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7 Th. If the base of a triangle be both increased and diminished by the difference of the two sides; four rectangles of the sides is to the rectangle of this sum and difference, as the square of the radius is to the square

of the sine of half the angle contained by the sides.

In the triangle ABC, let AB be greater than AC, and make AD equal to AB; join BD, and draw AE, CF perpendicular to it; with C, as centre, and CD radius, describe the semicircle GDH, cutting BD

B in K, BC in G, and meeting BC produced in H.

Because of the equal radii, BH=BC increased, and BG=BC diminished, by the difference of AB, AC. And since ABD is isosceles, and CF drawn from the centre, BD and KD are bisected in E, F. Wherefore DE-DF=EF; }(BD—DK)= BK. Wherefore DEDF={(BD-DK). And because in the triangle DAE, CF is drawn parallel to the base, AC : AD :: EF : FD (2, 6); and rectangles of the same altitude being as their bases, ACXAD : AD2 :: EFX ED : ED (1, 6); therefore 4ACXAD: AD2 : : 4EFXED: ED?, or alternately, 4ÁCXAD: 4EF XED :: AD2 : ED.

But since 4EF=2BK, 4EFXED=2BKXED, or 2EDXBK=BDX BK, or BHXBG (cor. 36, 3); therefore 4ACXAD:BHXBG :: AD2 : ED. Now AD: ED:: R: sin. EAC=sin. BAC (p. 1); therefore AD2 : ED2 :: R2 : (sin. JBAC). Therefore (11,5), 4ACXAD: BH XBG:: R2 : (sin. BAC). But 4ACXAD are 4 rectangles of the sides, and BHXBG is the rectangle of the base increased and diminished by the difference of the sides. Wherefore, if the base, &c.

Q. E. D. Cor. 2V ACXAD: VBGXBH :: R: sin. BAC.

8 Th. Four rectangles of two sides of a triangle is to the rectangle of the whole perimeter and the excess of the two sides above the base, as the square of radius is to the square of the cosine of half the angle contained by those sides.

Given the triangle ABC, of which BC is the base, NN and AB greater than AC.

With C, as centre, and radius CB, describe the circle BLM, meeting AC produced, in L, M; produce AL, so that AN=AB; let AD=AB; join BD, BN; draw CO perpendicular to BN, and AE to BD.

Now MN=AB+AC+CB, the perimeter; and LN=AB+AC—BC, the excess of the two sides above the base : and because BD is bisected in E, and DN in A, AE is parallel to BN, and perpendicular to BD; and the triangles DNB, DAE are equiangular. "Wherefore, since DN=2AD, BN=2AE, and BP 2B0=2RE, also PN=2AR.

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But since the triangles ARC, AED are equiangular, AC : AD:: AR: AE; and since rectangles of the same altitude are as their bases (1, 6) ACXAD: AD2 : : AŘXAE : AE” ; and alternately, ACXAD: ARXAE :: AD2 : AE2.

Moreover, 4ARXAE=QARX2AE=NPXNB=NMXNL. Therefore, 4ACXAD: MNXNL:: AD2 : AE?; but AD: AE::R: cos. DAE=cos. BAC: therefore 4ACX AD : MNXNL :: R2: (cos. BAC)2; and AD=AB, and MNVNL=the excess of the two sides above the base. Wherefore, four rectangles, 8

Q. E. D.

9 Th. In any triangle, the less of two given sides is to the greater, as radius is to the tangent of an angle greater than half a right angle; also, radius is to the tangent of this excess, as the sum is to the difference of the sides; which sum and difference have the ratio of the tangent of half the sum to that of half the difference of the angles at the base.

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1. In the triangle ABC, where AB is less than AC, AB : AC :: R: tangent of an angle exceeding half a right angle. Draw AD at right angles to BA, and make it equal to AC: join BD. If then BA be made radius, AD is the tangent of the angle ABD (a); which is the greater of the two acute angles, in the triangle ABD, and therefore exceeds half a right angle (b).

2. Radius is to the tangent of the excess just mentioned, as the bum is to the difference of the sides AB, AČ. Produce DA to E; make AE, AF each equal to AB; join BE, and parallel to it draw DG; through the point F join BG.

Then, because of the right angles at A, and that AF=AB, ABG is half a right angle; and DBG is the excess of ABD over that half

. And because of the parallels DG, BE (c), and that EBG is a right angle, the alternate angle BGD is also a right angle. Then, if BG be made radius, DG is the tangent of the angle DBG, which is the excess, &c. Also, since G is a right angle, and the verticals at F are halves, the sides GF and GD are equal: moreover, ED is the sum and DF is the difference of AB, AC. Then, in the similar triangles FBE. FGD, FB: FE :: FG : FD, and alternately, FB : FG:: FE: FD; also, jointly BG: GF, or GD :: ED: DF; therefore, radius is to the tangent, &c., as stated.

3. Which sum and difference have the ratio of the tangent of half the sum to that of half the difference of the angles at the base. This is demonstrated in prop. 4. Wherefore, in any triangle, &c.

Q. E. D. Recite (a) def. 6, P.T.; (6) P. 18, 1, and def. 3, P. T.; (c) p. 2, b. 6.

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10 Th. If from half the perimeter of any triangle the base be taken, and also each of the sides; the rectangle of the half and its excess above the base, is to the rectangle of the other two excesses, as the square of the radiu is to the square of the tangent of half the angle contained by the sides.

Given the triangle ABC: bisect its angles at A and B, by the straight lines AG, BG; B produce AG, AB, AC; bisect the exterior an н. gles at B and C by the lines BK, CK; draw the perpendiculars GD, GE, GF, which are equal radii of the inscribed circlé; draw also the perpendiculars KH, KM, KL; which are equal, because of the similar triangles.

Because the triangles AHK, ALK, are equiangular, and have AK common, AH is equal to AL; for the same reason BD is equal to BF, AD to AE, CE to CF, CM to CL, BH to BM.

And, because AD+BF+CE=AE+CF+BD=AE+CE+CM, taking away the parts shown to be equal, the remainders BF=BDE CM, ax. 3. Therefore AH+AL: perimeter, and AH=} p. BC therefore equals HD, and AC equals BD; and the excess of \ p. over BC=AD, over AB=HB, over AC=BD.

Hence the hypothesis is AHXAD : BHXBD:: R2 : (TIBAC)

Again, because the triangles BDG, KHB are similar, GD: BD :: BH: HK. Therefore GDXHK=BDXBH. But since AH:HK:: AD : DG. Therefore AHXAD: HKXDG :: ADXAD: DGXDG (p. 22 of b. 6); or AHXAD : BDXBH : : ADXAD: DGXDG. Now, if AD be made radius, DG will be the tangent of half the angle BAC. Wherefore, if from half, &c.

Q. E. D.

QUADRATURE OF THE CIRCLE.

Definition. A convex line is any arc of a circle, or any polygonal line, which has no re-entrant anglés, or inward inflections; and which a straight line cannot cut in more than two points.

Lemma. Any curve, or polygonal line, which passes round a convex line, from end to end, is longer than the convex line.

Let AMB be a convex line, enveloped from A to А B, by the polygonal line APMQB: the distance from A or B to M, through P or Q, is longer than the part of the convex line from A or B to M; because every point of the convex line, from A to M,

MD or from B to M, would fall within the triangle APM, or BQM, if the points M, P, A—M, Q, B, were join

E ed by straight lines (a).

B Q In like manner, it may be proved that the distance from P to Q, through CDE, is longer than the straight line PQ.

Therefore, any curve, or plolygonal line, which passes round, &c. Recite (a) p. 21 of b. 1.

Q E. D. Cor. 1. Hence the perimeter of any polygon inscribed in a circle is less than the circumference of the circle.

Cor. 2. And the perimeter of any polygon described about a circle exceeds the circumference of the circle.

1 Th. If from the greater of two unequal magnitudes its half be taken ; and from the remainder its half; and so on; there will remain at length, a magnitude less than the less of the two given magnitudes.

Let AB and C be two unequal magnitudes; of which AB is the greater : take DE a multiple of C, which is greater than AB; divide DE into parts DF, FG, GE, each equal to C. From AB take its half BH; and from the remainder take its half K+ F. HK; until AB be divided into as many parts as DE.

Then since DE is greater than AB; and EG is Ht not greater than the half of DE, but BH is the half of AB, the remainder DG is greater than the remainder AH. Again, since GF is not greater than the half of DG, but HK is the half of AH, the remainder DF is greater than the remainder AK.

Now DF is equal to C, the less of the two given magnitudes. There. fore, AK, the part left of the greater, is less than C. Q. E. D.

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2 Th. Equilateral polygons of the same number of sides, inscribed in circles, are similar, and are to one another as the squares of the diameters of the circles.

Let ABCDEF and GHIKLM, be two equilateral polygons of the same number of sides, inscribed in the circles ABD, GHK: A

K these two polygons are similar, and are to one another as the squares of the diameters AD, GK.

From the centres N, O, draw NB, OH. And because the sides of each polygon are all equal to one another, the arcs which they subtend in each polygon are also equal (p. 28 of b. 3); and the number of them in one circle is equal to the number of them in the other; therefore, whatever part any arc, AB, is of the circumfer ence ABD, the same part is the arc GH, of the circumference GHK

But the angle ANB is the same part of four right angles, that the arc AB is of the circumference ABD (p. 33 of b. 6); and the angle GOH is the same part of four right angles, that the arc GH is of the circumference GĦK (p. 33 of b. 6): therefore, the angles ANB, GOH, are each of them the same part of four right angles, and are therefore equal to one another. The isosceles triangles, ANB, GOH, are therefore equiangular, and the angle ABN equals the angle GHO. In the same manner, by joining NC, OI, it may be proved, that the angles NBC, OHI, are equal to one another, and to the angle ABN. Therefore, the whole angle ABC is equal to the whole GHI; and the same may be proved of the angles BCD, HIK; and so of the

The polygons, ABCDEF and GHIKLM, are therefore equiangular to one another; and since they are equilateral, the sides about the equal angles are proportionals: these two polygons are therefore similar to one another (def. 1 of b. 6.) And because similar polygons are as the squares of their homologous sides, the polygon ABČĎEF is to the polygon GHIKLM, as the square of AB to the square of GH: but because the triangles ANB, GOH are equiangular, the squares of AB, GH, are as the squares of AN, GO (p: 4 of b. 6); or as four times the square of AN to four times the square of GO (p. 15 of b. 5); that is, as the square of AD to the square of GK (cor. 2, p. 8, 2). Therefore also, the polygon ABCDEF is to the polygon GHIKLM, as the square of AD to the square of GK: and they have been shown to be similar.

Cor. Every equilateral polygon inscribed in a circle is also equiangular: for the isosceles triangles which have their common vertex in the centre, are all equal and similar; therefore, the angles at their bases are all equal, and the angles of the polygons are therefore also equal.

rest.

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