: ED × DF× DG :: AB x BC x CA: AF x CE x BG. K G N M D/ H 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, .. AC × AB × BC : CE × BG × AF :: BI× AH × CK : [EL 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, X .. DA × DB × DC: DE × DF × DG :: BI × AH × CK: EL × GM × FN :: AB × BC × CA : AF × CE ×: 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 O the centre of the centre, 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. (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. D is 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 = B 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 B E. From E let fall the perpendicular EF; F is the point required. For (Eucl. vi. 8. Cor.) AF : FE :: FE : FD, .. 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. Let A and B be the given points, and CD the given line. Join BA, and produce it to meet CD in D. Take DC a mean proportional between DA and DB. Cis the point required. Join AC, BC; and about the triangle ABC describe a circle; DC is a tangent at the point C (Eucl. iii. 37.), and .. the angle is the greatest (ii. 62.). (5.) To determine the position of a point, at which lines drawn from three given points, shall make with each other angles equal to given angles. A D B Let A, B, C be the three given points; join AB, and on it describe a segment of a circle containing an angle equal to that which the lines from A and B are to include. Complete the circle, and make the angle ABD equal to that which the lines from A and C are to include. Join DC, and produce it to the circumference in E. E is the point required. Join AE, BE. Then the angle AEC=ABD, and AEB is of the given magnitude, by construction. (6.) To divide a straight line into two parts such, that the rectangle contained by them may be equal to the square of their difference. Let AB be the given line; upon it describe a semicircle ADB From B draw BC at right angles and equal to AB. Take O the centre, and join OC; and from D draw DE perpendicular to AB; AB is divided in the point E, as was required. Since BC is double of BO, DE is double of OE (Eucl. vi. 2.), and OE OE B being half the difference between AE and EB, DE is equal to the difference. Also (Eucl. vi. 13.) the rectangle AE, EB is equal to the square of DE. (7.) If a straight line be divided into any two parts; to produce it, so that the rectangle contained by the whole line so produced, and the part produced may be equal to the rectangle contained by the given line and one segment. D B Let AB be the given line divided into two parts in the point C. On AB as a diameter describe a circle AD B. From B draw BE at right angles to AB, and .. a tangent to the circle; and make BE a mean proportional between AB and AC. Take O the centre; join EO, and produce it to F. Produce AB to G, making BG equal to ED. Then will the rectangle AG, GB be equal to the rectangle BA, AC. Since DE=BG, the rectangle BG, GA is equal to the rectangle DE, EF, i. e. to the square of EB, or to the rectangle AB, AC, by construction. COR. 1. If it be required to produce the line, so that the rectangle contained by the whole line produced and the part produced, may be equal to the rectangle con |