* 11. 1. * 3. 1. square of AC, is equal to the square of AD, that is of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D. PROP. IX. THEOR. If a straight line be divided into two equal, and also into two unequal parts, the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section. Let the straight line AB be divided into two equal parts at the point C, and into two unequal parts at the point D: the squares of AD, DB shall be together double of the squares of AC, CD. From the point C draw* CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; * 31. 1. through D draw* DF parallel to CE, and through F draw FG parallel to AB; and join AF. Then, because AC is equal to CE, the angle AEC is equal to the angle EAC; and because the angle ACE is a right angle, the two others AEC, EAC, together make one *32. 1. right angle; and they have been proved equal to one * 5. 1. * another; therefore each of them is half of a right angle. For the same 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 * 29. 1. angle, for it is equal to the in terior and opposite angle ECB; therefore the remaining angle EFG is half a right angle; therefore the angle GEF is equal to the angle EFG, and the side EG* to the side GF. Again, because the angle at B is half a right angle, and FDB a right angle, for it is 29. 1. equal to the interior and opposite angle ECB, therefore the remaining angle BFD is half a right angle: * 6. 1. therefore the angle at B is equal to the angle BFD, and the side DF to the side DB. And because AC 6. 1. 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 * 47. 1. the squares of AC, CE, because ACE is a right angle; therefore the square of EA is double of the square of AC. 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; * 47. 1. therefore the square of EF is double of the square GF; and GF is equal to CD; therefore the square of EF * 34. 1. 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 * 47. 1. AE, EF, because AEF has been proved to be 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: but DF is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. Therefore, if a straight line, &c. Q. E. D. PROP. X. THEOR. If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, 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 produced. 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. * 5. 1. From the point C draw* CE at right angles to AB, and make it equal to AC or CB, and join AE, EB; 31. 1. through E draw* EF parallel to AB, and through D draw DF parallel to CE. And because the straight line EF meets the parallels EC, FD, the angles CEF, 29. 1. 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 * 12 Ax. two right angles, will meet* if produced far enough: therefore EB, FD will meet, if produced towards B, D; let them meet in G, and join AG: then, because AC is equal to CE, the angle AEC is equal to the angle EAC; and the angle ACE is a right angle; therefore * 32. 1. each of the angles CEA, EAC is half a right angle;* for the same reason, each of the angles CEB, EBC is half a right angle; therefore AEB is a right angle. 15. 1. 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 angle, and is therefore equal to the angle DBG; therefore the side BD is equal to the side DG. Again, because EGF is half a right angle, and that the angle at F is a right an* 34. i. gle, because it is equal to the opposite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; therefore 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: *47. 1. but the square of EA is equal to the squares EC, CA; therefore the square of EA is double of the square of AC. Again, because GF is equal to FE, the square * 29. 1. * 6. 1. * 6. 1. B F 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 EF, * 47. 1. FG; therefore the square of EG is double of the square of EF: but EF is equal* to CD; therefore the square * 34. 1. 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 * 47. 1 square of AG is double of the squares of AC, CD: but the squares of AD, DG are equal to the square of AG; * 47. 1. 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. 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. * 10. 1. 3. 1. Upon AB describe the square ABDC; bisect* AC* 46. 1. in E, and join BE; produce CA to F, making * EF equal to EB, and upon AF describe the square ⋆ 46. 1. FGHA; then AB shall be divided in H, so that the rectangle AB, BH is equal to the square of AH. Produce GH 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 G * 6. 2. 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 * * 47. 1. squares of BA, AE are equal to the JA C G K square of EB, because the angle EAB PROP. XII. THEOR. In obtuse-angled triangles, 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. Let ABC be an obtuse-angled triangle, having the |