PROP. IX. THEOR. Fa ftraight line be divided into two equal, and alfo qual parts, are together double of the fquare of half the line, and of the fquare of the line between the points of fection. Let the straight line AB be divided at the point C into two equal, and at D into two unequal parts. the fquares of AD, DB are together double of the fquares of AC, CD. a Book II. From the point C draw CE at right angles to AB, and make it a. i. i: equal to AC or CB, and join EA, EB; thro' D draw b DF parallel b. 31. £. to CE, and thro' F draw FG parallel to AB; and join AF. then because AC is equal to CE, the angle EAC is equal to the angle e. s. t. AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle 4; and they are equal to d. 3i. ti one another; each of them therefore is half of a right angle. for the fame reafon each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a right angle. and because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the interior and oppo-A fite angle ECB, the remaining an gle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the fide EG equal f to the fide. GF. again, f. 6. 1. because the angle at B is half a right angle, and FDB a right an gle, for it is equal to the interior and opposite angle ÉCB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the fide DF to the fide DB. and because AC is equal to CE, the fquare of AC is equal to the fquare of CE; therefore the fquares of AC, CE are double of the fquare of AC. but the fquare of EA is equal to the fquares of AC, CE, be caufe ACE is a right angle; therefore the fquare of EA is double of the fquare of AC. again, because EG is equal to GF, the fquare of EG is equal to the fquare of GF; therefore the fquares of EG, GF are double of the iquare of GF; but the fquare of EF is e Book II. qual to the fquares of EG, GF; therefore the fquare of EF is double of the fquare GF. and GF is equal to CD; therefore the h. 34. 1. fquare of EF is double of the fquare of CD. but the fquare of of AC, CD. and the fquare of g. 47. 1. AF is equal to the fquares of AE, EF because AEF is a. right 2. 11. 1. E G A CD B angle; therefore the fquare of AF is double of the fquares of AC, CD. but the fquares of AD, DF are equal to the square of AF; because the angle ADF is a right angle; therefore the squares of AD, DF are double of the fquares of AC, CD. and DF is equal to DB; therefore the fquares of AD, DB are double of the squares of AC, CD. If therefore a straight line, &c. Q. E. D. Fa ftraight line be bifected, and produced to any point, the fquare of the whole line thus produced, and the fquare of the part of it produced are together double of the fquare of half the line bifected, and of the fquare of the line made up of the half and the part produced. Let the straight line AB be bifected in C, and produced to the point D; the fquares of AD, DB are double of the fquares of AC, CD. From the point C draw CE at right angles to AB, and make it b. 31. 1. equal to AC or CB, and join AE, EB; thro' E draw b EF parallel to AB, and thro' D draw DF parallel to CE. and because the straight line EF meets the parallels EC, FD, the angles CEF, EFD e. 19. 1. are equal to two right angles; and therefore the angles BEF, EFD are less than two right angles. but straight lines which with another ftraight line make the interior angles upon the fame fide lefs d. 11. Ax. than two right angles, do meet if produced far enough. therefore EB, FD fhall meet, if produced, toward BD. let them meet in G, and join AG. then because AC is equal to CE, the angle CEA is equal to the angle EAC; and the angle ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle *. for e. s. 1. • 32. 1. e the the fame reason, each of the angles CEB, EBC is half a right Book II. angle; therefore AEB is a right angle. and because EBC is half a c right angle, DBG is alfo f half a right angle, for they are vertical- f. 15. 1 ly oppofite; but BDG is a right angle, because it is equal to the c. 29. 1. alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore alfo the fide BD is equal to the fide DG. again, because EGF is g. 6. 1. cqual to the fide FE. And because EC is equal to CA, the square of EC is equal to the fquare of CA; therefore the fquares of EC, CA are double of the fquare of CA. but the square of EA is equal i i. 47. 1 to the fquares of EC, CA; therefore the square of EA is double of the fquare of AC. again, because GF is equal to FE, the square of GF is equal to the fquare of FE; and therefore the squares of GF, FE are double of the fquare of EF. but the fquare of EG is equal i to the fquares of GF, FE; therefore the square of EG is double of the fquare of EF. and EF is equal to CD, wherefore the fquare of EG is double of the fquare of CD. but it was demonstrated that the square of EA is double of the square of AC; therefore the squares of AE, EG are double of the fquares of AC, CD. and the fquare of AG is equal to the fquares of AE, EG; therefore the fquare i. 47. x. of AG is double of the fquares of AC, CD. but the fquares of AD, DG are equal to the square of AG; therefore the fquares of AD, DG are double of the squares of AC, CD. but DG is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. Wherefore if a straight line, &c. Q. E. D. L Book II, 46. I. the PROP. XI. PROB. O divide a given ftraight line into two parts, fo that the rectangle contained by the whole, and one of fhall be equal to the fquare of the other part. parts, a Let AB be the given ftraight line; it is required to divide it into two parts, fo that the rectangle contained by the whole, and one of the parts, fhall be equal to the fquare of the other part. Upon AB defcribe the fquare ABDC, bifect AC in E, and 10. 1. join BE; produce CA to F, and make EF equal to EB; and upon C. 3. 1. AF defcribe the fquare FGHA, and produce GH to K. AB is divided in H fo, that the rectangle AB, BH is equal to the square of AH. c Because the straight line AC is bifccted in E, and produced to the point F, the rectangle CF, FA, together with the fquare of 4. 6. 2. AE, is equal to the fquare of EF. but EF is equal EB; therefore the rectangle CF, FA, together with d the fquare of AE is equal to the fquare of EB. and the fquares of BA, AE are e F G H R e. 47. 1. equal to the fquare of EB, because the K D PROP. PROP. XII. THEOR. IN obtufe angled triangles, if a perpendicular be drawn from any of the acute angles to the oppofite fide produced, the fquare of the fide fubtending the obtufe angle, is greater than the squares of the fides containing the obtufe angle, by twice the rectangle contained by the fide upon which when produced the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtufe angle. a Book II. Let ABC be an obtufe angled triangle, having the obtufe angle ACB, and from the point A let AD be drawn perpendicular to BC a. 11. 1. produced. the fquare of AB is greater than the squares of AC, CB by twice the rectangle BC, CD. Because the straight line BD is divided into two parts in the point b C, the fquare of BD is equal to the fquares of BC, CD, and twice the rectangle BC, CD. to each of thefe equals add the fquare of DA; and the fquares of BD, DA are equal to the fquares of BC, CD, DA, and twice the rectangle BC, CD. but the fquare of BA is equal to the fquares of BD, DA, because the B angle at D is a right angle; and the fquare of CA is equal to the fquares of CD, DA. c fquare of BA is equal to the fquares of BC, CA, rectangle BC, CD; that is, the fquare of BA is greater than the fquares of BC, CA, by twice the rectangle BC, CD. Therefore in obtufe angled triangles, &c. Q. E. D. |