and the square of AF is equal (I. 47.) to the squares of AE, EF, because AEF is a right angle; therefore 14. 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 15. The squares of AD, DF, are double of the squares of AC, CD; and DF is equal to DB; therefore 16. The squares of AD, DB, are double of the squares of AC, CD. Therefore, if a straight line, &c. Q E.D. 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 C draw (I. 11.) CE at right angles to AB, and make (I. 3.) it equal to AC or CB, and join AE, EB; through E draw (I. 31.) EF parallel to AB, and through Ď 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 1. 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 2. 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 (I. 5.) to the angle EAC; and the angle ACE is a right angle; therefore (I. 32.) 3. Each of the angles CEA, EAC, is half a right angle. For the same reason, 4. Each of the angles CEB, EBC, is half a right angle; therefore 5. The whole AEB is a right angle. And because EBC is half a right angle, also (I. 15.) 6. DBG is half a right angle, for they are vertically opposite; but 7. BDG is a right angle, because it is equal (I. 29.) to the alternate angle DCE; therefore 8. The remaining angle DGB is half a right angle, and is therefore equal to the angle DBG; wherefore also (I. 6.) 9. 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 angle, because it is equal (I. 34.) to the opposite angle ECD, 10. The angle FEG is half a right angle, and equal to the angle EGF; wherefore also (I. 6.) 11. 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 12. The squares of EC, CA, are double of the square of CA; but the square of EA is equal (I. 47.) to the squares of EC, CA; therefore 13. 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 14. The squares of GF, FE, are double of the square of EF; but the square of EG is equal (I. 47.) to the squares of GF, FE; therefore The square of EG is double of the square of EF; 15. and EF is equal (I. 34.) to CD; wherefore 16. 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 17. The squares of AE, EG, are double of the squares of AC, CD; and the square of AG is equal (I. 47.) to the squares of AE, EG; therefore 18. The square of AG is double of the squares of AC, CD; but the squares of AD, DG, are equal (I. 47.) to the square of AG; therefore 19. The squares of AD, DG, are double of the squares of AC, CD; but DG is equal to DB; therefore 20. AC, CD. The squares of AD, DB, are double of the squares of Wherefore, if a straight line, &c. Q.E.D. PROP. XI.-PROBLEM. 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 AB describe (I. 46.) the square ABDC; bisect (I. 10.) AC in E, and join BE; produce CA to F, and make (I. 3.) EF equal to EB; and upon AF describe (I. 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. Then because the straight line 40 is bisected in E, and produced to the point F, (II. 6.) 1. The rectangle CF, FA, together with the square of AE, is equal to the square of EF; but EF is equal to EB; therefore 2. The rectangle CF, FA, together with the square of AE, is equal to the square of EB; and the squares of BA, AE, are equal (I. 47.) to the square of EB, because the angle EAB is a right angle; therefore 3. 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 and the figure FK is the rectangle contained by CF, FA, for AF is equal (Def. 30.) to FG; and AD is the square of AB; therefore 5. The figure FK is equal to AD. Take away 6. the common part AK, and The remainder FH is equal to the remainder HD; and HD is the rectangle contained by AB, BH, for AB is equal to BD; and FH is the square of AH; therefore 7. 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.-THEOREM. 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 obtuse angle ACB; and from the point A let AD be drawn (I. 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 BD is divided into two parts in the point C, (II. 4.) 1. 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; and 2. 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 (I. 47.) to the squares of BD, DA, because the angle at D is a right angle; and the square of CA is equal (I. 47.) to the squares of CD, DA; therefore 3. 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. PROP. XIII.—THEOREM. In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing 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 (I. 12.) the perpendicular AD from the opposite angle: the square of AC opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB, BD. B D First, let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D, (II. 7.) 1. The squares of CB, BD, are equal to twice the rectangle contained by CB, BD, and the square of DC; to each of these equals add the square of AD; therefore 2. The 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 (I. 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 3. 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, (I. 16.) and therefore (II. 12.) 2. The square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD; |