EXAMPLES.-1. To find a mean proportional between 1 and 16.

Here a=1, b=16, and x=√ab=√16=4, the mean required.

2. To find a mean proportional between 15 and 11.

Here a=15, b=11, and x=/ab= √15 x 11=√165= 12.845232578, the required mean.

Make BC

231. Prop. 19. By the help of this useful proposition we are enabled to construct similar triangles, having any given ratio to each other; thus, let it be required to make two similar triangles, one of which shall be to the other as m to n. =m, BG=n, and between BC and BG find a mean proportional ·EF (13. 6.) upon BC and EF make similar triangles ABC, DEF (18.6.) then by the present proposition m:n :: ABC : DEF.

EXAMPLES.-1. Let the side of a triangle ABC, viz. BC=8, it is required to make a similar triangle, which shall be only half as large as ABC.

Bisect BC in G (10. 1.) and between BC and BG find a mean proportional EF (13. 6.); if a triangle be made on EF similar to ABC, it will be half of ABC. Thus BC being=8, BG will=4, and BC× BG= √8×4=√32=5.656854=EF.

2. Let EF=8, required the side of a triangle five times as large as DEF, and similar to it? Ans. √8×40√320= 17.88854382 the side required.

232. Prop. 20. Hence, if the homologous sides of any two similar rectilineal figures be known, the ratio of the figures to one another may be readily obtained, namely, by finding a third proportional to the two given sides: for then, the first line will be to the third, as the figure on the first, to the similar and similarly described figure on the second, as is manifest from the

tween two given straight lines a and b; this may however be done algebraically by the following theorems.

second cor. to the proposition. Hence also any rectilineal figure may be geometrically increased, or decreased in any assigned ratio. Thus, let it be required to find the side of a pentagon one fifth as large as ABCDE, and similar to it; find a mean proportional between AB and AB (13. 6.) let this be FG, and upon FG describe the pentagon FGHKL similar and similarly situated to ABCDE (19. 6.) then will the former be of the latter. Again, let it be required to find the side of a polygon 3 times as large as ABCDE, and similar to it?

Thus ABXAB the side required.

233. Prop. 22. By means of this proposition, the reason of the algebraic rule for multiplying surd quantities together, may be readily shewn. Thus, let it be required to prove that ax

bab, First, since unity: the multiplier: the multiplicand the product; therefore, in the present case, 1: √ɑ:: √b:

ax b the product, but by the proposition (12: a2:: √b2: √a2× √b, that is) 1:a :: b: ab=the square of the product, wherefore ab=the product.

234. Prop. 23. Hence, if two triangles have one angle of the one equal to one angle of the other, they will have to each other the ratio compounded of the ratios of the sides about their equal angles; this will appear by joining DB and GE; for the triangles DBC, GEC have the same ratio to one another, that the parallelograms DB and GE have (1.6.). Also it appears from hence, that parallelograms and triangles have to one another respectively, the ratio compounded of the ratios of their bases and altitudes.

235. Prop. 30. This proposition has been introduced under a different form in another part of the Elements, (viz. 11. 2.) there, we have merely to divide a straight line, so that the rectangle contained by the whole and the less segment, may equal the square of the greater; we have to determine the properties of a figure, but the idea of ratio does not occur; here we are to divide a line, so that the whole may be to the greater segment, as the greater segment is to the less, and the idea of figure has no place; but our business is solely with the agreement of certain ratios. I do not recollect a single reference to this proposition in any subsequent part of the Elements, except in some of the books which are omitted.

236. Prop. 31. What was proved of squares in prop. 47. b. 1.

is here shewn to be true of rectilineal figures in general; and the same property belongs likewise to the circle, and to all similar curvilineal and similar mixed figures, with respect to their diameters or similar chords; but the six former books of Euclid's Elements do not furnish us with sufficient principles to extend the doctrine beyond what is proved in this proposition. We are here taught how to find the sum and difference of any two similar rectilineal figures, that is, to find a similar figure equal to the said sum or difference. See the observations on 47. 1.

237. Prop. 33. This useful proposition is the foundation of Goniometry, or the method of measuring angles. If about the angular point as a centre with any radius, a circle be described, it is here shewn, that the arc intercepted between the legs of the angle will vary as the angle it subtends varies; thus, if the angle be a right angle, the subtendiug arc will be a quadrant (or quarter of a circle); if it be half a right angle, the subtending arc will be half a quadrant; if it be equal to two right angles, the subtending are will be a semi-circle; and if it equal four right angles, the subtending arc will be the whole circumference. Now if two things vary directly as each other, it is plain that the magnitude of one, will always indicate the contemporary magnitude of the other; that is, it will be a proper measure of the other. Such then is the intercepted arc described about an angle, to that angle; and therefore if the whole circumference be divided into any number of equal parts, the number of those parts intercepted between the legs of the angle, will be the measure of that angle. It is usual to divide the whole circumference into 360 equal parts called degrees, to subdivide each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds, &c. wherefore, if an angle at the centre be subtended by an arc which consists of suppose 30 degrees, that angle is said to be an angle of 30 degrees, or to measure 30 degrees; if it be subtended by an arc of 45 deg. 54 min. the angle is said to measure 45 deg. 54 min. &c.

238. Hence the whole circumference which subtends four right angles at the centre (Cor. 1. 15. 1.) being divided into 360 degrees, a semicircle which subtends two right angles will contain 180 degrees, and a quadrant which subtends one right angle will contain 90 degrees, wherefore two right angles are said to measure 180 degrees, one right angle 90 degrees, &c. and note,

degrees, minutes, and seconds, are thus marked ",",", thus 12 degrees, 3 minutes, 4 seconds, are usually written 12o, 3′, 4", &c.

H

238. B. Hence, if about any angular point C as a centre, several concentric circles be described, cutting CA and CB in the points X, Z, A, B, the arc AB, will be to the whole circumference of which it is an arc, as the arc XZ is to the whole circumference of which it is an arc. Produce BC to D, and through C draw HK at right angles to DB (11. 1.); then BA:: BH:: angle BCA: angle BCH (13. 6.) ... BA : 4x BH:: angle BCA: 4x angle BCA, (15. 5.); that is, BA is to the whole circumference BHDK, as the angle BCA, is to four right angles; in the same manner it is shewn, that XZ is to the whole circumference ZXE as the same angle BCA to four right angles; wherefore AB : the whole

D

E

K

N

B

circumference BHDK :: XZ: the whole circumference ZXE. Q. E. D.

239. Hence also, if the circumferences of these two circles be each divided into 360 degrees, as above (Art. 236.) AB will contain as many degrees of the circumference BHDK, as XZ does of the circumference ZXE.

AN APPENDIX TO THE FIRST SIX BOOKS OF

EUCLID.

Containing some useful propositions which are not in the Elements.

240. If one side of a triangle be bisected, the sum of the squares of the two remaining sides is double the square of half the side bisected, and of the square of the line drawn from the point of bisection to the opposite angle.

Let ABC be a triangle, having BC bisected in D, and DA drawn from D to the opposite angle A; then will BA)2+AC)2= 2.BD+DA.

Let AE be perpendicular to BC, then because BEA is a right angle, AB=BE)2+EA2; and AC)2=CE)2+EA)2, (47. 1.)

2

AC2=2.BD+DE+EA. But DE2+EA2=DA2, (47.1.)

2

2. DE2 + EA2 = 2. DA2, ··· AB2 + AC)2 = (2. BD)2 + 2. DE)2+EA2= 2.BD)2 + 2.DA2=) 2.BD2+DA; and the same may be proved if the angle at C be obtuse, by using the 10th proposition of the second book instead of the 9th. Q.E.D. 241. In any parallelogram, the sum of the squares of the diameters, is equal to the sum of the squares of the sides.

Let ABCD be a parallelogram, AC and BD its diameters, then will AC2+BD2=AB)2+BC)2+CD\2+DA2.

and a side opposite to the equal angles in each, equal, viz. AD= BC (34. 1.) ·.· AE=EC and DE=EB (26.1.); and because BD is bisected in E, BA2+AD2=2.BE2+EA2, and BC2+CD2 =(2.BE+EC (Art. 239.) =) 2.BE+EA; BA+AD® +DC+CB24.BE+EA=(since 4.BEBD, and 4.EA

=AC. by 4.2) BD2+AC2. Q. E. D.

2

Cor, Hence the diameters of a parallelogram bisect each

other.

242. If the sum of any two opposite angles of a quadrilateral figure be equal to two right angles, its four angles will be in the circumference of a circle.

Let ABCD be a quadrilateral figure, having the sum of any two of its opposite angles equal to two right angles, and let a circle be described passing through the three points, A, B, D, (5. 4. and Art. 194.) I say the circumference shall likewise pass