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From the centre of CD, with a radius equal to the difference or sum of the radii of AB and CD, describe a circle; and from the centre of EF, with a radius equal to the difference or sum of the radii of AB and EF, describe a circle : then (App. II. 5.) find the centre of a circle touching the circles now described, and passing through the centre of AB: it will also be the centre of the required circle. The proof is obvious.

PROP. IX. PROB. Through a given point A, to describe a circle, touching one given straight line BC, and having its centre in another DC.

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If the lines be parallel, the radius will evidently be their perpendicular distance asunder, and the construction is obvious.

But, if they be not parallel, let them meet in C, and draw a straight line through and A; through A draw BD perpendicular to BC; from D as centre, at the distance DB describe a circle cutting CA in E; join ED; and through A draw AF parallel to ED meeting CD in F: F is the centre of the required circle.

Draw FG perpendicular to BC. Then, because the triangles CDE, CFA, and CDB, CFG are similar, ED:AF :: DC:FC, and BD : GF::DC: FC; whence ED: AF::BD: GF; and (V. 14.) AF, GF are equal, because ED is equal to BD. A circle therefore described from F as centre, with FG as radius, passes through A; and it touches BC, because the angles at G are right angles.

Schol. It is plain, that with the relations of the data in the diagram here employed, there may be two solutions, as either of the points E may be used. Should the circle described with DB as radius, merely touch AC, there would be but one solution : should it not meet CA, the solution would be impossible.

PROP. X. THEOR. If a chord of a given circle have one extremity given in position, and if a segment terminated at that extremity, be taken on the chord, produced if neces

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sary, such that the rectangle under the segment and chord, may be equal to a given space ; the locus of the point of section is a straight line given in position.

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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 ACB, AED right angles: therefore AB : AC : : AD : AE; 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 Ď 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 D; 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 C were 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.

* 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 equiangular ; 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

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 be a

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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 of Euclid, 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: then 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 (I. 5.) the angles D and DAB are equal, and therefore each of them is half of ABC, because (1. 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, 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 half of another; draw AB, AC, making the angle DAB equal to D, and EAC to E: ABC is the triangle required.

For (I. 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 angle 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 Appendix, both in the first book, and in the others,

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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 B as 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 Das 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.*

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 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 ADC is also given in magnitude : therefore (schol

. 1. page 272, No. 2.) the locus of D is a given circle.

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PROP. XIII. THEOR. IF from two given points, two straight lines be

The analysis of this proposition is easily obtained from proposition A. of the second book. For, since AB? + BC%, or AC?, or (II. 4. cor. 2.) 4CD : 2CD + 2BD? ; take away 2BD’, and halve the remainder : then CD2 = BD? ; 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.

drawn meeting one another, such that the square of one of them is to the difference between the

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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.

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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 (VI. 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: BB:: 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 272, 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 AC, BC is equal to the given space; and therefore (Schol. page 273, 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

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