DEMONSTRATION. Bisect the base AB in E (a), and draw the straight line CE. The rectangle under AD and DB is equal in area to the difference between the squares on AE and DE (b); add to both these squares the square on the perpendicular CE, and the rectangle under AD and DB is equal in area to the difference between the sum of the squares on AE and CE, or the square on AC (c), and the sum of the squares on DE and CE, or the square on CD (c). A A PROPOSITION VII. I. 10. THEOREM.—If a straight line (AB) be divided into any two parts, the sum of the square on the whole line (AB) and the square on either segment (CB) is equal in area to double the rectangle under the whole line and that segment, together with the square of the other segment (AC) CONSTRUCTION. On AB construct the square ADEB (a), and join DB; through C draw CF parallel to AD (b), and through G draw HK parallel to AB (6). H D DEMONSTRATION. The square on AB is equal in area to the rectangles AK and GE together with the square HF; add to both the square CK, and the sum of the squares AE and CK is equal in area to the rectangles AK and CE together with the square HF (c). But because CB is equal to BK (d), AK is the rectangle under AB and CB; and because BE is equal to AB (e), CE is also equal to the rectangle under AB and CB; and because HG is equal to AC (f), HF is the square on AC. Therefore the sum of the square on AB and the square on CB is equal in area to double the rectangle under AB and CB together with the square on AC. a) I. 46. (b) I. 31. Ax. 1. (d) II. 4, cor. 4. (e) I. def. 28. (ƒ) I. 34. SCHOLIUM. This proposition, algebraically expressed, is as follows: THEOREM. If a number (a) be divided into two parts (m, n), the sum of the second power of the whole number (a) and the second power of either of the parts (m) is equal to twice the product of the whole number and that part, together with the second power of the other part (n). DEMONSTRATION. By the hypothesis m = a - N, COROLLARY. If we consider AB and CB as two independent lines, and AC as their difference, the foregoing proposition may be thus expressed:— THEOREM. The sum of the squares on any two lines is equal in area to twice the rectangle under them, together with the square on their difference. PROPOSITION VIII. THEOREM.—If a straight line (AB) be divided into any two parts, the square on the sum of the whole line (AB) and either segment (BC) is equal in area to four times the rectangle under the whole line and that segment (AB and BC), together with the square on the other segment (AC). CONSTRUCTION. Produce AB until BD is equal to CB (a). On AD construct the square AEFD (b), and join ED; through C and B draw CH and BL parallel to AE (c), and through K and P draw MN and XO parallel to AD (c). M X G K N Р R 0 H F (a) I. 3 and Post. 2. (b) I. 46. (c) I. 31. (d) I. 34. (e) Constr. (ƒ) Ax. 1. (g) I. 43. (h) I. 46, cor. 1. (i) Ax. 2. (k) II. 4, cor. 4. DEMONSTRATION. Because GK is equal to CB (d), CB to BD (e), and BD to KN (d), GK is equal to KN (f), and therefore the rectangle GL is equal to the rectangle KF; and because AK and KF are complements, they are equal (g), therefore AK is equal to GL (f). Because GK is equal to KN, therefore GR is equal to BN (h); and because MP and PL are complements, they are equal (g); adding these equals together, the rectangle MP together with the square BN is equal in area to the rectangle GL (i), and therefore to its equal the rectangle AK (ƒ). Therefore AK, GL, and KF, together with MP and BN, are equal in area to four times AK; but AK, GL, and KF, together with MP and BN, and the square XH, make up the whole square AEFD; therefore the square AEFD is equal in area to four times AK together with XH (f). But because BK is equal to BD (k), and BD to CB (e), BK is equal to CB (f), and there fore AK is the rectangle under AB and CB; and because XP is equal to AC (d), XH is the square on AC. Therefore the square on the sum of AB and BD is equal in area to four times the rectangle under AB and CB, together with the square on AC. SCHOLIUM. The foregoing proposition, algebraically expressed, is as follows: THEOREM. If a number (a) be divided into two parts (m, n), the second power of the sum of the whole number and either of the parts (a+m) is equal to four times the product of the whole number (a) and that part (m), together with the second power of the other part (n).` DEMONSTRATION. By the hypothesis therefore m = a - n, (a + m)2 = (2 a — n)2 = 4 a2 — 4 a n + n2 = 4 a (a − n) + n2, and substituting for (an) its equal m, we have (a + m)2 = 4 am + n2, PROPOSITION IX. THEOREM.-If a straight line (AB) be bisected (in C), and also cut into two unequal parts (in D), the sum of the squares on the unequal parts (AD and DB) is equal in area to double the sum of the square on half the line (AC) and the square on the line between the points of section (CD). CONSTRUCTION. From the point C draw CE perpendicular (a) and equal to AC (b), and join EA and EB; through D draw DF parallel to CE (c), and through F draw GF parallel to AB (c); and join AF. (a) I. 11. I. 31. DEMONSTRATION. Because in the triangle CAE the sides AC and CE are equal and the angle ACE a right angle (e), the angles CAE and CEA are each equal to half a right angle (ƒ); and in the same manner it may be shown that the angles CEB and CBE are likewise each equal to half a right angle; therefore AEB is a right angle. Then because GF is parallel to CD (e), the angle EGF is equal to the interior and opposite angle ECB (g), therefore EGF is a right angle; and in the triangle EGF I. 32, cor. 4. (9) I. 29. the angle EGF being a right angle, and the angle GEF half a right angle, the other angle GFE is also equal to half a right angle (h), and these angles being equal, the sides opposite to them, GE and GF, are also equal (i); and in the same manner it may be shown that FD and DB are also equal to each other. Again, because in the triangle CAE, AC and CE are equal, and the angle ACE is right, the square on AE is double the square on AC (k); and because in the triangle GEF, GE and GF are equal, and the angle EGF is right, the square on EF is double the square on GF (k); but CD is equal to GF (7), therefore the square on EF is double the square on CD; and therefore the squares on AE and EF, taken together, are double the squares on AC and CD taken together. But because in the triangle EAF the angle AEF is a right angle, the square on AF is equal in area to the squares on AE and EF taken together (k); therefore the square on AF is double the squares on AC and CD taken together. But because in the triangle DAF the angle ADF is a right angle, the square on AF is equal in area to the squares on AD and DF taken together (k); therefore the sum of the squares on AD and DF is double the sum of the squares on AC and CD; but DF and DB are equal, therefore the sum of the squares on AD and DB is equal in area to double the sum of the square on the line AC, and the square on the line CD. SCHOLIA. 1. The foregoing proposition, algebraically expressed, is as follows: THEOREM. If a number (a) be divided into two equal parts (m, n), and also into two unequal parts (p, q), the sum of the second powers of the unequal parts (p and q) is equal to twice the sum of the second powers of half the line (m), and of the difference between either of the unequal parts and equal parts. DEMONSTRATION. Let p be the greater of the unequal parts, and d the difference between p and m; then by the hypothesis therefore pm+d, and q = n - - d, p2 + q2 = (m + d)2 + (n − d)2 = m2 + 2 m d + d2 + n2 — 2 nd + d2. But by the hypothesis m=n, therefore p2 + q2 = 2 m2 + 2 d2. 2. The foregoing proposition holds true if, instead of squares, we construct any similar figures on the respective lines, such as similar segments A B D H of circles, or any similar irregular figures; thus the sum of the segments C and D is double the sum of the segments A and B, and the sum of the figures G and H is double the sum of the figures E and F. COROLLARY 1. THEOREM. In a triangle (ABC) if a line (AD) be drawn from its vertex to bisect its base (BC), the sum of the squares on the two sides (AB and AC) is equal in area to double the sum of the squares on half the base (BD) and on the bisecting line (AD). DEMONSTRATION. Draw AE perpendicular to BC (a). Then because AEB is a right angle, the square on AB is equal in area to the sum of the squares on BE and AE (b); and because AEC is a right angle, the square on AC is equal in area to the sum of the squares on EC and AE (b); therefore the squares on AB and AC, taken together, are equal in area to the squares on BE and EC and twice the square on AE taken together (c). But because the line BC is cut equally in D and unequally in E, the squares on BE and EC, taken together, are equal to twice the squares on BD and DE taken together (d); therefore the squares on AB and AC, taken together, are equal in area to double the sum of the squares on BD, DE, and AE. But the square on AD is equal in area to the squares on DE and AE taken together (b); therefore the sum of the squares on AB and AC is equal in area to double the sum of the squares on BD and AD. COROLLARY 2. THEOREM. If from the middle point (D) of a finite straight line (BC) as a center, a circle be described, and lines be drawn from any point (A) in its circumference to the extremities of the line (BC), the sum of the squares on those lines is always the same, and equal in area to double the sum of the squares on the radius (AD) and half the given line (BC). For the base of the triangle ABC is bisected by the radius AD (a), and therefore the sum of the squares on its two sides AB and AC is equal in area to double the sum of the squares on half the base BD, and on the bisecting line AD (b). D E (a) I. 12. (b) I. 47. (c) Ax. 2. (d) II. 9. Constr. (b) II. 9, cor. 1. |