SCHOLIA. 1. It should be remarked that the line AL is a perpendicular from the point A on to the hypotenuse BC, dividing it into two segments Be and eC, and that the square described on either side of the triangle ABC is equal in area to the rectangle under the whole hypotenuse and the segment of the hypotenuse adjacent to the same side. 2. This proposition may be demonstrated in a variety of ways, but that adopted by Euclid has the preference of the others. The following mode of proving it is given because it is capable of being generalised, as is done in the third scholium. CONSTRUCTION. Upon the sides AB, AC, and BC, construct the squares BG, CH, and BE; produce FG and KH to meet in N; draw NA, and produce it to meet DE in L; and produce DB to meet FG in M. F DEMONSTRATION. The angles MBC and FBA, being both right angles, are equal (a); take from each the angle MBA, and the remaining angles FBM and ABC are equal (b); also the angles F and CAB are equal, being both right angles (a). Then, because in the triangles MFB and CAB the angles F and FBM are equal to the angles CAB and ABC, and the side FB is equal to AB, therefore the remaining sides are equal (c), FM to AC and MB to CB. Then, because FN is parallel to BH, and GC to NK, therefore GH is a parallelogram, and GN is equal to AH or AC (d); but FM is equal to AC, therefore GN is equal to FM; add to each the MG, and FG is equal to MN. But FG equal BA, therefore MN equal BA, and they are parallel; therefore the straight lines BM and AN joining their extremities are equal and parallel (e), and MBAN is a parallelogram. Then, because the parallelogram MBAN and the square BG are upon the same base AB and between the same parallels FN and BA, therefore they are equal in area (f). Also, because the parallelograms MBAN and BL are upon equal bases MB and BD, and between the same parallels MD and NL, they are equal in area (g); therefore the parallelogram BL is equal to the square BG (h); and in like manner it may be proved that the parallelogram CL is equal in area to the square CH, and therefore that the whole square BE is equal in area to the squares BG and CH. (a) Ax. 11. (b) Ax. 3. (c) I. 26. (d) I. 34. (e) I. 33. (f) I. 35. (g) I. 36. (4) Ax. 1. 3. The forty-seventh proposition is only a particular case of the following theorem, as will be seen by comparing the demonstration given in the foregoing scholium with that given below. THEOREM. If parallelograms (FBAG and HACK) be constructed upon two of the sides (AB and AC) of any triangle (ABC), and their sides (FG and KH) parallel to the sides of the triangle be produced to meet in a point (N); if a straight line (NA) be drawn from that point to the vertex of the triangle, and if a parallelogram (BDEC) be constructed upon the base of the triangle whose other sides are equal F D and parallel to that straight line, then the last parallelogram (BDEC) is equal in area to the two former (FBAG and HACK). CONSTRUCTION. Produce DB to meet FG in M, and NA to meet DE in L. DEMONSTRATION. The parallelograms FA and BN are between the same parallels FN and BA and upon the same base AB, therefore FA is equal in area to BN (a); and because the parallelograms BN and BL are between the same parallels DM and LN, and upon equal bases BD and NA, therefore BN is equal in area to BL (b); and therefore the parallelogram FA is equal in area to BL (c); and in the same manner it may be proved that HC is equal in area to CL, therefore the whole parallelogram BDEC is equal in area to the parallelograms FA and HC. F B M (a) I. 35. COROLLARY 1; THEOREM. If a perpendicular (CD) be drawn from the vertex of a triangle (ABC) cutting the base, the difference of the squares on the sides (AC and CB) is equal to the difference of the squares on the segments of the base (AD and DB). DEMONSTRATION. For the square on AC is equal in area to the squares on AD and CD (a), and the square on CB is equal in area to the squares on DB and CD (a); therefore the difference of the squares on AC and CB is equal to the difference of the sum of the squares on AD and CD, and the sum of the squares on DB and CD (b); or, taking from each the common square on CD, equal to the difference of the square on AD and the square on DB. A A (a) I. 47. SCHOLIUM. This theorem applies whether the base is cut internally, as in the left-hand figure, or the base produced is cut externally, as in that on the right hand. See Scholium to Definition 2. COROLLARY 2; THEOREM. If a perpendicular (CD) be drawn from the vertex of a triangle (ABC) cutting the base, the sum of the squares on one side and the alternate segment (AC and DB) is equal to the sum of the squares on the other side and the alternate segment (BC and AD). DEMONSTRATION. For the square on AC is equal in area to the sum of the squares on AD and CD (a), and the sum of the squares on DB and CD is equal in area to the square on CB (a); adding these equals together, we have the sum of the squares on AC, DB, and CD, equal to the sum of the squares on CB, AD, and CD (b); or, taking away the common square on CD, the sum of the squares on AC and DB is equal to the sum f the squares on CB and AD A A I. 47. COROLLARY 3; PROBLEM. To construct a square equal in area to the sum of two or more given squares. SOLUTION. Let the straight lines A, B, and C be sides of the given squares. Draw ED and DF at right angles, and make GD equal to A and DH equal to B; join GH, and draw GI perpendicular to it; make GK equal to C, and join KĤ, then a square constructed on the line KH shall be the square required. DEMONSTRATION. For the square on KH is equal in area to the sum of the squares on GK or C, and on GH (a), but the square on GH is equal in area to the squares on GD and DH, or on A and B (a); therefore the square on KH is equal in area to the sum of squares on A, B, and C. the (a) I. 47. COROLLARY 4; PROBLEM. To construct a square equal in area to the difference of two given squares. SOLUTION. Let A and B be sides of the given squares. Draw CD equal to the side of the greater square B, and produce it until the produced part DE is equal to the side of the other square A. From the center D, at the distance CD, describe the circle CFG, and through E draw EF perpendicular to CE, to meet the circle in F; then FE is a side of the required square. (a) I. 47. DEMONSTRATION. Join DF. The square on DF, or its equal B, is equal in area to the sum of the squares on DE or A, and on EF (a); therefore if the square on DE be taken from the square on DF, the difference is equal in area to the square on EF. &c. COROLLARY 5; PROBLEM. To find geometrical values of 1, √2, √3, SOLUTION. Take AB equal to 1; draw CB perpendicular to AB, and also equal to 1; join AC, and draw DC perpendicular to it and equal to 1; join DA, and draw DE perpendicular to it and equal to 1, fc.; then the lines AC, AD, AE, &c., are respectively equal to ✅ 1, √2, √3, &c. DEMONSTRATION. The square on AC equals the sum of the squares on AB and BC (a); the squares on AB and BC each equal 1 (b); therefore the square on AC equals 2, and AC equals ✔2. Again, the square on AD equals the sum of the squares on AC and DC (a); the square on AC equals 2, and the square on (a) I. 47. DC equals 1 (6); therefore the square on AD equals 3, and AD equals √3. Therefore the square roots of the natural numbers 1, 2, 3, 4, &c., are represented geometrically by the lines AB, AC, AD, AE &c. D COROLLARY 6. By means of the 47th proposition, any two sides of a rightangled triangle being given, the third side may be readily found; for if the two sides given are those which form the right angle, the sum of the squares on those two sides will be equal to the square on the third side; and if one of the given sides subtends the right angle, then the square on the third side is equal to the difference of the squares on the two given sides. This application of the 47th proposition to the purposes of finding the parts of right-angled triangles is of immense use, and is really the foundation of Trigonometry. COROLLARY 7. The 47th proposition holds true, if, instead of squares, we construct any similar figures on the sides of the triangle, such as circles, equilateral triangles, &c., and may therefore be generalized as follows: THEOREM. If a triangle be right-angled, any figure which is constructed upon the side subtending the right angle is equal in area to the sum of the similar figures constructed upon the sides which form the right angle. SCHOLIUM. If three semicircles be described on the three sides of a right-angled triangle, the area of ABC will therefore be equal to the sum of the areas of ADB and BFC; if, now, from each we deduct the common segments AEB and BGC, we have the triangle ABC equal in area to the sum of two lunes ADBE and BFCG. This proposition was discovered by Hippocrates, and was the first instance of the determination geometrically of the area of a space entirely bounded by curved lines; it led him to believe that, by means of this proposition, he would be able to solve the problem, to determine geometrically the area of the circle. PROPOSITION XLVIII. THEOREM.—If the square constructed upon one side (BC) of a triangle (ABC) be equal in area to the sum of the squares constructed upon the other two sides (AC and AB), the angle BAC opposite to that side is a right angle. CONSTRUCTION. From the point A draw AD at right angles to one of the sides AC (a), and equal to the other AB (b); and join DC. DEMONSTRATION. Because DA equals AB (c), therefore the square on DA equals the square on AB (d); and adding to each the square on AC, therefore the sum of the squares on DA and AC is equal to the sum of the squares on AB and AC (e). But the square on DC is equal in area to the sum of the squares on DA and AC (ƒ), because DAC is a right angle (c); and the square on BC is equal in a) I. 11. (b) I. 2. (c) Const. (d) I. 46, cor. 1. (e) Ax..2. (ƒ) I. 47. area to the sum of the squares on AB and AC (g); therefore the square on DC is equal to the square on BC, and therefore DC is equal to BC (h). Then, because in the triangles DAC and BAC the sides AB and AD are equal, AC common to both, and the base DC equal to BC, therefore the angle DAC is equal to BAC (i); but DAC is a right angle (c), therefore BAC is also a right angle. SCHOLIUM. This proposition is the converse of the preceding one. Const. (g) Hypoth. |