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angles BEF, EFD are less than two right angles : therefore
PROP. XI. PROB. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, may be equal to the square of the other part.
Let AB be the given straight line ; it is required to divide it
* The ten foregoing propositions may all be proved very easily by means of algebra, in connexion with the principles of mensuration, already established in the corollaries to the 46th proposition of the first book. Thus, to prove the fourth proposition, let AC=a, CB=b, and, consequently, AB=a+b. Now, the area of the square described on AB will be found (I. 46. cor. 4.) by multiplying a tb by itself; and this product is found, by performing the actual operation, to be a? +2ab+62 ; an expression, the first and third parts of which are, by the same corollary, the areas of the squares of AC and CB, and the second is twice the rectangle of those lines.
In like manner, to prove the eighth, adopting the same notation, we have the line which is made up of the whole and CB=a+2b; and multiplying this by itself, we get for the area of the square of that line, a? + 4ab + 462, or a? + 4(a + b)b, the first part of which is the area of the square of AC, and the second four times the area of the rectangle under A B and CB.
It will be a useful exercise for the student to prove the other propositions in a similar manner. He will also find it easy to investigate various other relations of lines and their parts by means of algebra.
All the properties delivered in these propositions hold also respecting numbers, if products be substituted for rectangles. Thus, 7 being equal to the sum of 5 and 2, the square, or second power, of 7, is equal to the squares of 5 and 2 and twice their product; that is, 49= 25+4+20.
into two parts, so that the rectangle under the whole and one of the parts, may be equal to the square of the other.
Upon AB describe (I. 46.) the square AD ; bisect (I. 10.) AC in E, and join E with B, the remote extremity of AB; produce CA to F, making (I. 3.) EF equal to EB, and cut off AH equal to AF: AB is divided in H, so that the rectangle AB.BH is equal to the square of AH.
Complete the parallelogram AG, and produce GH to K. Then, since BAC is a right angle, FAH is also (I. 13.) a right angle; and (I. def. 27.) AG is a square, AF, AH being equal by construction. Because the straight line AC is bisected in E, and produced to F, the rectangle CF.FA and the square of AE are together equal (II. 6.) to the square of EF or of EB, since (const.) EB, EF are equal. But the squares of BA, AE are equal (I. 47.) to the square of EB, because the angle EAB is a right angle : therefore the rectangle CF.FA and the square of AE are equal (I. ax. 1.) to the squares of BA, AE. Take away
square of A E, which is common to both; therefore the remaining rectangle CF.FA is equal (1. ax. 3.) to the square of AB. But the figure FK is the rectangle contained by CF, FA, for AF is equal to FG; and AD is the square
of AB: therefore FK is equal to AD. Take away the common part AK, and (I. ax. 3.) the remainders FH and HD are equal. But HD is the rectangle AB.BH, for AB is equal to BD; and FH is the square of AH. Therefore the rectangle AB.BH is equal to the square of AH: wherefore the straight line AB is divided in H, so that the rectangle AB.BH is equal to the square of AH:* which was to be done.
Schol. The line CF is equal to BA and AH; and since it has been shown that the rectangle CF.FA is equal to the square of BA or CA, it follows, that if any straight line AB (see the next
* In the practical construction in this proposition, and in the 30th of the sixth book, which is virtually the same, it is sufficient to draw A E perpendicular to A B, making it equal to the half of AB, and producing it through A ; then, to make E F equal to the distance from E to B, and AH equal to AF. It is plain that BD might be bisected instead of AC, and that in this way another point of section would be obtained.
While the enunciation in the text serves for ordinary purposes, it is too limited in a geometrical sense, as it comprehends only one case, excluding another. The following includes both :
In a given straight line, or its continuation, to find a point, such that the rectangle contained by the given line, and the segment between one of its extremities and the required point, may be equal to the square of the segment between its other extremity and the same point.
The point in the continuation of BA, will be found by cutting off a line on EC and its continuation, equal to E B, and describing on the line composed of that line and AE, a square lying on the opposite side of AC from AD; as the angular point of that square in the continuation of BA is the point required. The proof is the same as that given above, except that a rectangle corresponding to AK is to be added instead of being subtracted.
diagram) be divided according to this proposition in C, AC being the greater part, and if AD be made equal to AB, DC is similarly divided in A.
So also if DE be made equal to DC, and EF to EA, EA is
divided similarly in D, and FD in E: and the like additions may be continued as far as we please. Conversely, if any straight line FD be divided according to this proposition in E, and if EA be made equal to EF, DC to DE, &c.; EA is similarly divided in D, DC in A, &c. It follows also, that the greater segment of a line so divided, will be itself similarly divided, if a part be cut off from it equal to the less; and that by adding to the whole line its greater segment, another line will be obtained, which is similarly divided.
PROP. XII. THEOR. In an obtuse-angled triangle, the square of the greatest side exceeds the squares of the other two by twice the rectangle contained by either of the last-mentioned sides, and its continuation to meet a perpendicular drawn to it from the opposite angle.
Let ABC be a triangle, having the angle ACB obtuse ; and let AD be perpendicular to BC produced; the square of AB is equal to the squares of AC, CB, and twice the rectangle BC.CD.
Because the straight line BD is divided into two parts in the point C, the square of BD is equal (II. 4.) to the squares of BC, CD, and twice the rectangle BC.CD. To each of these equals add the square of DA; and the squares
of DB, DA are equal to the squares of BC, CD, DA, and twice the rectangle BC.CD. But, because the angle D is a right angle, the square of BA is equal (I. 47.) to the squares of BD, DA, and the square of CA is equal to the squares of CD, DA ; therefore the square of BA is equal to the squares of BC, CA, and twice the rectangle BC.CD. Therefore, in an obtuse-angled triangle, &c.
PROP. XIII. THEOR. In any triangle, the square of the side subtending an acute angle, is less than the squares of the other sides, by twice the rectangle contained by either of those sides, and the straight line intercepted between the acute angle and the perpendicular drawn to that side from the opposite angle.
Let ABC (see this figure and that of the foregoing proposition)
be any triangle, having the angle B acute ; and let AD be perpendicular to BC, one of the sides containing that angle: the square of AC is less than the squares of CB, BA, by twice the rectangle CB.BD.
The squares of CB, BD are equal (II. 7.) to twice the rectangle contained by CB, BD, and the square of DC. To each of these equals add the square of AD; therefore the squares of CB, BD, DA are equal to twice the rectangle CB.BD, and the squares of AD, DC. But, because AD is perpendicular to BC, the square of AB is equal (I. 47.) to the squares of BD, DA, and the B square of AC to the squares of AD, DC: therefore the squares of CB.BA are equal to the square of AC, and twice the rectangle CB.BD; that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB.BD.
If the side AC be perpendicular to BC, then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest that the squares of AB, BC are equal (I. 47.) to the square of AC and twice the square of BC. Therefore, in any triangle, &c. *
PROP. XIV. PROB. + To describe a square that shall be equal to a given rectilineal figure.
Let A be the given rectilineal figure; it is required to describe a square that shall be equal to it.
By means of this or the foregoing proposition, the area of a triangle may be computed, if the sides be given in numbers. Thus, let AB=17, BC= 28, and AC= 25. From AB? + BC? take AC?; that is, from 172 +282 take 252; the remainder 448 is twice CB.BD. Dividing this by 56, twice BC, the quotient 8 is BD. Hence, from either of the triangles ABD, ACD, we find the perpendicular to be 15; and thence the area is found, by taking half the product of BC and AD, to be 210.
The segments of the base are more easily found by means of the 4th corollary to the 5th proposition of this book, in connexion with the principle, that if half the difference of two magnitudes be added to half their sum, the result is the greater; and if half the difference be taken from half the sum, the remainder is the less. Hence if 42, the sum of AB and AC, be multiplied by 8, their difference, and if the product 336 be divided by 28, the sum of the segments of the base, the quotient 12 is their difference. The half of this being added to the half of 28, the sum 20 is the greater segment CD; and being subtracted from it, the remainder 8 is BD.
To prove the principle mentioned above, let AB be the greater, and BC the less of two magnitudes. Bisect AC in D, and make AE equal to BC. Then AD or DC is half the A_E D 모 sum, and ED or DB half the difference of A B and BC; and A B the greater is equal to the sum of AD and DB, and BC the less is equal to the difference of DC and DB.
† This is a particular case of the 25th proposition of the sixth book.
Describe (I. 45.) the rectangle BD equal to A. Then, if the sides of it, BE, ED, be equal to one another, it is a square, and what was required is done. But if they be not equal, produce one of them BE to F, and make EF equal to ED; bisect BF in G, and from the centre G, at the distance GB, or GF, describe (I. post. 3.) the semicircle BHF: produce DE to H, and join GH. Therefore, because the straight line BF is divided equally in G, and unequally in E, the rectangle BE.EF, and the square of EG, are equal (II. 5.) to the square of GF, or of GH, because GH is equal to GF. But the squares of HE, EG are equal (1. 47.) to the square of GH: therefore the rectangle BE.EF and the square of EG are equal to the squares of HE, EG. Take away the square of EG, which is common, and the remaining rectangle BE.EF is equal to the square of EH. But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED ; therefore BD is equal to the
square of EH: but BD is equal to the figure A ; therefore the square of EH is equal to A. The square described on EH, therefore, is the required square.
PROP. A. THEOR.
The squares of two sides of a triangle are together equal to twice the square of half the remaining side, and twice the square of the straight line drawn from its point of bisection to the opposite angle.
Let ABC be a triangle, of which the side BC is bisected in D, and let DA be drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA.
If DA be perpendicular to BC, the squares of BA and AC are equal (1. 47.) to the squares of BD and DC and twice the square of DA, or to twice the square of BD and twice the square of DA, because BD, DC are equal.
But if AD be not perpendicular to BC, draw AE perpendicular to it. Then (II. 12.) the square of AB is equal to the
squares of BD and AD, together with twice the rectangle BD.DE; and (II. 13.) the square of AC and twice the rectangle CD.DE, or BD.DE, are together equal to the squares of CD and AD, or of BD and AD. Add these equals together, and the squares of AB and AC, together with twice the rectangle BD.DE, are equal to twice the squares of BD and AD, together with twice the rectangle BD.DE. Take away twice the rectangle BD.DE, and