Sidebilder
PDF
ePub

с

D

D

A

E

B

E

Through a given point A in the circumference of the given circle ABC, let a chord AC be drawn, and in AC or its continuation, let D be taken, so that the rectangle AC.AD may be equal to a given space: the point D is always in a straight line given in position, however the point C may vary its position on the circumference.

Draw the diameter AB, and to AB and the side of a square equal to the given space, take AE a third proportional, so that the rectangle AB.AE may be equal to the given space ;-through E draw DE perpendicular to AB: DE is the locus of D.

Join BC. The triangles ABC, ADE have the angle A common, and the angles A.CB, AED right angles: therefore AB : AC :: AD : AĒ; and hence (VI. 16) the rectangle AC.AD is equal to AB.AE, that is (const.) to the given space. In the same manner, it may be shown, that wherever D is taken in the straight line DE, the rectangle AC.AD is equal to the given space.

Therefore (APP. II. def. 1) DE is the locus of and its position is given from that of the point E.* Schol. If, on the contrary, the point A and the straight line DE were given, and if, in a revolving straight line AD, or its continuation, the point Cwere taken, so that the rectangle DA.AC should be always equal to the same given space; it would be shown in a similar manner, that the locus of C would be a given circle ACB.

D;

* In discovering loci, as well as in other investigations in geometry, the student is assisted by what is termed geometrical analysis ; of the nature of which it may be proper here to give some explanation.

If, in the foregoing proposition, instead of being informed that the locus is a straight line, we were required to find what the locus is, we might proceed in the following manner: Let D be any point in the required line, so that the rectangle AC. AD is equal to the given space; and having drawn the diameter AB, find E so that the rectangle AB.AE may be equal to AC. AD, and therefore E a point in the required line; and join DE, BC. Then (VI. 16) AB : AC :: AD : AE. Hence (VI. 6) the triangles DAE, BAC, having the angle A common, are equi. angular; and therefore AED is equal to ACB, which is a right angle. The point D is therefore in a perpendicular passing through E; and in the same manner it would be shown, that any other point in the required line is in the perpendicular; that is, the perpendicular is its locus.

The investigation now given is called the analysis of the proposition, while the solution in the text is called the synthesis or composition. In analysis we commence by supposing that to be effected, which is to be done, or that to be true, which is to be proved; and, by a regular succession of consequences founded on that supposition, and on one another, we arrive at something which is known to be true, or which we know the means of effecting. The synthesis then commences with the conclusion of the analysis, and retraces its several steps, making that precede which before followed, till we arrive at the required conclusion. From this it appears that analysis is the instrument of investigation; while synthesis affords the means of communicating what is already known; and hence, in the Elements Cuclid, the synthetic method is followed throughout. What is now said will receive farther illustration from the solution of the following easy problem. Given the perimeter and angles of a triangle, to construct it.

Analysis.—Suppose ABC to be the required triangle, and produce BC both ways, making BD equal to BA, and CE to CA: tben DE is given, for it is equal to the sum of the three sides AB, BC, CA; that is, it is equal to the given perimeter. Join AD, AE. Then (1. 5) the angles D and DAB are equal, and therefore, each of them is half of ABC, because (I. 32) ABC is equal to both. The angle D therefore is given; and in the same manner it may be shown that E is given, being half of ACB. Hence the triangle ADE is given, because the base DE, and the angles D,

Prop. XI. THEOR.— The locus of the middle point of a straight line given in magnitude, and subtending a right angle given in position, is the circumference of a circle given in magnitude and position.

Let ABC be a right angle given in position, and let there șe a straight line of given length subtending it: then, while that straight line changes its position in such a manner that its extremities continue always on AB, BC, its point of bisection, describes the circumference of a circle.

From Boas centre, with the half of the given straight line as radius, describe a circle: it is the locus required.

Draw BD, any radius of that circle: from D as centre, with DB as radius, describe an arc cutting BA in A, and draw ADC. Then the angles BAD, ABD are equal ; and (I. ax. 3) their complements DBC, DCB are equal. Hence DC is equal to DB, and therefore to DA; and AC is equal to the given line, since DB is equal to its half. Its point of bisection, also, is on the circumference, since AD, DC are equal.*

A

D

B

[ocr errors]

B E are given: and ADE being given, ABC is also given, the angle DAB being equal to D, and EAC equal to E.

Composition.---Make DE equal to the given perimeter, the angle D equal to the half of one of the given angles, and E equal to the half of another; draw AB, AC, making the angle DAB equal to D, and EAC to E, ABC is the triangle required. * For (1. 6) AB is equal to BD, and AC to CE. To these add BC, and the three AB, BC, CA are equal to DE, that is to the given perimeter. Also (I. 32) the anglo ABC is equal to D and DAB, and is therefore double of D, since D' and DAB are equal. But D is equal to the half of one of the given angles ; therefore, ABC is equal to that angle: and, in the same manner, ACB may be proved to be equal to another of the given angles. ABC therefore is the triangle required, since it has its perimeter equal to the given perimeter, and its angles equal to the given angles.

It is impossible to give rules for effecting analyses, that will answer in all cases. It may be stated, however, in a general way, that when sums or differences are concerned, the corresponding sums or differences should be exhibited in the assumed figure; that in many cases remarkable points should be joined ; or that through them lines may be drawn perpendicular or parallel to remarkable lines, or making given angles with them: and that circles may be described with certain radii, and from certain points as centres; or touching certain lines; or passing through certain points. Some instances of analysis will be given in subsequent notes : and the student will find it useful to make analyses of many other propositions, such as several in this ppendix, both in the first book, and in the others.

* The analysis of this proposition is easily obtained from proposition A of the second book. For, since AB + BC, or AC2, or (II. 4, cor. 2) 4CD'=2CD2 + 2BD%; take away 2CD-, and halve the remainder: then CD' =BD2; and therefore BD=CD, half the given line. Hence the distance from the point of bisection D to the point B is equal to half the given line. D, therefore, is always in the circumference of a circle described from B as centre, with half the given line as radius.

Schol. If the extremity C pass to the other side of the centre B, the middle point will describe one quadrant, while the extremity A descends to B, another while it continues to descend below B, and a third while it re-ascends to B.

PROP. XII. THEOR. If the base of a triangle be given in magnitude and position, and the vertical angle in magnitude, the locus of the centre of the inscribed circle is a given circle.

Let ABC be a triangle of which the base BC is given in magnitude and position, and the angle A in magnitude; the locus of D, the centre of the inscribed circle, is a given circle.

Join DB, DC: these (IV. 4) bisect the angles at B and C. Then, by drawing a straight B line through A, D, which would also bisect the angle A, it would appear that the angle BDC is equal to A and half the sum of the angles at the base, or to A and half its supplement. But A, and therefore its supplement, are given in magnitude; and hence BDC also given in magnitude; therefore (schol. 1, page 272, No. 2) the locus of D is a given circle.

PROP. XIII. THEOR. - If from two given points, two straight lines be drawn meeting one another, such that the square of one of them is to the difference between the square of the other and a given space, in a given ratio of inequality, the locus of their point of intersection is the circumference of a given circle.

Let A, B be the given points, and let C be another point such that AC, BC being drawn, the difference between the square of AC and a given space, has to the square of BC a given ratio of inequality: the locus of C is a given circle.

In AB produced, take D (VI. 10, schol.) so that AD has to BD the given ratio; and in AB find E so that the rectangle AB.AE may be equal to the given space; take DF a mean proportional between ED, BD : a circle described from D as centre, with DF as radius, is the locus of C.

Join CE, and through B, C, E describe a circle cutting AC in G; draw BG, and through ABG describe a circle meeting CB produced in H, and join AH, CD. Then because FD or CD is a mean proportional between ED, BD, ED : DC :: DC : DB. Hence (VÍ. 6) since the triangles EDC, BDC have a common angle, the angles DEC, DCB are equal.

But (III. 21) the angle DEC or BEC is equal to BGC, and this again (III. 22, cor.) to H: therefore the alternate angles H and DCB are equal, and AH, CD are parallel; wherefore (VI.2) AD: DB :: HC : CB, or (VI. 1) AD : DB :: HC.CB, or AC.CG : BC%. But AC.AG is equal to AB.AE, and therefore to the given space; and AC.CG is the difference between the square of AC and the rectangle AC.AG: the difference, therefore, between the square of AC and the given space, has to the square of BC the ratio of AD to DB, which is the given ratio.

Schol. 1. The circle FC is (Schol. page 265, No. 3) the locus of straight lines drawn from E, B, in the ratio of EF to FB, or, as is easily shown, of ED to BD. It is plain also, that if the given ratio be that of equality, the difference of the squares of XC, BC is equal to the given space; and therefore (Schol. page 266, No. 4) the locus of C is a straight line.

Schol. 2. Hence we have the means of determining the locus of the intersection of tangents AC, BC to two given circles, having to one another a given ratio, that of m to n. For AD, BE being radii drawn to the points of contact, we have the square of AC to the square of BC, that is, CD— AD? : CE - EB? :: ma : n?. Take a space S such that S : EB? :: m: no. Then CD2 -ADP+S: CE? :: m : na; that is, the difference between the square of CD and the given space ADP-S, has to the square of CE a given ratio; and, therefore, if m, n be unequal, the locns of C is a given circle; if they be equal, it is a straight line. It is plain that if the circles be diminished to points, this becomes the same as the third locus mentioned in the scholium, page 265.

: Prop. XIV. THEOR. If two straight lines in a given ratio, and containing a given angle, be drawn from a given point, and if one of them terminate in a given circumference, the other will terminate in another given circumference.

Let A be a given point, and AB a straight line drawn from it to any point B in the circumference of a given circle: if another straight line be drawn through A, making with AB a given angle, and having to it a given ratio, its other extremity is in a given circumference.

Join A and B with D, the centre of the given circle, and make the angle DAE equal to the given angle; make ÅD to AE in the given ratio; and make the angles EAC, AEC respectively equal to BÅD, ADB : the circle of which is the centre, and EC the radius, is the required locus. For since AD, which is given, has to AE a given ratio, AE is given; and (VÍ. 4) AD : AB :: AE : AC; and, alternately, AD: ÅE :: AB : AC; therefore, AB has to AC the given ratio. But

the angles BAD, CAE being equal, add DAC to each ; then BAC is equal to DAE, that is, (const.) to the given angle. In the same manner, it would be shown, that any other straight lines drawn through A, one to each circumference, and inclined at the given angle, would be in the given ratio. Therefore the circle of which E is the centre, and EČ the radius, is the locus of C.

Prop. XV. THEOR.—Two concentric circles being given, if from any point A in the circumference of one of them, and from another point D, straight lines be drawn to B, C, the extremities of any diameter of the other circle, so that the squares of AB, AC, DB, DC are together double of the square of DA, the locus of D is a straight line given in position.

Draw AEF through the common centre, and to the radii AE, EF take EG a third proportional; draw GD perpendicular to GÀ: it is the required locus.

Join ED. Then (II. A) AB'+AC2 + DB2+DCP=2AE +2DEP+4ECʻ= 2AE? + 2DE? +4AE.EG. But (II. 12) AD? = AEP+ DEP+ 2AE.EG; wherefore AB'+ AC? + DB2+DCP=2AD”; and the same may be shown to hold, wherever D is taken in DG. Therefore, DG is the required locus.

D

A

E E

F

PROP. XVI. PORISM.—A point may be found such that if, to any straight line whatever passing through it, perpendiculars be drawn from three given points, one of these perpendiculars is equal to the other two.

Let A, B, C be three given points: join BC and bisect it in D; join AD, and take DE a third of it: if any straight line whatever FEGH 'be drawn through E, the perpendicular AG, drawn from one of the points A to FH, is equal to the two perpendiculars BF, CH, drawn from the points on the other side o FH.

Draw DK perpendicular to FH, and through D draw LDM parallel to FH, and meeting the perpendiculars LB, HC in L, M. Then, in the similar triangles AGE, DKE, AE :-AG :: DE : DK; therefore, because AE is double of DE, AG is double of DK. Now, in the equiangular triangles DLB, DMC, BD, DC are equal, and therefore BL, CM are also equal. To each add CH: then LB, CH are equal to HM or KD. Add again FL; then BF, CH are equal to twice DK, that is, to AG.

K

L

EG

B

[ocr errors]
« ForrigeFortsett »