with LG. But the gnomon CMG, and LG make up the figure CEFB, which is the square of CB: also AH is the rectangle under AD and DB, because DB is equal (II. 4. cor. 1.) to DH, and LG is the square of CD. Therefore the rectangle AD.DB, together with the square of CD, is equal to the square of CB. * Wherefore, if a straight line, &c. Cor. 1. Hence the difference of the squares of CB and CD is equal to the rectangle under AD and DB. But since AC is equal to CB, AD is equal to the sum of CB and CD, and DB is the difference of these lines. Hence the difference of the squares of two straight lines is equal to the rectangle under their sum and difference. Cor. 2. Since the square of CB, or which is the same, the rectangle AC.CB, is greater than the rectangle AD.DB by the square of CD, it follows, that to divide a straight line into two parts, the rectangle of which may be the greatest possible, or, as it is termed, a maximum, the line is to be bisected. Cor. 3. Hence also the sum of the squares of the two parts into which a straight line is divided, is the least possible, or is, as it is termed, a minimum, when the line is bisected. For (II. 4.) the square of the line is equal to the squares of the parts and twice their rectangle ; and therefore the greater the rectangle is, the less are the squares of the parts: but, by the foregoing corollary, the rectangle is a maximum when the line is bisected. Cor. 4. Since (I. 47. cor. 4.) the difference of the squares of the sides of a triangle is equal to the difference of the squares of the segments of the base, it follows, from the first corollary above, that the rectangle under the sum and difference of the sides of a triangle, is equal to the rectangle under the sum and difference of the segments, intercepted between the extremities of the base and the point in which the perpendicular cuts the base, or the base produced. Cor. 5. Hence also, if a straight line AD be drawn from the vertex A of an isosceles triangle ABC, to any point D in the base, or its continuation, the difference of the squares of that line and either of the equal sides is equal to the rectangle under BD and DC, the segments intercepted between the extremities of the base and the point D. To prove this, bisect BC in E, and join , A B E DC B E C D Otherwise. Since, as is proved in the deinonstration in the text, AL is equal to DF, take these separately from the entire figure, and there remain AH and LG equal to the square CF, as before. The proof may also be as follows: AD.DB = CD.DB + AC.DB (II. 1.), or AD.DB = CD.DB + CB.DB, because CB = AC. Hence (II. 3.) AD. DB CD.DB + CD. DB + DB2 = 2CD. DB + DBP. To each of these add CD2 : then AD.DB + CD: = 2CD.DB + DBP + CDs, or (II. 4.) AD. DB + CD:= CB. AE: thén (1. 8.) AE is perpendicular to BC. But (cor. 4.) the difference of the squares of AB and AD is equal to the rectangle under the sum and difference of BE and ED: that is, to the rect. angle under BD and DC, since BE is equal to EC. Cor. 6. Since (I. 47.) the square of one of the legs of a rightangled triangle is equal to the difference of the squares of the hypotenuse and the other leg, it follows (II. 5. cor. 1.) that the square of one leg of a right-angled triangle is equal to the rectangle under the sum and difference of the hypotenuse and the other. 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 D; the rectangle AD.DB, and the square of CB, are equal to the square of CD. A с B D K M E GF Upon CD describe (I. 46.) the square CF, and join DE; through B draw (1. 31.) BHG parallel to CE or DF; through H draw KLM parallel to AD or EF, and through A draw AK parallel to CL or DM. Then, because AC is equal to CB, the rectangle AL is equal (I. 36.) to CH; but (I. 43.) the complements CH, HF are equal ; therefore also AL is equal to HF. To each of these add CM and LG; therefore AM and LG are equal to the whole square CEFD. But AM is the rectangle under AD and DB, because (II. 4. cor. 1.) DB is equal to DM: also LG is the square of CB, and CEFD the square of CD. Therefore the rectangle AD.DB and the square of CB are equal to the square of CD. * Wherefore, if a straight line, &c. OTHERWISE. Produce CA to N, and make CN equal to CD. To these add the equals CB and CA ; therefore NB is equal (I. ax. 2.) to AD. But (II. 5.) the rectan BD gle NB.BD, or AD.BD, together with the square of CB, is equal to the square of CD: which was to be proved. NA * After it has been proved, as above, that HF is equal to AL, the rest of the proof will be simplified by taking these separately from the whole figure, as there will remain the rectangle AM and the square LG, together equal to the square CF. A с B H D PE 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 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 (I. 46.) the square AE, and construct the figure as in the preceding propositions. Then, because (I. 43.) the complements AG, GE are equal, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE are together double of AK. But AK, CE are the gnomon AKF, together with the square CK; therefore the gnomon AKF, and the square CK are double of AK, or double of the rectangle AB.BC, because BC is equal (II. 4. cor. 1.) to BK. To each of these equals add HF, which is equal (II. 4. cor. 1.) to the square of AC; therefore the gnomon AKÈ, and the squares CK, HF, are equal to twice the rectangle AB.BC and the square of AC. But the gnomon AKF, and the squares CK, HF, make up the whole figure, AE, together with CK; and these 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 of AC:* wherefore, if a straight line, &c. Cor. 1. Since AC is the difference of AB and BC, it follows that the square of the difference of two straight lines is equal to the sum of their squares, wanting twice their rectangle. Cor. 2. Since (II. 4.) the square of the sum of two lines exceeds the sum of their squares, by twice their rectangle, and since, by the foregoing corollary, the square of their difference is less than the sum of their squares, by twice their rectangle, it follows that the square of the sum of two lines, is equal to the square of their difference, together with four times their rectangle. PROP. VIII. THEOR. If a straight line be divided into any two parts, four times the rectangle contained by the whole 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 the former part. Let the straight line AB be divided into any two parts in C; "Otherwise. Because (II. 4.) ABP = AC? + BC? + 2AC.BC, add BC to both ; then AB: + BC = AC: + 2BC2 + 2 AC.BC. But (II. 3.) BC? + AC.BC = AB. BC, and therefore 2BC2 + 2AC.BC= 2A B.BC; wherefore AB' + BC = AC? + 2 AB.BC, : A с вр GK M N х PR o E H LF : four times the rectangle AB.BC, and the square of AC, are equal to the square of the line made up of AB and BC together. Produce AB to D, so that BD may be equal to CB, and upon AD describe the square AF: join DE, and draw BKL, CPH parallel to AE or DF, and NKGM, XPRO parallel to AD or EF. Because CB is equal (const.) to BD, and that CB is equal (I. 34.) to GK, and BD to KN; therefore GK is equal to KN. For the same reason PR is equal to RO: and because CB is equal to BD, and GK to KN, the rectangle CK is equal (I. 36.) to BN, and GR to RN. But CK is equal (I. 43.) to RN; therefore also BN is equal to GR; and the four rectangles BN, CK, GR, RN are therefore equal to one another, and so are quadruple of one of them CK. Again, because CB is equal to BD, and that BD is equal (II. 4. cor. 1.) 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 (1. 43.) 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 are quadruple of one of them AG It was also demonstrated, that the four CK, BN, GR, and RN are quadruple of CK: therefore the eight rectangles which make up the gnomon AOH, are quadruple of AK: and because AK is contained by AB, BK, or AB, 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 AOH. To each of these add XH, which is equal (II. 4. cor. 1.) to the square of AC: then four times the rectangle AB.BC, and the square of AC are equal to the gnomon AOH and the square XH, that is, the square of AD: and AD is equal to AB and BC added together in one straight line. * Wherefore, if a straight line, &c. PROP. IX. THEOR. If a straight line be divided into two equal, and also into two unequal parts; the squares of the 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 equally in C, and unequally * This proposition is virtually the same as the second corollary to the seventh proposition, AD being the sum, and AC the difference of AB and BC. Should any person wish to demonstrate it independently of that corollary, the following proof may probably be preferred to the tedious one given in the text. AD2 = AC? + CD + 2AC.CD (II. 4.) But (II. 4. cor. 2.) CD = 4CB?: also 2AC.CD= 4AC.CB: wherefore AD? = AC? + 4CB: + 4AC.CB. But (II. 3.) AB.BC=CB? + AC.CB, and therefore 4A B. BC= 4CB: + 4AC. CB. Hence AD' = AC: + 4AB.BC. E F G in D: the squares of AD, DB are together double of the squares of AC, CD. From C draw (1. 11.) CE at right angles to AB, and make it equal to AC or CB, and join EA, EB: through D draw (I. 31.) DF parallel to CE, and through F draw FG parallel to AB; and join AF. Then, because (const.) the triangles ACE, BCE are right-angled and isosceles, the angles CAE, AEC, CEB, EBC, are (1. 32. cor. 4.) each half a right angle. So also (I. 29. part 2.) are EFG, BFD, because FG is parallel to AB, and FD to EC: and, for the same reason, EGF, ADF are right angles. The angle AEB is also a right angle, its parts being each half a right angle. Hence (I. 6.) EG is equal to GF or CD, and FD to DB. Again (1. 47.) the square of AE is equal to the squares of AC, CE, or to twice the square of AC, since AC and CĘ are equal. In like manner, the square of EF is equal to twice the square of GF or CD. Now, (I. 47.) the squares of AD and DF, or of AD and DB, are equal to the square of AF; and the squares of AE, EF, that is, twice the square of AC and twice the square of CD, are also to the square of AF: therefore (I. ax. 1.) the squares of AD, DB are equal to twice the square of AC and twice the square of CD. If, therefore, a straight line, &c. C D B equal OTHERWISE. DB' +2BC.CD=BC+CD: (11. 7.), or DB° +2AC.CD= AC+CD2: also (II. 4.) AD=AC+CD+2 AC.CD. Add these equals together, and from the sums take 2 AC.CD; then AD'+DBP=2AC° +2CD. PROP. X. THEOR. If a straight line be bisected, and produced to any point, the squares of the whole line thus produced, and 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 D; the squares of AD, DB are double of the squares of AC, CD. From C draw (1. 11.) CE perpendicular to AB; and make it equal to AC or CB; join AE, EB, and through E and D draw (1. 31.) EF parallel to AB, and DF parallel to CE. Then 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 F с B A |