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 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.

F

G

H

A

B

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 the square of AE, which is common to both; therefore the remaining rectangle CF.FA is equal (I. 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

E

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.

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+26; and, multiplying this by itself we get for the area of the square of that line, a2+4ab4b2, or a 2 +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 AB 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.

*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 AE perpendicular to AB, making it equal to the half of AB, and producing it through A; then, to make EF equal to the distance from E to B, and AH equal to AF. It is plain that BD

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 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 lastmentioned 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 and 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 and 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 B equal (I. 47) to the squares of BD, DA, and the

C D

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 EB, 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.

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. of a

PROP. XIII. THEOR.-In any triangle, the square 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 AB, BC, 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 of AD: therefore the squares square 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 (1. 47) to the squares of BD, DA, and the square of AC to the squares of AD, DC: therefore B 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.

D

C

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 B square of BC. Therefore, in any triangle, &c.*

* 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 AB2+BC take AC2; 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 AD 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. Thus, 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 A E D B C equal to BC. Then AD or DC is half the sum, and ED or DB half the difference of AB and BC; and

AB the greater is equal to the sum of AD and DB, and BC the less is equal to the difference of DC and DB.

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.

H

B

F

G E

C

D

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 (I. 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 BEEF 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.

C

Let ABC be a triangle, of which the side AB is bisected in D, and let DC be drawn to the opposite angle; the squares of AC and CB are together double of the squares of AD and DC.

A D B

If DC be perpendicular to AB, the squares of AC and CB are equal (I. 47) to the squares of AD and DB and twice the square of DC, or to twice the square of AD and twice the square of DC, because AD, DB are (L. 47, cor. 5) equal.

But if CD be not perpendicular to AB, draw CE perpendicular to it. Then (II. 12) the square of AC is equal to the squares of AD and DC, together with twice the rectangle AD.DE; and (II. 13) the square of CB and twice the rectangle BD.DÉ, or AD.DE, are together equal to the squares of BD and CD, or of AĎ and DC. Add these equals together, and the squares of AC and CB, together with twice the rectangle AD.DE, are equal to twice the squares of AD and DC, together

A DE B

* This is a particular case of the 25th proposition of the sixth book.

with twice the rectangle AD.DE. Take away twice the rectangle AD.DE, and the squares of AC and CB are equal to twice the squares of AD and DC. Therefore the squares, &c.*

PROP. B. THEOR.-The squares of the diagonals of a parallelogram are together equal to the squares of its sides.

Let ABCD be a parallelogram, of which the diagonals are AC and BD: the sum of the squares of AC and BD is equal to the sum of the squares of the sides AB, BC, CD, DA.

E

B

Let AC and BD intersect each other in E. Then, in the triangles AED, CEB, the angles AED, ADE are respectively equal (I. 15, and 29) to CEB, CBE, and (I. 34) the side AD is equal to CB: therefore the sides AE, ED are respectively equal (I. 26) to CE, EB. Since, therefore, BD is bisected in E, the squares of AB, AD are together equal (II. A) to twice the squares of AE, EB. But BC and CD are respectively equal (I. 34) to AD and AB; and therefore the squares of AB, BC, CD, and DA are together equal to four times the squares of AE and EB, that is (II. 4, cor. 2), to the squares of AC and BD, since AC is double of AE, and BD of BE. Therefore, the squares, &c.t

D

Cor. The diagonals of a parallelogram bisect one another.

C

PROP. C. THEOR.-The sum of the squares of the sides of a trapezium is equal to the sum of the squares of the diagonals, together with four times the square of the straight line joining the points of bisection of the diagonals.

B

Let ABCD be a trapezium, having its diagonals AC, BD bisected in E and F, and let EF be joined: the squares of AB, BC, CD, DA are together equal to the squares of AC, BD, together with four times the square of EF.

Join AF, FC. The squares of AB, AD are together equal (II. A) to twice the sum of the A squares of DF and AF; and the squares of BC, CD are equal to twice the sum of the squares of DF and CF. Add these equals together,

E

F

and the sum of the squares of AB, BC, CD, DA is equal to four times the square of DF, together with twice the sum of the squares of AF and CF. But twice the squares of AF and CF are equal

*Hence, if the sides of a triangle be given in numbers, the line AD can be computed. Thus, if AC=11, AB=14, and BC=7, we have AC+BC-121+49= 170, and 2AD-98. Then 170-98-72, the half of which is 36; and 6, the square root of this, is CD.

Hence, if we have the sides and one of the diagonals of a parallelogram in numbers, we can compute the remaining diagonal. Thus, if AB, DC be each =9, AD, BC each=7, and AC=8, we have AB +BC2+CD2+DA2-81+49+81+49 260, and AC2=64. Taking the latter from the former, and extracting the square root, we find BD=14.