= EG, and therefore HK = HF – EG = DG + AF (DH - AE) = AF + AE - (DH-DG); whence HK and the rectangle HF, FK are given, and consequently (VI. 20. El.) the point F is given. If DF'E' intersect the straight lines AB and AC on the other side of their vertex A, the triangles EDG and DFH are still similar, and EG: DG :: DH: HF'; wherefore E'G.HF', being equal to DG.DH, is given. Make F'K' = E'G, and thence HK' = EG — HF = AE' + DH (DG - AF) = AF + AE' + (DH — DG); consequently HK' and the rectangle HF.F'K' are given, and therefore (VI. 20. El.) the point F is given. COMPOSITION. Make OP or OP' equal to the difference of the parallels DH and DG, from H place likewise towards opposite parts HK = PN and HK' = P'N, on HK and HK' describe semicircles, from H erect the perpendicular HI equal to DG, and, from K and K', the perpendiculars KL and K'L', each equal to DH, join IL and IL, and, at right angles to these, from the points of section M and M', draw MF and M'F'; the straight lines DEF and DF'E' will cut off segments from AB and AC, which are together equal to ON. For (VI. 20. El.) HF.FK = HI.KM = DG.DH; but DG.DH = HF.EG, and consequently HF.EG = HF.FK, or EG = FK. Wherefore HK = HF - EG = AF + AE - (DH - DG); and since HK = PN = ON (DH - DG), it follows that AF + AE = ON. In like manner, it is shown that E'G = F'K', and hence HK' = E'G HF = AF + AE' + (DH – DG); but HK' = PN' = ON + (DH-DG), and consequently AF + AE = ON. PROP. XVIII. PROB. From one of the corners of a given square, to draw a straight line, such that its portion, intercepted between the opposite sides of the figure, shall be equal to a given straight line. Let ABCD be a square, and from the point A let it be required to draw AEF, so that the part EF, intercepted between CD and BC, or their extension, may be equal to a given straight line. ANALYSIS. Draw FG perpendicular to AF, meeting AD produced in G, from G let fall the perpendicular GH upon BC produced, and join EG. The angle EFH is (I. 34. El.) equal to ECF and FEC, and it is also equal to EFG and GFH; consequently, ECF and EFG being right angles, the remaining E GH or CD,-are (I. 23. El.) equal, and therefore the side AE is equal to FG. But EFG and EDG being rightangled triangles, EF + FG = EG' = ED + DG', (II. 14. El.), or EF + AE = ED + DG*; but AE = AD + ED, and hence EF + AD + ED = ED + DG', or EF + AD = DG. Wherefore, since EF and AD are both given, DG is also given, and consequently AG; but the right angle AFG being contained in a semicircle described upon AG, the point For F', its contact or intersection with BC, is given, and consequently the straight line AEF. COMPOSITION. Make Al equal to the given straight line, join DI, and, equal to this, produce AD to G, upon AG describe a semicircle meeting the extension of BC in F or F', and join AEF or AFE'; EF, the external part of that straight line, is equal to AI. For join EF, FG, EG, and let fall the perpendicular GH upon BF. It is evident that EF + FG* = ED + DG; and FG being equal to AE, EF + AE = ED + DG. But AE = AD + ED, and DG' = DI' = AD + AI; whence EF + AD + ED' = ED + AD + AI', and therefore EF = AI', and EF = AI. PROP. XIX. PROB. Given the base of a triangle, its altitude, and the rectangle under its two sides, to determine the triangle. ANALYSIS. About the triangle ABC describe (III. 11. cor. El.) a circle, and draw the diameter BF and the radii AE and CE. Because the given rectangle AB.BC is (VI. 22. El.) equal to BD.BF, this rectangle is likewise given; and since the perpendicular BD is given, the diameter BF, and therefore the radii AE, CE, are given. But the base AC being given, the triangle AEC is hence given, and consequently the centre E and the B B E A C circle ABCF are given. Again, because BD, the distance of the D F vertex of the triangle from its base, is given, that point must occur in the parallel BB', and, being thus placed in the contact or intersection of a given straight line with a given circle, is itself given. COMPOSITION. On AC construct (II. 11. El.) a rectangle equal to the given space, also form on AC the triangle AEC, having AE and CE each equal to half the side of that rectangle, from E with the radius EA describe a circle, on AC erect a perpendicular DB equal to the altitude of the triangle, and through B draw a parallel meeting the circumference in B or B'; ABC is the triangle required. For ABC has evidently the given altitude BD, and the rectangle AB.BC, being equal (VI. 22. El.) to BF.BD, is therefore equal to the given space. PROP. XX. PROB. Given the hypotenuse of a right-angled triangle, and the sum or difference of the base and perpendicular, to construct the triangle. ANALYSIS. In the base AB, or its production, make BD or BE equal to the perpendicular BC, and join CD or CE. The triangles CBD and CBE are right-angled and isosceles, and therefore the angles at D and E are each of them half a right angle. If AD, the sum of AB and BC, be given, the point D is given, and consequently the straight line DC, making a given angle with C AE, the difference between the DA, is given in position; or if base and perpendicular, be gi A E ven, the point E is given, and the straight line EC is given in position. But the hypotenuse AC being given, the point C must, therefore, occur in the contact or intersection of a circle described from A with that radius and the straight line CD or CE. Consequently C is given, the perpendicular CB, and thence the right-angled triangle ABC. COMPOSITION. Make AD or AE equal to the sum or difference of AB and BC, draw (I. 5. and 4. El.) DC or EC at an angle CDE or CED equal to half a right angle, from A with the radius AC describe a circle meeting DC or EC in the point C, and from C (I. 6. El.) let fall the perpendicular CB; ACB is the triangle required. For the right-angled triangles CBD and CBE are evidently isosceles, and therefore AD is equal to the sum, and AE to the difference, of AB and BC. |