and having the angle K equal to the given angle E (a); and upon the straight line GH construct a parallelogram GM equal in area to the triangle DBC, and having. the angle GHM equal to the given angle E (b); then FKML will be the required parallelogram. DEMONSTRATION. Because the straight lines FK and GH are parallel, therefore the internal angles K and GHK are together equal to two right angles (c); but the angles K and GHM being both equal to the given angle E (d), are equal to one another (e); therefore the angles GHK and GHM are together equal to two right angles (e), and therefore KH and HM are in the same straight line (f). Because the straight line_GH intersects the parallels FG and KM, the alternate angles FGH and GHM are equal (c); but the internal angles GHM and HGL are together equal to two right angles (c), therefore the angles FGH and HGL are together equal to two right angles (e), and therefore FG and GL are in the same straight line (f). Then, because FK and LM are both parallel to GH, therefore they are parallel to each other (g); and FL being parallel to KM, FKML is a parallelogram. And because the parallelogram FH is equal in area to the triangle ABD, and the parallelogram GM to the triangle DBC, therefore the whole parallelogram FKML is equal in area to the whole figure ABCD (h), and has the angle K equal to the given angle E. COROLLARY. By means of this proposition, and that immediately preceding it, a parallelogram can be constructed on a given line equal in area to a given rectilineal figure, and having an angle equal to a given rectilineal angle, by constructing on the given line a parallelogram equal in area to the first triangle ABD. PROPOSITION XLVI. PROBLEM.-Upon a given finite straight line (AB) to construct a square. SOLUTION. From the point A draw AC perpendicular to AB (a), and make AD equal to AB (b); through the point D draw DE parallel to AB (c), and through the point B draw BE parallel to AD (c); then DABE is the required square. DEMONSTRATION. The side DE being parallel to the opposite side AB, and the side BE to ÂD (d), the figure DABE is a parallelogram (e); and therefore its opposite sides are equal (f), that is DE to AB, and BE to AD; and because AD is equal to AB (d), therefore all the sides are equal (g); but the angle A is a right angle, wherefore the four-sided figure DABE has all its sides equal and one of its angles a right angle; therefore it is a square (h), and it is constructed on the line AB. (d) Constr. Def. 26. f) I. 34. Ax. 1. Def. 28. SCHOLIUM. The definition of a square, as given by Euclid, viz. "a four-sided figure, which has all its sides equal, and all its angles right angles," is more than sufficient, and really involves a theorem. It is only necessary to state in the definition that one of its angles is a right angle, and the proposition that its remaining angles are right may be demonstrated as follows. THEOREM. If a four-sided figure (DABE) be a square, all its angles are right angles. DEMONSTRATION. The opposite sides AB and DE are parallel (a), and they are met by the line AD, therefore the angles A and D are together equal to two right angles (b); but A is a right angle (a), therefore D is a right angle. And because, in the parallelogram DABE, the angles E and B are respectively opposite to A and D, which are right angles, therefore and B are right angles; and therefore all the angles of the square DABE are right angles. COROLLARY 1. If two squares are constructed on equal straight lines AB and CD, they are equal. DEMONSTRATION. Draw the diagonals EB and GD. Because, in the triangles EAB and GCD, the sides EA and AB are respectively equal to GC and CD (a), and the angle A to the angle C, therefore the triangles are equal (b). And because the squares AF and CH are doubles of the triangles EAB and GCD (c), therefore they are equal (d). COROLLARY 2. If two squares AF and CH are equal. DEMONSTRATION. For, if it be possible, let one of them AF be the greater; take AK equal to CD, and AI equal to CG (a), and join IK. Then the triangles IAK and GCD are equal (b); but the triangles EAB and GCD are equal, being halves of the equal squares AF and CH (c), therefore the triangle IAK is equal to EAB (d), a part to the whole, which is absurd; therefore neither of the sides AB or CD is greater than the other, but they are equal. COROLLARY 3. To construct a rectangle under two given finite straight lines F and AB. SOLUTION. From the point A draw AC perpendicular to AB (a), and make AD equal to F(b); through the point D draw DE parallel to AB (c), and through the point B draw BE parallel to AD (c); then DABE is the rectangle required. E (a) Constr. I. 29. H (a) Hypoth. and Def. equal, their sides are (a) I. 3. Ax. 7. Ax. 1. (a) I. 11. E B H I DEMONSTRATION. The side DE being parallel to the opposite side AB, and the side BE to AD (d), the figure DABE is a parallelogram (e); and the angle A is a right angle (d), therefore it is a rectangle (f), and it is constructed under the two straight lines F and AB. (d) Constr. PROPOSITION XLVII. THEOREM.-If a triangle (ABC) be right-angled, the square which is constructed upon the side (BC) subtending the right angle is equal in area to the sum of the squares constructed upon the sides (AB and AC) which form the right angle. CONSTRUCTION. On the sides AB, BC, and AC, construct the squares BG, DEMONSTRATION. Because the angles FBA and CBD are both right angles (a), therefore they are equal (c); add to both the angle ABC, and the angle FBC is equal to ABD (d). Because the sides FB and BC are respectively equal to AB and BD (e), and the angle FBC to the angle ABD, therefore the triangle FBC is equal to the triangle ABD (f). Because the angles GAB and BAC are both right angles, therefore GA and AC are in the same straight line (g). Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD and between the same parallels BD and AL (h); and the I. 46. (b) I. 31. Def. 28. I. 4. I. 14. I. 41. Ax. 6. square GB is double of the triangle FBC, being on the same base FB and between the same parallels FB and GC (h). But the doubles of equals are equal to one another (i), and therefore the parallelogram BL is equal in area to the square GB. And in the same manner, by joining AE and BK, it may be proved that the parallelogram CL is equal in area to the square CH. Therefore the whole square BDEC is equal in area to the two squares BG and CH. 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. M K 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. I. 36. 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 M 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 (a) I. 35. (b) I. 36. Ax. 1. 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 I. 47. Ax. 3. 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 of the squares on CB and AD. A4 (a) I. 47. |