PROP. XXVII. THEOR. Book VI. O F all the rectangles contained by the fegments of a given ftraight line, the greatest is the fquare which is defcribed on half the line. Let AB be a given straight line, which is bifected in C; and let D be any point in it, the fquare on AC is greater than the rectangle AD, DĚ. A C D B N For, fince the ftraight line AB is divided into two equal parts in C, and into two unequal parts in D, the rectangle contained by AD and DB, together with the fquare of CD, is equal to the fquare of AC. The fquare of AC is there- a 5. 2. fore greater than the rectangle AD, DB. Therefore, &c. Q. E. D. PROP. XXVIII. PROB. O divide a given ftraight line, fo that the rectangle contained by its fegments may be equal to a given space; but that fpace muft not be greater than the fquare of half the given line. Let AB be the given ftraight line, and let the fquare upon the given straight line C be the space to which, the rectangle contained by the fegments of AB muít be equal, and this fquare, by the determination, is not greater than that upon half the ftraight line AB. Book VI. a 5. 2. E C Bifect AB in D, and if the fquare upon AD be equal to the fquare upon C, the thing required is done: But if it be not equal to it, AD must be greater than C, according to the determination: Draw DE at right angles to AB, and make it equal to C; produce ED to F, fo that EF be equal to AD or DB, and from the centre E, at the distance EF, defcribe a circle meeting AB in G. Join EG; and because AB is divided equally in D, A. GB F and unequally in G, AG.GB+DG2 a DB2=EG2. But b b 47. I. ED2+DG2=EG2; therefore AG GB + DG2=ED2+ DG2, and taking away DG, AG.GB=ED2. Now EDC, therefore the rectangle AG.GB is equal to the square of C and the given line AB is divided in G, so that the rectangle contained by the fegments AG, GB is equal to the fquare upon the given ftraight line C. Which was to be : done. T PROP. XXIX. PROB. produce a given ftraight line, fo that the rectangle contained by the fegments between the extremities of the given line, and the point to which it is produced, may be equal to a given fpace. Let AB be the given ftraight line, and let the fquare upon the given straight line C be the space to which the rectangle under the fegments of AB produced, muft be equal. Bifect AB in D, and draw BE at right angles to it, so that BE be equal to C; and having joined DE, from the centre D at the distance DE defcribe a circle meeting AB produced in G G. And because AB is bifected in D, and produced to G, a AG.GB+DBDG DE2. But b DE DB2+BE2, therefore AG.GB+ DB2 DB2+ BE2, and AG.GB=BE2. Now, BE C; wherefore the ftraight line AB is produced to G, so that the rectangle contained by the fegments AG, GB, of the line produced, is equal to the fquare of C. Which was to be done. 189 Book VI. a 6. 2. b 47. I. PROP. XXX. PROB. O cut a given ftraight line in extreme and mean Let AB be the given straight line; it is required to cut it in extreme and mean ratio. Upon AB defcribe a the fquare BC, and produce CA to D, a 46. 1. fo that the rectangle CD.DA may be equal to the fquare D CB. Take AE equal to AD, and complete the rectangle b 29. 6. DF under DC and AE, or under DC and DA. Then, because the rectangle CD.DA is equal to the fquare CB, the rectangle DF is equal to CB. Take away the common part CE from each, and the remainder FB is equal to the remain. A. der DE. But FB is the rectangle contained by FE and EB, that is, by AB and BE; and DE is the fquare upon AE; therefore AE is a mean proportional between AB and BE, E B C 17.6. or AB is to AE as AE to EB. C -F But AB is greater than AE; wherefore AE is greater than EBe: Therefore the ftraight e 14. 5. Book VI. line AB is cut in extreme and mean ratio in Ef. Which f3. def. 6. was to be done. h 17.6. Otherwife. Let AB be the given ftraight line; it is required to cut it in extreme and mean ratio. Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the fquare of C B g 11 2. AC: Then, becaufe the rectangle AB.BC is equal to the fquare of AC, as BA to AC, fo is AC to CB h: Therefore AB is cut in extreme and mean ratio in C. Which was to be done. a 8. 6. PROP. XXXI. THEOR. Icribed upon the fide oppofite to the right angle, is equal to the fimilar, and fimilarly defcribed fi- Let ABC be a right angled triangle, having the right angle BAC: The rectilineal figure described upon BC is equal to the fimilar, and fimilarly defcribed figures upon BA, AC. Draw the perpendicular AD; therefore, because in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the bafe BC, the triangles ABD, ADC are fimilar to the whole triangle ABC, and to one another, and because the triangle ABC is fimilar to ADB, as CB to BA, fo is BA to BD b; and because these three ftraight lines are proportionals, as the firft to the third, fo is the figure upon the first to the fimilar, and fimilarly defcribed figure upon the c2. Cor. fecond c; Therefore, as CB to BD, fo is the figure upon b 4. 6. 20. 6. CB figures, upon BA and on AC, together, to the figure upon e 24. 5. BC; therefore the figures on BA, and on AC, are together equal to that on BC; and they are fimilar figures. Wherefore, in right angled triangles, &c. Q. E. D. IF F two triangles which have two fides of the one proportional to two fides of the other, be joined at one angle, fo as to have their homologous fides. parallel to one another; the remaining fides fhall be in a ftraight line. Let ABC, DCE be two triangles which have two fides BA, AC proportional to the two ČD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE; BC and CE are in a straight line. Because AB is parallel to DC, and the ftraight line AC meets them, the alternate angles BAC, ACD are equal; for the same reason, the angle CDE is equal to the angle ACD; wherefore alfo BAC is equal to CDE: And because the triangles ABC, DCE have one angle at A equal to one at D, and the fides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangularb to DCE: Therefore the angle ABC is equal to the angle DCE: And the angle BAC was pro A B D C 29. I. b 6. 6. ved |