is half of a right angle. For a like reason 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 #29. 1. #6. 1. #29. 1. #6. 1. #47. 1. C D equal to the interior and opposite angle ECB, the remaining angle EFG is half a right angle: therefore the angle GEF is equal to the angle EFG, and the side EG equal to the side GF. Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal to the interior and opposite angle ECB, the remaining angle BFD is half a right angle: therefore the angle at B is equal to the angle BFD, and the side DF to the side DB. And because AC is equal to CE, the square of AC is equal to the square of CE; therefore the squares of AC, CE are double of the square of AC: but the square of AE is equal to the squares of AC, CE, because ACE is a right angle; therefore the square of AE is double of the square of AČ: again, because EG is equal to GF, the square of EG is equal to the square of GF; therefore the squares of EG, GF are double of the square of GF: but the square of EF is equal to the squares of EG, GF; therefore the square of EF is double of the square of GF; and GF is equal to CD; therefore the square of EF is double of the square of CD: but the square of AE is likewise double of the square of AC; therefore the squares of AE, EF are double of the squares of AC, CD: but the square of AF is equal to the squares of AE, EF, because AEF is a right angle; therefore the square of AF is double of the squares of AC, CD: but the squares 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 squares of AC, CD: and DF is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. If therefore a straight line, &c. Q. E. D. +47. 1. #34. 1. #47. 1. 1 PROP. X. THEOR. If a straight line be bisected, and prolonged to any point, the square of the whole line thus prolonged, and the square of the part of it prolonged, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part prolonged. Let the straight line AB be bisected in C and produced to the point D: the squares of AD, DB shall be double of the squares of AC, CD. *11. 1. +3. 1. #29. 1. From the point C draw* CE at right angles to AB, and maket it equal to AC or CB, and draw AE, EB; through E draw* EF parallel to AB, and through D draw DF parallel to CE. Then because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal* to two right angles; and therefore the angles BEF, EFD are less than two right angles: but straight lines which with another straight line make the interior angles upon the same side less than two right angles *12 Ax. will meet* if prolonged; therefore EB, FD will meet, if prolonged towards B, D: let them meet in. G, and join A, G. Then, because AC is equal to CE, the angle CEA 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 a like reason, each of the angles CEB, EBC, is half a right angle: therefore AEB is a right angle. And because EBC is half a right angle, DBG is also* half a right angle, for they are vertically opposite; but BDG is a right angle, because it is equal to the alternate angle DCE; therefore the remaining angle DGB is half a right *5. 1. *32. 1. #15. 1. #29. 1. #6. 1. #34. 1. #6. 1. #47. 1. #47. 1. angle, and is therefore E B angle ECD, the remaining angle FEG is half a right angle, and therefore equal to the angle EGF; wherefore also the side GF is equal to the side FE. And because EC is equal to CA, the square of EC is equal to the square of CA; therefore the squares of EC, CA are double of the square of CA: but the square of EA is equal to the squares of EC, CA; therefore the square of EA is double of the square of AC: again, because GF is equal to FE, the square of GF is equal to the square of FE; and therefore the squares of GF, FE are double of the square of EF: but the square of EG is equal to the squares of GF, FE; therefore the square of EG is double of the square of EF; and EF is equal to CD; wherefore the square of EG is double of the square 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 squares of AC, CD: but the square of AG is equal* to the squares of AE, EG; therefore the square of AG is double of the squares of AC, CD: but the squares of AD, GD are equal to the square of AG; therefore the squares 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. +34. 1. #47. 1. #47. 1. Q. E. D. PROP. XI. PROB. To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part. Let AB be the given straight line: it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part. *11. and 3. From the point A draw* AC perpendicular and equal to AB; bisect* AC in E, and join B, E; prolong CA to F, making* EF equal to EB, and from AB cut off AH equal to AF: AB shall be divided in H, so that the rectangle AB, BH is equal to the square of AH. *6. 2. #47. 1. F G A HB Complete the square* whose adjacent sides are *46. 1. BA, AC, and that whose adjacent sides are HA, AF, and prolong the side GH of the square AG to K: and because the straight line AC is bisected in E, and produced to the point F, the rectangle CF-FA together with the square of AE, is equal to the square of EF; but EF is equal to EB; therefore the rectangle CF-FA, together with the square of AE, is equal to the square of EB: but the squares of BA AE are equal to the square of EB, because the angle EAB is a right angle; therefore the rectangle CF. FA, together with the square of AE, is equal to the squares of BA, AE: take away the square of AE, which is common to both, therefore the remaining rectangle CF FA is equal to the square of AB: but the figure FK is the +30 Def. rectangle contained by CF, FA, for AF is equalt to FG; and AD is the square of AB; therefore FK is equal to AD: take away the common part AK, and the remainder FH is equal to the remainder HD: but HD is the rectangle contained by AB, +3 Ax. +1 Ax. 13 Ax. E C KD +26 Def. BH, for AB is equal to BD; and FH is the square of AH: therefore the rectangle AB⚫BH is equal to the square of AH. Wherefore the straight line AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. Which was to be done. PROP. XII. THEOR. In an obtuse angled triangle, if a perpendicular be drawn from either of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle, is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle. #12. 1. Let ABC be an obtuse angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn* perpendicular to BC produced: the square of AB shall be greater than the squares of AC, CB, by twice the rectangle BC.CD. *4. 2. +2 Ax. B Because the straight line BD is divided into two parts in the point C, the square of BD is equal to the squares of BC, CD, and twice the rectangle BC-CD: to each of these equals add the square of DA; therefore the squares of BD, DA are equal to the squares of BC, CD, DA, and twice the rectangle BC-CD: but the square of BA is equal to the squares of BD, DA, because the angle at D is a right angle; and the #47. 1. square of CA is equal to the squares of CD, DA; therefore the square of BA equal to the squares of BC, CA, and twice the rectangle BC-CD; that is, the square of BA is greater than the squares of BC, CA, by twice the rectangle BC-CD. Therefore, in obtuse angled triangles, &c. Q. E. D. #47. 1. |