PROPOSITION II. THEOREM.-If a straight line (AB) be divided into any two parts (in C), the rectangles under the whole (AB), and each of the parts (AC, CB), are together equal in area to the square on the whole line (AB). CONSTRUCTION. On AB construct the square A ADEB (a), and through C draw CF parallel to AD (b). DEMONSTRATION. It is evident that the square AE is equal to the sum of the rectangles AF, CE; but because AD is equal to AB (c), the rectangle AF is the rectangle under AD, or AB and AC; and because CF is equal to AD (d), the rectangle CE is the rectangle under CF, or AB and CB. Therefore the square AE on the whole line is equal in area to the rectangles AF and CE under the whole line and its parts. D (a) I. 46. SCHOLIUM. This proposition, algebraically expressed, is as follows:THEOREM. If a number (a) be divided into any two parts (m, n), the products of the whole number and each of the parts are together equal to the number multiplied by itself, or to the second power of that number. DEMONSTRATION. By the hypothesis a = m + n. Multiply these equal quantities by a, then a. a = (m + n). a, or a2 = a man. PROPOSITION III. THEOREM.-If a straight line (AB) be divided into any two parts (in C), the rectangle under the whole line (AB) and one of those parts (CB) is equal in area to the square on that part (CB) together with the rectangle under the two parts (AC and CB). CONSTRUCTION. On CB construct the square CDEB (a), and through A draw AF parallel to CD (b) until it meet DE produced in F. (a) I. 46. (c) Constr. and I. Def. 28. Constr. DEMONSTRATION. It is evident that the rectangle AE is equal to the square CE, together with the rectangle AD; but the rectangle AE is the rectangle under AB and CB; for BE is equal to CB (c), and the square CE is on CB (d); and the rectangle AD is the rectangle under AC and CB, for CD is equal to CB (c). Therefore the rectangle under AB and CB is equal in area to the square on CB, together with the rectangle under AC and CB. SCHOLIUM. This proposition, algebraically expressed, is as follows: THEOREM. If a number (a) be divided into any two parts (m, n), the product of the whole number and one of the parts is equal to the product of that part multiplied by itself, together with the product of the two parts. DEMONSTRATION. By the hypothesis a = m + n. Multiplying these equal quantities by m, we have a. m = (m + n) . m, or am = m2 + m n. PROPOSITION IV. THEOREM.-If a straight line (AB) be divided into any two parts (in C), the square on the whole line is equal in area to the sum of the squares on the parts (AC and CB), together with twice the rectangle under the parts. CONSTRUCTION. On AB construct the square ADEB (a), and join BD; through C draw CF parallel to AD (b), and through G draw HK parallel to AB (¿). H D C I. 46. (b) I. 31. (ƒ) I. 6. (g) I. 34. (h) I. Def. 28. (i) I. 43. K DEMONSTRATION. Because the line BD intersects the parallel lines CF and AD, it forms the external angle BGC equal to the interior opposite angle BDA (c); and because the triangle ADB is isosceles (d), the angle ABD is equal to the angle BDA (e); therefore the angle BGC is equal to ABD, and the side CG to CB (f). And because CGKB is a parallelogram (d), the side GK is equal to CB, and BK to CG (g), it is therefore equilateral. Also because the line AB intersects the parallel lines CF and AD, it forms the alternate angles A and BCG equal (c), therefore the angle BCG is a right angle, and CGKB is a square on CB (h). same manner it may be shown that the parallelogram HDFG is also a square on AC, because HG is equal to AC (g). Further AG and GE are together equal in area to double the rectangle under AC and CB, because AG is equal to GE (i), and AG is the rectangle under AC and CG, which is equal to CB. Therefore the square ADEB on the whole line is equal in area to the sum of the squares HF and CK on the parts of the line, together with twice the rectangle under those parts. And in the SCHOLIUM. This proposition, algebraically expressed, is as follows: THEOREM. If a number (a) be divided into any two parts (m, n), the second power of the whole number is equal to the sum of the second powers of the parts, together with twice the product of the parts. DEMONSTRATION. By the hypothesis a = m + n Squaring both sides of the equation— a2 = (m + n)2, or a2 = m2 + 2 m n + n2. COROLLARY 1. If a straight line (AB) be divided into any number of segments (AC, CD, DB), the square on the whole line is equal in area to the sum of the squares upon the segments, together with twice the rectangle under each pair of segments. F C D B M DEMONSTRATION. On AB construct the square AEFB, and join EB; through C and D draw CG and DH parallel to AE, and through P and O draw KM and IL parallel to AB. It may then be shown, by similar reasoning to that in the foregoing proposition, that the squares KG, NQ, and DL are respectively constructed on AC, CD, and DB; and further, that the rectangles IP and PH are equal to twice the rectangle under AC and CD, that the rectangles CO and OM are equal to twice the rectangle under CD and DB, and that the rectangles AN and QF are equal to twice the rectangle under AC and DB. It is therefore evident that the square on AB is equal to the sum of the squares upon the segments AC, CD, DB, together with twice the rectangles under each pair AC and CD, CD and DB AC and DB. COROLLARY 2. If a straight line AB be divided into three parts, the squares on the double segments AD, CB, together with twice the rectangle under the extreme segments AC and DB, are equal in area to the squares on the whole line AB and the middle segment CD. For it is evident that the squares IH and CM, together with the rectangles AN and QF, are equal in area to the square AEFB, together with the square NQ. D B H COROLLARY 3. If from either end of the hypotenuse (AB) of a right-angled triangle parts be cut off equal to the adjacent sides, the square on the middle segment (CD) thus formed is equal in area to twice the rectangle under the extreme segments (AC and DB). DEMONSTRATION. For the straight line AB being divided into three parts, the squares on AD and CB, together with twice the rectangle under AC and DB, are equal in area to the squares on AB and CD (a). But the squares on AE and EB, or their equals AD and CB, are equal in area to the square on AB (b). Therefore, taking these equals from the former equals, we have twice the rectangle under AC and DB equal in area to the square on CD (c). (a) Cor. 2. COROLLARY 4. From the demonstration, it is manifest that the parallelograms about the diagonal of a square are also squares. PROPOSITION V. THEOREM.-If a straight line (AB) be bisected (in C), and also cut into two unequal parts (in D), the rectangle under the unequal parts (AD and DB), together with the square on the line (CD) between the points of section, is equal in area to the square on half the line (CB). CONSTRUCTION. On CB construct the square CEFB (a), and join EB; through D draw DG parallel to CE (b), through H draw KM parallel to AB (b), and through A draw AK parallel to CL (b). I. 46. H M G (b) I. 31. DEMONSTRATION. In the parallelogram CF the complements CH and HF are equal (c); and because AC is equal to CB (d), AL is equal to CM; therefore the rectangle AH is equal in area to the gnomon CMG (e). To each of these add the square LG, and the rectangle AH together with the square LG is equal in area to the whole square CEFB (e). But, because DH is equal to DB (ƒ), AH is the rectangle under AD and DB; and because LH is equal to CD (g), LG is the square on CD, and CEFB is the square on CB. Therefore the rectangle under AD and DB, together with the square on CD, is equal in area to the square on CB. SCHOLIA. 1. This 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 product of the unequal parts, together with the second power of the difference between either of the unequal parts and equal parts, is equal to the second power of either of the equal parts. DEMONSTRATION. Let p be the greater of the unequal parts, and d the difference between p and m; then by the hypothesis Then if we multiply the first member of the first equation by q, and the second member by its equal (m- d), we have p. q = (m + d). (m — d) = m2 — d2, and transposing, we obtain Pq + d2 = m2. 2. The fifth proposition may be otherwise expressed, as follows:THEOREM. The square on half the sum of two lines is equal in area to the rectangle under them, together with the square on half their difference. For if AD and DB be considered the two lines, AC is half their sum, CD half their difference, and AH the rectangle under them. COROLLARY 1. The rectangle under the sum and difference of two lines (CB and CD) is equal in area to the difference of the squares on those lines. DEMONSTRATION. For the gnomon LBG is the difference between the square on CB and the square on CD. But the gnomon LBG is equal in area to the rectangle under AD and DB (a); and because AC is equal to CB, therefore AD is equal to the sum of CB and CD, and DB is equal to their difference. SCHOLIA. 1. The foregoing corollary, expressed algebraically, forms a theorem of the utmost importance. E (a) II. 5. The product of the sum and difference of two numbers (a and b) is equal to the difference of their second powers. For (a+b). (a − b) = a2 + ab. ab b2 a2 — b2. 2. When a line is divided into two parts, the rectangle under those parts is a maximum, or the greatest possible, when the line is bisected; and the sum of the squares on those two parts is then a minimum, or the least possible. COROLLARY 2. If a perpendicular (CD) be drawn from the vertex of a scalene triangle (ABC), the difference of the squares on the sides (AC and CB) is equal in area to twice the rectangle under the base (AB) and the distance (DE) of its middle point from the perpendicular. ДА DEMONSTRATION. For the difference of the squares on the sides AC and CB is equal in area to the difference of the squares on the segments of the base AD and DB (a), and therefore to the rectangle under the sum and difference of the segments AD and DB (b); but when the perpendicular falls within the triangle, the base AB is equal to their sum, and the distance DE of its middle point from the perpendicular to half their difference; therefore the difference of the squares on the sides AC and CB is equal in area to twice the rectangle under AB and DE. And when the perpendicular falls without the triangle, the base AB is equal to the difference of the segments AD and BD, and the distance DE to half their sum; and therefore, in this case, the difference of the squares on the sides AC and CB is equal in area to twice the rectangle under AB and DE. (a) I. 47, cor. 1. SCHOLIUM. From the foregoing corollary we derive the following rule for finding the area of a triangle of which the three sides are given. Divide the difference between the second power of the two sides of the triangle by twice the base; add half the base to the quotient, and subtract the second power of the sum from the second power of the greater side; the remainder is the second power of the perpendicular CD. Then half the product of the perpendicular CD multiplied by the base AB is the area of the triangle. |