" 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.” Otherwise : * Because AD is divided any how in C (4. 2.), AD2=AC? + CD2 “ +2CD.AC. But CD=2CB : and therefore CD =CB2 +BD2+ * 2CB.BD (4. 2.)=4CB2, and also 2CD.AC=4CB.AC ; therefore, "ADP=AC2 +4BC3 +4BC.AC. Now BC? +BC.AC=AB.BC(3.2.) " and therefore ADP=AC+4AB.BC. 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 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 of AC, CD. From the point C draw (11. 1.) CE at right angles to AB, and make it equal to AC or CB, and join EA, EB ; through D draw (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 (32. 1.); and they are equal to one another ; each of them therefore is half of a right ' angle. For the same reason each of E the angles CEB,EBC is half a right angle : and therefore the whole AEB is a right angle : And because the angle G GEF is half a right angle, and EGF a right angle, for it is equal (29. 1.). to the interior and opposite angle ECB, theremaining angle EFG is half a right A C D B 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=CEP, and AC: +CE=2AC2. But (47. 1.) AE =AC2+CE; therefore AE2=2ACH. Again, because ÈG=GF, EG SGF2, and EGP +GF2 =2GF2. But EF EGI+GF ; therefore, EF=2GF2=2CD2, because (34. 1.) CD=GF. And it was shown that AES=2AC2; therefore AE EF2 =2AC: +2CD2, But (47. 1.) AF%=AE: + EF, and AD2 +DF2 ==AF2, or ADP +DB2 =AF2; therefore, also Otherwise ; taking away the equal rectangles 2BC.CD and 2AC. CD, there re. " mains AD2 +DB2 =2AC: +2CD2." 2 PROP. X. THEOR. the whole line thus produced, and the square of the part of it produ- Let the straight line AB be bisected in C, and produced to the point From the poiot C draw (11. 1.) CE at right angles to AB, and make it equal to AC or CB; join AE, EB ; through E draw (31. 1.) EF parall i to Av. ad through D draw DF parallel to CE. And because the straight line EF meets the parallels EC, FD, the angles CEFEFD 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, (Cor. 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 ACE is a right angle ; therefore each of the angles CEA, EAC is half a right angle (32. 1.): For the same reason, each of the angles CEB, EBC is balf a right angle; therefore AEB is a right angle : And because EBC is half a right angle, DBG is also (10. 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 T B because EF=FG, EF2=FG2, and EF2 +FG2=2EF. But EGP = (47. 1.) EF: _FG2 ; therefore EG2 =2EF? ; and since 'EF=CD, ÈG2=2CD2. And it was demonstrated, that AES=2ACP ; therefore, AE2 + EG?=2AC2 +2CD2. Now, AGP=AE: +EG, wherefore AG?=2AC: +2CD. But AG? (47. 1.)=AD? + DGʻ =AD? + DB, because DG=DB : Therefore, ADP +DB2=2AC2 +2CD2. 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 con tained 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 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. 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 equal to the square of EB: And the squares of BA, AE are equal (47. 1.) to the square of EB, be. I F G cause 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 H -В 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 E 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 : К. D 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 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 betweeen 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 (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.) BC: +CD2+ 2BC.CD; add AD2 to both: Then BD2 +AD?=BC+CD2 + AD2 +2BC.CD. But AB2=BD2 + AD (47. 1.), and AC2=CD + AD2 (47. 1.); therefore, AB2= BC2 + AC: + 2BC.CD; that is, AB2 is greater than BC2 +AC2 by 2BC.CD. Therefore, in obtuse angled triangles, &c. Q. E. D. B С D PROP. XIII. THEOR. In every triangle the square of the side subtending any of the acute an gles, 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 ucute 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 trian А. gle ABC, and because the straight line CB is divided into two parts in the point D (7. 2.), BC2 +BD2= 2BC.BD+C). Add to each AD2; then BC + BD +AD2 S2BC.BD + CD? + AD2, But BD? + AD2 = AB2, and CD2 + DA? AC2 (47. 1.); therefore BC2 + AB2= 2BC.BD+AC2; that is, AC2 is less B D than BC2 + AB2 by 2BC.BD. 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 AB = (12. 2.) AC? +BC2 + 2BC.CD. Add BC2 to each ; then AB: + BC=AC: +2BC2 + 2BC.CD. But because BD is divided into two parts in C, BC3 + BC.CD= (3. 2.) BC.BD, and 2BC: +2BC.CD= 2BC.BD: therefore ABP + BC = AC? + 2BC.BD ; and ACs is less than AB:+BC, by 2BD.BC. Lastly, Let the side AC be perpendicular to BC; then is BC the straight Α. line between the perpendicular and the acute angle at B; and it is manifest that (47. 1.) ABS +BC=AC+2BC=AC2 +2BC.BC. Therefore in every triangle, &c. Q. E. D. B 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 RCDE 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 of HE and EG are equal (47. 1.) to the square of GH: Therefore also the rectangle BE.EF to H gether with the square of ĒG,is equalto the squares A of HE and EG. Take away the square of EG, which is common to both, B F and the remaining rect 6 E angle BE.EF is equal to the square of EH : But C D BD is the rectangle con * See figure of the last Propositio. |