the square of EA is equal (1. 47.) to the square 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; therefore the square of EF is double of the square of GF; and GF is equal (1. 34.) to CD: therefore the square of EF is double of the square of cp: But the square of AE is likewise double of the square of ac: therefore the squares of A E, EF are double of the squares of AC, CD: And the square of AF is equal (1. 47.) to the squares of A E, EF, because A E F 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. 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 are double of the squares of AC, CD. From the point o draw (1. 11.) CE at right angles to AB: And make it equal to ac or CB, and join A E, EB; through E draw (1. 31.) 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, EFD are equal (I. 29.) 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, do meet (Ax. 12.) if produced far enough: Therefore EB, FD shall meet, if produced towards B, D: Let them meet in G, and join AG: Then, because Ac is equal to CE, the angle cea is equal (1. 5.) to the B G angle EAC; and the angle ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle (1. 32.): For the same reason, each of the angles CEB, EBC is half a right angle ; therefore A EB is a right angle: And because EBC is half a right angle, DBG is also (1.15.) half a right angle, for they are vertically opposite ; but BDG is a right angle, because it is equal (I. 29.) to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also the side BD is equal (1. 6.) to the side DG: Again, because EGF is half a right angle, and that the E F angle at F is a right angle, because it is equal (1. 34.) to the opposite angle ECD, the remaining angle FEG is half a right angle, and A D equal to the angle EGF; wherefore also the side GF is equal (1.6.) 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 (1. 47.) 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 GF, FE are double of the square of EF: But the square of EG is equal (1. 47.) 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 A E, EG are double of the squares of AC, CD: And the square of ag is equal (1. 47.) to the squares of A E, EG: therefore the square of AG is double of the squares of AC, CD: But the squares of AD, DG are equal (1. 47.) 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. 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. Upon A B describe (1. 46.) the square ABDC; bisect (1. 10.) AC in E, and join BE; produce ca to F, and make (1. 3.) EF equal to E B, and upon A F describe (1. 46.) the square FGHA; AB is divided in H, so that the rectangle AB, Bh is equal to the square of AH. Produce GH to K: Because the straight line Ac is bisected in E, and produced to the point F, the rectangle CF, FA, together with the square of A E, is equal (11. 6.) to the square of EF: But EF is equal to EB; therefore the rectangle CF, FA, together with the square of A E, is equal to the square of EB: And the squares of BA, A E are equal F G (1. 47.) to the square of EB, because the angle EAB is a right angle; therefore the rectangle CF, FA together with the square А of a E, is equal to the squares of BA, AE: Take away the square of AE, which is common to both, therefore the remaining E rectangle CF, FA is equal to the square of AB: and the figure rk is the rectangle contained by CF, FA, for AF is equal to FG; С K 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: And he is the rectangle contained by AB, BH, for AB is equal to BD; and Fh is the square of AH. Therefore the rectangle A B, Bh is equal to the square of Ah: Wherefore the straight line A B is divided in 1, so that the rectangle A B, BH is equal to the square of AH. Which was to be done. H Н B PROP. XII. THEOR. In obtuse angled triangles, if a perpendicular be drawn from any 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 obtuse angle ACB, and from the point A let AD be drawn (1. 12.) perpendicular to Bc produced : the square of AB is greater than the squares of AC, CB by twice the rectangle BC, CD. Because the straight line Bp is divided into two parts in the point c, the square of BD is equal (11. 4.) to the squares of BC, CD, and twice the rectangle BC, CD: To each of these equals add the square of DA; and the squares of DB, DA are equal to the squares of BC, CD, DA, and twice the rectangle BC, CD: But the square of Ba is equal (1. 47.) to the squares of BD, DA, because the angle at D is a right angle; and the square of ca is equal (1. 47.) to the squares of CD, DA: Therefore the square of Ba is 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. А B C PROP. XIII. THEOR. In every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides con. taining that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon bc, one of the sides containing it, let fall the perpendicular (1. 12.) AD from the opposite angle: A The square of Ac opposite to the angle B, is less than the squares of CB, BA by twice the rectangle CB, BD. First, let ad fall within the triangle ABC; and because the straight line cb is divided into two A parts in the point D, the squares of CB, BD are equal (11. 7.) to twice the rectangle contained by CB, BD, and the square of Dc: To each of these equals add the square of Ad; therefore the B В. squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of AD, DC: But the square of AB is equal (1. 47.) to the squares of BD, DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC: Therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD, that is, the square of Ac alone is less than the squares of CB, Ba by twice the rectangle CB, BD. Secondly, let AD fall without the triangle ABC: Then, because the angle at D is a right angle, the angle ACB is greater (I. 16.) than a right angle; and therefore the square of AB is equal (11. 12.) to the squares of AC, CB, and twice the rectangle BC, CD: To these equals add the square of BC, and the squares of AB, BC are equal to the square of Ac, and twice the square of BC, and twice the rectangle BC, CD: But because BD is divided into two parts in C, the rectangle DB, BC is equal (11. 3.) to the rectangle BC, CD and the square of BC: And the doubles of these are equal: Therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: Therefore the square of ac alone is less than the squares of AB, Bo by twice A the rectangle DB, BC. Lastly, let the side ac be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB, BC are equal (1. 47.) to the square of AC, and twice the square of BC:. Therefore in every triangle, &c. Q.E.D. B С |