PROP. VI. THEOR. If a straight line be bisected, and produced to any point; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is 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 rectangle AD, DB, together with the square of CB, is equal to the square of CD. Upon CD describe (46. 1.) the square of CEFD, join DE, and through B draw (31. 1.) BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK parallel to CL or DM: and because AC is equal A С B D to CB, the rectangle AL is L H equal (43. 1.) to CH; but CH is K к M equal (36. 1.) to HF; therefore also AL is equal to HF: to each of these add CM; therefore the whole AM is equal to the gno E G F mon CMG: and DM is the rectangle contained by AD, DB, for DM is equal (Cor. 4. 2.) to DB : therefore the gnomon CMG is equal to the rectangle AD, DB, add to each of these LG, which is equal to the square of CB, therefore the rectangle AD, DB, together with the square of CB, is equal to the gnomon CMG and the figure LG: but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D. PROP. VII. THEOR. If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in 17 d DE in the preceding. Because CB is equal to BD, and that CB the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC. Upon AB describe (46. 1.) the square ADEB, and construct the figure as in the preceding propositions : and because AG is equal (43. 1.) to GE, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE are double of AK; but AK, A с B CE are the gnomon AKF, together with the square CK; therefore the gnomon AKF, G together with the square CK, is double H K of AK: but twice the rectangle AB, BC is double of AK, for BK is equal (Cor. 4. 2.) to BC: therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC: to each of D F E these equals add HF, which is equal to the square of AC: therefore the gnomon AKF, together with the D squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC : but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC: therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square AC. Wherefore, if a straight line, &c. Q. E. D. T PROP. VIII. THEOR. If a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part. Let the straight line AB be divided into any two parts in the point C ; four times the rectangle AB, BC, together with the square of AC, is equal to the square of the straight line made up of AB and BC together. Produce AB to D, so that BD be equal to CB, and upon AD describe the square ÅEFD; and construct two figures such as is equal (34. 1.5 to GK, and BD, to KN ; therefore GK is Te 70 and lian equal to KN; for the same reason, PR is equal to RO; and be- CB D fore CG is equal to GP: and because GIK N 'P R rectangle AG is equal to MP, and PL to RF: but MP is equal (43. 1.) to PL, X 0 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 E H L F one another, and so are quadruple of one of them AG. And it was demonstrated that the four CK, BN, GR, and RN are quadruple of CK: therefore the eight rectangles which contain the gnomon AOH are quadruple of AK: and because AK is the rectangle contained by AB, BC, for BK is equal to BC, four times the rectangle AB, BC is quadruple of AK: but the gnomon AOH was demonstrated to be quadruple of AK: therefore four times the rectangle AB, BC is equal to the gnomon AOH. To each of these add XH, which is equal (Cor. 4. 2.) to the square of AC: therefore four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH and the square XH: but the gnomon AOH and XH make up the figure AEFD, which is the square of AD: therefore four times the rectangle AB, BC, together with the 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 at the point C into two equal, and at D into two unequal parts: the squares of AD, , ᎠᏴ 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 E of a right angle. For the same reason each of the angles CEB, EBC is half a right angle; and therefore the whole. F AEB is a right angle: and because the angle GEF is half a right angle, and EĞF a right angle, for it is equal A (29. 1.) to the interior and opposite C D B 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 half 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: and because AC 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 EA is equal (47. 1.) to 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 FG, GF are double of the square of GF; but the square of EF is equal to the squares H of EG, GF; therefore the square of EF is double of the square of GF; and GF is equal (34. 1.) to CD; therefore the square of EF 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 : and the square of AF is equal (47. 1.) to the squares of AE, EF, because AEF 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, DA 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 C draw (11. 1.) CE at right angles to AB: and make it equal to AC or CB, and join AE, EB; through E draw (31. 1.) .F 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 mcet (12. Ax.) if produced far enough : therefore EB, FD shall meet, if produced towards B, D: let them mect 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 half a right angle : therefore AEB is a right angle: and because EBC is half a right angle, DBG is also Y LIF |