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 arc 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 are 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 : 4× 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 circumference BHDK :: XZ: the whole circumference ZXE. Q. E. D.

D

E

C

K

X

IN

B

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+DA2.

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

2

2

:: BA2 + AC2=BE? +EC2+2.EA. But since BC is divided into two equal parts in D, and into two unequal parts in E, BE

+ EC22. BD2 +

DE (9.2.) BA2+

B

D

E

AC2=2.BD2+DE+EA2. But DE+EA2=DÃ2, (47. 1.)

=

2. DE2 + EA2 = 2. DA3, ·· AB2 + AC)2 = (2. BD)2 + 2. DE+EA 2.BD2 + 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 AC)2 + BD2=AB)2+BC¦2+CD2+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+EД2, and BC2+CD2 =(2.BE+EC (Art. 239.) =) 2.BE)2+EA; BA+AD +DC+CB 4.BE+EA=(since 4.BEBD, and 4.EA

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

Cor. Hence the diameters of a parallelogram bisect each other.

2

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

angles (22.3.) ·.· BAD+BGD=BAD+BCD, take away the common angle BAD, and BGD=BCD, the interior and opposite equal to the exterior which is impossible (16. 1.) ·.· the fourth point cannot fall without the circle, in the same manner it may be shewn that it cannot fall within it, it must fall on the circumference at. C. Q. E. D.

Cor. If one side BC of a quadrilateral figure inscribed in a circle be produced, the exterior angle DCG=the interior and opposite BAD; for DCG+DCB=two right angles (13. 1.) and BAD+DCB=two right angles (22.3.) ·.· DCG+DCB=BAD +DCB, take away DCB, and DCG=BAD.

243. If the vertical angles of several triangles described on the same base, be equal to each other, and the circumference of a circle pass through the extremities of the base, and one of the vertical angles, it shall likewise pass through all the others.

E

Let ACB, ADB, AEB, &c. be the several equal vertical angles of triangles described on the common base AB, if a circle pass through A, B, and C, it shall likewise pass through the remaining points D, E, &c. Take any point K in the circumference on the other side of AB, and join AK, KB, then will ACB+ AKB==2 right angles, (22.3.); but ADB=AEB ACB by hypothesis, each of the angles AEB, ADB together with AKB =2 right angles, (Art. 241.) the angles E and D are in the circumference. Q. E. D.

K

B

243. If two straight lines cut one another, and the rectangle

contained by the segments of one of them, be equal to the rectangle contained by the segments of the other, the circum. ference which passes through three of the extremities of the two given straight lines, shall likewise pass through the fourth. Let AB and CD cut each other in E, so that AEX EB= CEX ED, the circumference ACB, which passes through the three points A, C, and B, shall likewise pass through the fourth D.

For if not, let the circumference, if possible, cut CD in some other point G; then since A, C, B, and G, are in the circumference, the rectangle AEX EB= CEX EG (35.3.) but AE× EB= CEX ED by hypothesis; CE× EG=CEXED, ·.· EG=ED, the less the greater, which is absurd; therefore G is not in the

=

C

A

P

E

D

B

K

H

L

F

circumference; and in the same manner it may be shewn, that no other point in CD, except D, can be in the circumference. Q. E. D.

• Join CB, and through K draw KP parallel to FC; then since the angle AEC=ABC+ DCB (32, 1.) if the angular point E were in the circumference, it is plain that it would be subtended by an arc equal to AC+ DB; and consequently, if E were at the centre, it would be subtended by an arc equal to AC+ DB

to

2

(20.3.) Again, if KC be joined, it may be proved (29.1. and 26. 3.) that CP and HK are equal, but the arc BDP=(CPB—CP=) CPB-HK; and since the angle BKP=BFC, and BKP is subtended by the arc BDP, if BKP were in the circumference, it would be subtended by an arc equal to BDP; but if it were at the centre, BKP would be subtended BDP CPB-HK by an arc = 2 2

(20. 3.) that is)

by what has been shewn. And since an angle is measured by the subtending arc described about the angular point as a centre (Art. 262.) it follows, that if two straight lines AB, CD cut one another within a circle, the angle AEC is measured by half the sum of the subtending arcs AC and BD, and (by similar reasoning) the angle AED is measured by half the sum of the arcs APD, CKB. But if two straight lines CF, FB cut one another without the circle, the angle BFC is measured by half the difference of the intercepted arcs CPB and HK; this is connected with Art, 261, 262.