ich is the same thing," the aggregate of the

in and difference of two fright es e ad a

the squares of those nes" QED.

9. Let AC=CB=4, BD=2, then was 40=
)=a+.

- =4+4arin

> this, and the sum is 4 -4 -2

7)2=aa+2 ar+17, add as to this, and a bergue rx; now the former of these soms is dui f at is 4 aa+4 ar+2zz=220s-2u-z: a. the eproduced line 2a+1, together with the spare of luced 1, is double the square of a half the ane, and of a+d the line made up of the half and the part Q. E. D.

p. 11. This proposition is impossible by numbers, fue number that can be so divided, that the product of to one part, shall equal the square of the other part, may however be approximated to as follows. B=2a, AH=1, HB=y, then by the prociez z+3= y=xx; from the first equation y=24—z, this valve stituted for y in the latter equation, we shall have =x1, or xx+2a1=4 aa, this solved by Art. 97. part. 3. 1 /5aa-a, and y=(2 a—1=2 a—/5 a—e=,34— which is the same z=1.936065, &c. xe, and y=

c. xa.

op. 12. Let AB=a, BC=b, CD=z. and AD=z; (47. 1.) AB=BD.'+DA2=b+2,2+22=

is, the square of AB, the side subtending the obture greater than the sum of the squares of CB and AC, the aining the obtuse angle, by (2 bz, twice the rectangle Q. E. D.

rop. 13. Let AB=a, CB=b, AC=c, AD=d, BD=m, then the first case of this proposition is proved as

, bb+mm=2bm+nn (7.2.) To each of these equals add

168. Prop. 6. Let AC=CB=a, BD=x, then will AB=2a, and AD=2a+x; then the rectangle contained by AD and DB will be 2a+x.x=2ax+xx, to these equals let aa (the square of half AB) be added, and 2 a+x.x+aa==(aa+2 ax+xx=) a +x)2 ; that is, the rectangle contained by the line produced and part produced, together with the square of half the line bisected, is equal to the square of the line made up of the half, and part produced. Q. E. D.

Cor. Hence, if three lines x, a+ x, and 2a+x be arithmetically proportional, the rectangle contained by the extremes (x.2a+x) together with the square of the common difference a, (or aa) is equal to (a+x) the square of the middle term.

169. Prop. 7. Let AB=s, AC-a, CB=b, then s=a+b, and ss=(a+b)2=aa+2 ab+bb=) 2 ab+bb+aa, to these equals add bb, and ss+bb= (2 ab+2 bb+aa=2.a+b.b+aa=) 2sb+aa ; that is, the square of the whole line, (or ss) and the square of one part b (or bb,) is equal to twice the rectangle contained by the whole s, and that part b, (or 2 sb,) together with (aa) the square of the other part. Q. E. D.

Cor. Hence, because 2 sb+au=ss+bb, by taking 2 sb from both, we have aa≈ss—2 sb+bb ; that is, the square of the difference of two lines (s) AB and (b) CB, is less than the sum of the squares of (s) AB and (b) CB, by twice the rectangle (2sb) 2.4B.CB contained by those lines.

170. Prop, 8. Let AB=s, AC=a, CB=b, then s=a+b, or a=s—b, ·: aa=(s—b)2==) ss — 2 sb+bb, to each of these equals add 4 sb, and 4 sb+aa=ss+2sb+bb=s+b2; that is, (4 sb, or) four times the rectangle contained by the whole s, and one part b, together with (aa) the square of the other part a, is equal to (s+bor) the square of the straight line made up of the whole s, and the part b. Q. E. D.

2

171. Prop. 9. Let AC=CB=a, CD=x, then will the greater segment AD=a+x, and the less segment DB=a-x.

Then a+x=aa+2 ax+xx

And a-xl-aa-2 ax + xx

The sum of both =2aa+2xx=2.aa+xx

That is, aa)2+a-x2=2.aa+xx, or the sum of the squares of the unequal parts (a+x and a-x) is equal to double the square of the half a, and of the part x between the points

of section; or, which is the same thing, "the aggregate of the squares of the sum and difference of two straight lines a and x is equal to double the squares of those lines." Q. E. D.

172. Prop. 10. Let AC=CB=a, BD=x, then will AD= 2 a+x, and CD=a+x.

Now 2a+x=4 aa+4 ax + xx

Add xx to this, and the sum is 4 aa+4 ax + 2 xx

Also a+xaa+2 ax+xx, add aa to this, and it becomes 2 aa+2 ax + xx; now the former of these sums is double of the latter, that is 4 aa+4 ax+2xx=2.2 au + 2 ax + xx; or, the square of the produced line 2a+x, together with the square of the part produced x, is double the square of a half the line, and the square of a+d the line made up of the half and the part produced. Q. E. D.

173. Prop. 11. This proposition is impossible by numbers, for there is no number that can be so divided, that the product of the whole into one part, shall equal the square of the other part; the solution may however be approximated to as follows:

Let AB=2 a, AH=x, HB=y, then by the problem x+y= 24, and 2 ay=xx; from the first equation y=2a-x, this value being substituted for y in the latter equation, we shall have 4 aa-2 ax=xx, or xx+2 ax=4 aa, this solved (by Art. 97. part. 3.) gives x=±√5 aa—a, and y=(2a-x=2a- /5 aa—a=) 3a√5 aa, or which is the same x=1.236068, &c. ×a, and y= .763931, &c. x a.

174. Prop. 12. Let AB=a, BC=b, CD=x. and AD=z; Then (47. 1.) AB'=BD\'+DA)2=b+x)2+zz=

2

2

2 bx * *

That is, the square of AB, the side subtending the obtuse angle, is greater than the sum of the squares of CB and AC, the sides containing the obtuse angle, by (2 bx) twice the rectangle BC, CD. Q. E. D.

175. Prop. 13. Let AB=a, CB=b, AC=c, AD=d, BD=m, DC=n; then the first case of this proposition is proved as follows:

First, bb+mm=2bm+nn (7.2.) To each of these equals add

dd, and bb+mm+dd=2 bm+dd+nn. But aa=mm+dd, and .cc=dd+nn (47. 1.) ·.·' if aa and cc be substituted for their equals in the preceding equation, we shall have bb+aa=2 bm+cc, or cc=bbaa-2 bm.

Second case. Because aa=cc+bb+2 bn (12. 2.) add bb to both sides, and aa+bb=cc+2 bb+2 bn, but bm=bn+bb (3. 2.) ... 2 bm=2bn+2 bb; substitute 2 bm for its equal in the preceding equation, and aa+bb=cc+2 bm, or cc==aa+bb−2 bm.

Third case. Here the points C and D coincide, ·.· b=m; wherefore since cc+bb=aa (47. 1.) to each of these equals add bb, and cc+2 bb=aa+bb, or cc=aa+bb-2 bb, which corresponds with the former cases since 2 bb here answers to 2 bm there. Wherefore cc is less than aa+bb by 2 bm, or AC)2{ AB)2+BC)2 by 2, CB, BD. Q. E. D.

176. Prop. 14. By help of this problem any pure quadratic equation may be geometrically constructed. To construct an equation is to exhibit it by means of a geometrical figure, in such a manner, that some of the lines may express the conditions, and others the roots of the given equation.

EXAMPLES.-1. Let xab be given to find x by a geome trical construction. See Euclid's figure.

Make BE-a, EF-b, then if BF be bisected in the point G, (10. 1.) and from G, as a centre, with the distance GF, a circle be described, and EH be drawn perpendicular to BF from the point E, (11. 1.) it is plain that EH will be the value of x, For by the proposition EH)2=BE× EF=ab, but by hypothesis x2= ab, ··· EH)2=x2, and EH=x; which was to be shewn.

2

But the root of x' is either +x or -x, now both these roots may be shewn by the figure, for if EH=+x, and EH be produced through D till it meet the circumference below BF, the line intercepted between E and the circumference will =-x, for in this case BEX EF=—x× -x=+x2, as before.

2. Let x=36 be given, to find the value of x.

Here, because 36=9×4, make BE=9, EF=4; then pro

ceeding as before, EH2=9×4=36, and EH=6.

3. Let x2=120=12 × 10 be given.

Make BE=12, EF=10, then EH120, and EH= (120) 10.95445=x.

4. Let x=3 be given.

Here 3=3×1; make BE=3, EF=1, then EH1=3, and EH 1.73205=x.

ON THE THIRD BOOK OF EUCLID'S ELEMENTS.

177. This book demonstrates the fundamental properties of circles; teaching many particulars relating to lines, angles, and figures inscribed; lines cutting them; how to draw tangents; describe or cut off proposed segments, &c.

178. Def. 1. "This," as Dr. Simson remarks, "is not a definition, but a theorem;" he has shewn how it may be proved: and it may be added, that the converse of this theorem is proved in the same manner.

179. Def. 6 has been already given in the first book, and might have been omitted here, (see Art. 74.) Def. 7 is of no use in the Elements, and might likewise have been omitted. In the figure to def. 10 there is a line drawn from one radius to the other, by which the figure intended to represent a sector of a circle is redundant: that line should be taken out.

180. Prop. 1. Cor. To this corollary we may add, that if the bisecting line itself be bisected, the point of bisection will be the centre of the circle.

181. Prop. 2. This proposition is proved by reductio ad absurdum. The figure intended to represent a circle is so very unlike one, that it will hardly be understood, the part AFB of the circumference being bent in, in order that the line whichjoins the points A and B may fall (where it is impossible for that line to fall) without the circle.

The demonstration given by Euclid is by reductio ad absurdum. Commandine has proved the proposition directly; his proof depends on the following axiom which we have already given, viz. "If a point be taken nearer the centre than the circumference is, that point is within the circle." Thus,

182. Let AB be two points in the circumference ACB, join AB, this line will fall wholly within the circle. Find the centre