(Cor. 4. 2.) to BK, that is, to CG; and CB equal to GK, that is, to GP; therefore CG is equal to GP; and because CG is equal to GP, and PR to RO, the rectangle AG is equal to MP, and PL to RF: but MP is equal (43. 1.) to PL, because they are the complements of the parallelogram ML; wherefore AG is equal also to RF. Therefore the four rectangles AG, MP, PL, RF, are equal to one another, and so AG+MP+PL+RF =4AG. And it was demonstrated, that CK+BN+GR+RN=4CK; wherefore, adding equals to equals, the whole gnomon AOH=4AK. Now AK-AB.BK AB.BC, and 4AK=4AB.BC; therefore, gnomon AOH=4AB.BC; and adding XH, or (Cor. 4. 2.) AC2, to both, gnomon AOH+XH=4AB.BC+AC2. But AOH+XH=AF= AD2; therefore AD2-4AB.BC+AC2. "COR. 1. Hence, because AD is the sum, and AC the difference of "the lines AB and BC, four times the rectangle contained by any two "lines, together with the square of their difference, is equal to the square "of the sum of the lines." "COR. 2. From the demonstration it is manifest, that since the square "of CD is quadruple of the square of CB, the square of any line is quadruple of the square of half that line." SCHOLIUM. In this proposition, let the line AB be denoted by a, and the parts AC and CB by c and b; then AD=c+26. Now, since a=b+c, multiplying both members by 4b, we shall have 4ab=4b2+4bc; and adding c2 to each member of this equality, we shall have, 4ab+c2=c2+4bc+462, or 4ab+c2=(c+2b)2. PROP. IX. THEOR. If a straight line be divided into two equal, and also into tun 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 at the point C into two equal, and at D into two unequal parts; The squares of AD, DB are together double of the squares AC, CD. E F From the point C draw (Prop. 11.1.) CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; through D draw (Prop. 31. 1.) DF parallel to CE, and through F draw FG parallel to AB; and join AF. Then, because AC is equal to CE, the angle EAC is equal (5. 1.) to the angle AEC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle (Cor. 4. 32. 1.); and they are equal to one another; each of them therefore is half of a right angle. For the same reason each A C D B 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 (29. 1.) 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 (6. 1.) to the side GF; Again, because the angle at B is half a right angle, and FDB a right angle, for it is equal (29. 1.) 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 (6. 1.) the side DB. Now, because AC=CE, AC2-CE2, and AC2+CE2-2ÁC2. But (47. 1.) AE2= AC2+CE2; therefore AE2=2AC2. Again, because EG=GF, EG2=GF2, and EG2+GF2-2GF2. But EF2-EG2+GF2; therefore, EF2=2GF2 =2CD2, because (34. 1.) CD=GF. And it was shown that AE2=2AC2; therefore AE2+EF2=2AC2+2CD3. But (47. 1.) AF2=AE2+ EF2, and AD2+DF2=AF2, or AD2+DB2=AF2; therefore, also, AD2+DB2= 2AC2+2CD2. SCHOLIUM. This property is evident from the algebraical expression, (a+b)2+(ab)2=2a2+262; where a denotes AC, and b denotes CD; hence, a+b=AD, a—b=DB. 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 C draw (Prop. 11.1.) CE at right angles to AB, and make it equal to AC or CB; join AE, EB; through E draw (Prop. 31. 1.) 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 (29. 1.) 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 (29. 1.), 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 CEA is equal (5. 1.) to the angle EAC; and the angle E F right angle, DBG is also (15. 1.) half a right angle, for they are vertically opposite but BDG is a right angle, because it is equal (29. 1.) 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 DB is equal (6. 1.) to the side DG. Again, because EGF is half a right angle, and the angle at F a right angle, being equal (34. 1.) to the opposite angle ECD, the remaining angle FEG is half a right angle, and equal to the angle EGF; wherefore also the side GF is equal (6. 1.) to the side FE. And because EC=CA, EC2 + CA2 : 2CA2. Now AE2 (47. 1.) AC2+ CE2; therefore, AE2-2AC2. Again, because EF FG, EF2=FG2, and EF2+FG2=2EF2. But EG2= (47. 1.) EF2+FG2; therefore EG2-2EF2; and since EF=CD, EG2-2CD2. And it was demonstrated, that AE2=2AC2; therefore, AE2+ EG2=2AC2 +2CD2. Now, AG2=AE2+ EG2, wherefore AG2-2AC2+2CD2. But AG2 (47. 1.)=AD2+DG2=AD2+DB2, because DG=DB: Therefore, AD2+DB2 2AC2+2CD2. SCHOLIUM. Let AC be denoted by a, and BD, the part produced, by b; then AD= 2a+b, and CD=a+b. Now, (2a+b)2+b2=4a2+4ab+263; but 4a2+4ab+2b2=2a2+2 (a+ b)2; hence, (2a+b)2+b2=2a2+2(a+b)2, and the proposition is evident from this algebraical equality. 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, may 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 F the whole, and one of the parts, shall be equal to the square of the other part. Upon AB describe (46. 1.) the square ABDC; bisect (10. 1.) AC in E, and join BE; produce CA to F, and make (3. 1.) EF equal to EB, and upon AF describe (46. 1.) the square FGHA, AB is divided in H, so that the rectangle AB, BH is equal to the square of AH. E 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 AE, is equal (6. 2.) to the square of EF: But EF is equal to EB; therefore the rectangle CF. FA, together with the square of AE, is C G H B equal to the square of EB; And the squares of BA, AE are equal (47. 1.) 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. Now the figure FK is the rectangle CF.FA, for AF is equal 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 AB.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. 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 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 AĎ be drawn (12. 1.) 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, BD2=(4. 2.) BC2+CD2+2BC.CD; add AD2 to both: Then BD2+AD2 BC2+ CD2+ AD2+ 2BC.CD. But AB2-BD2+AD2 (47. 1.), and AC2 CD2+ AD2 (47. 1.); therefore, AB2=BC2+AC2+2BC.CD; that is, AB2 is greater than BC2+AC2 by 2BC.CD. B C D 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 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 the perpendicular (12. 1.) 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. First, let AD fall within the triangle ABC; and because the straight line CB is divided into two parts in the point D (7. 2.), BC2+ BD2=2BC. BD+CD2. Add to each AD2; then BC2+ BD2+AD2=2BC.BD+CD2+ AD2. But BD2+AD2-AB2, and CD2+ DA2 AC2 (47. 1.); therefore BC2+AB2=2BC.BD+AC2; that is, AC2 is less than BC2+AB2 by 2BC.BD. B A D Secondly, let AD fall without the triangle ABC:* Then because the angle at D is a right angle, the angle ACB is greater (16. 1.) than a right angle, and AB2= (12. 2.) AC2+ BC2+2BC.CD. Add BC2 to each; then AB2+BC2= AC2+2BC2+2BC.CD. But because BD is divided into two parts in C, BC2+ BC.CD (3. 2.) BC.BD, and 2BC2+2BC.CD =2BC.BD: therefore AB2+ BC2=AC2+ 2BC.BD; and AC2 is less than AB2+BC2, by 2BD.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 (47. 1.) AB2+BC2= AC2+2BC2 AC2+2BC.BC. A B C PROP. XIV. PROB. To describe a square that shall be equal to a given rectilineal figure. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. Describe (45. 1.) the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them, BE to F, and make EF equal to ED, and bisect BF in G; and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF, together with the square of EG, is equal (5. 2.) to the square of GF: but GF is equal to GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: But the squares * See figure of the last Proposition. of |