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E

EHN are the same multiples of the sectors BGC, EHF, that the arcs BL, EN are of the arcs BC, EF.

If then the arc BL be equal to, greater, or less than the arc EN: the sector BGL is equal to, greater, or less than the sector EHN. BL and BGL being therefore equimultiples of BC the first, and BGC the third ; and EN and EHN, being equimultiples of EF the second, and EHF the fourth; it follows, that BC: EF :: BGC : EHF (6): therefore the sectors have the same ratio as their arcs. Wherefore, in equal circles, &c.

Q. E. D.
Recite (a) p. 27, 3; (b) def. 5, 5; (c) p. 20, 3;

(d) p. 15, 5; (e) p. 11, 5; (f) p. 4, 1;
(8) p. 24, 3;

(h) p. 28,

3. B Th. If a straight line bisect an angle of a triangle, and likewise cut the base ; the rectangle contained by the sides is equal to the rectangle contained by the segments of the base, together with the square of the bisecting line.

Given ABC a triangle, and AD a straight line bisecting the angle BĂC: then BAXACEBDX DC+AĎ..

Describe the circle ACB about the triangle (a); produce AD to the circumference in Ě ; join EC.

Then because the angles BAD, CAE were made equal; and ABD equals AEC, in the same segment (6); the triangles ABD, AEC are equiangular: therefore BA : AD : : EA : AC (C); hence BAXAC= EXXAD (d), that is, EDXDA+AD2 (e). But EDXDA=BDXDC (f). Therefore BAXAC=BDXDC+A1), as stated. Wherefore, if a straight line bisect, &c.

Q. E. D. Recite (a) p. 5, 4; (b) p. 21, 3; (c) p. 4, 6;

(d) p. 16, 6; le) p. 3, 2; (f) p. 35, 3. C Th. If from any angle of a triangle a perpendicular be drawn to the base; the rectangle contained by the sides of the triangle equals that contained by the perpendicular and the diameter of the circle described about the triangle. Given ABC a triangle, AD a perpendicular

А from the vertex A to the base, also AE the diameter of the described circle : then BAXAC= ADX AE. Join CE.

BA Because the angle ACE, in a semicircle, equals the right angle ADB (a); and the angles ABC, AEC, in the same segment, are equal (6); the triangles ABD, AEC are equiangular: therefore

BA: AD :: EA : AČ (c); and it follows that BAXAC=EAXAD, (d), as stated. Wherefore, if from any angle, &c.

Q. E. D. Recite (a) p. 31, 3;

(b) p. 21, 3; 'c) p. 4, 6;

(d) p. 16, 6.

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3

D Th. The rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.

Given ABCD any quadrilateral inscribed in a circle; also AC, BĎ, diagonals joining its opposite angles: ACXBD=(BCXAD)+(ABXCD).

Make the angle ABE equal to CBD; to each of which add DBE: the sums ABD, CBE are equal: and as angles in the same segment BDA equal BCE(a). Therefore the triangles ABD, CBE are equiangular; and BC:EC :: BD: DA (6); from which it comes that BCXDA=BDXEC (c).

And because the angles ABE, DBC are inade equal; and BAE equals BDC of the same segment (a); therefore the triangles ABE, DBC are equiangular; and BA : AE :: BD: DC (6); from which it comes that BAXDC=AEXBD (c).

Therefore, the two rectangles BDXEC and BDXAE are equal to the two BOXDA and BAXDC. But AE and EC are the parts of the diagonal AC; therefore BDXAC=(BDXAE)+(BDXEC)(d)= (BCXĂD)+ABXCD). Therefore, the rectangle, &c.

Q. E. D. Recite (a) p. 21, 3;

(6) p. 4, 6; (c) p. 16, 6; d) p. 1, 2.

E Th. In a circle, the chord of an arc is to the chord of its half, as the sum of two straight lines drawn from the ends of the chord to any point in the opposite arc, is to the line which joins that point to the bisectional point of

the arc.

Given the arc ACB bisected in C; also the chords AB, BC, and the straight lines AD, DB, DC: it is to be proved that AB is to BC or AC as AD+ DB is to DC.

Since AB, DC are diagonals of a quadrilateral inscribed in a circle, the rectangle ABXDC equals

B the two rectangles ADXBC and DBXAC (a).

But since AC=BC (6), these two rectangles are equal to ADXAC and DBXAC, or (ADHDB) XAC (c). Therefore, ABXDC=(AD+DB)XAC: and since the sides of equal rectangles are reciprocally proportional (d), it follows that AB. is to AC, or BC as AD+DB is to DC, as stated.

Wherefore, in a circle, the chord of an arc, &c. Q. E. D.
Recite (a) p. D, 6; (b) p. 29, 3; (c) p. 1, 2;

(d) p. 14, 6.

F Th. If three straight lines be proportionals; namely, 1, a line drawn from a point without a circle to the cen

tre ; 2, the radius ; 3, a segment measured from the cen. tre to a point in the first : then, any two lines drawn from the two points to the circumference shall have the same ratio as those two segments of the first line which meet the circumference, beginning at the said points.

Let F be the outer point, D the centre, E the sectional point, and A, B, C, points in the circumference: Draw FB and produce it to G; join BA, BC, BD, BE.

Now, since FD: DB :: DB: DE, by hyp., and that these proportionals are about the same angle D, the triangles FBD, BED are equiangular (a).

Therefore FB : BD :: BÊ: EĎ (6), and alternately FB : BE:: BD, or AD: ED (c).

But since FD: AD:: AD: ED, by hyp. by division FA: AD :: AE; ED, (d); and alternately FA : AE:: AD, or BD: ED (c):

Therefore FA: AE :: FB : BE (e).

But FA, AE are the segments of the first line, which meet the circumference; and FB, BE are any two lines drawn from the points to the circumference. Wherefore, if three straight lines, &c.

Q. E. D. Recite (a) p. 6, 6; () p. 4, 6; (c) p. 16, 5;

(d) p. 17,5; (e) p. 11, 5.

Cor. The chord AB bisects the angle FBE: for, in the triangle FBE, the segments of the base FA, AĚ have the same ratio as the sides FB, BĚ. Recite p. 3, 6.

And for the same reason the chord BC bisects the exterior angle EBG, (p. A, 6).

For since by hyp. FD: DC :: DC : DE
by composition FC: CD::CE: DE (p. 18, 5);.
and by division FA : AD or CD :: AE: DE (p. 17, 5:

Therefore, ex æquo FA : AE :: FC:CE (p. 22, 5).
But it proves that FA: AE :: FB : BE;
Therefore FB : BE :: FC:CE (p. 11, 5).

G Th. If a chord, drawn through the extreme point of the diameter of a circle, meet a straight line in the circle or out of it, cutting the diameter or its exterior production, at right angles; the rectangle of the diameter and its segment, between the extreme point and perpendicular, is equal to the rectangle of the chord and its segment, between the same point and the cutting line.

Given the chord AB and its segment AF; also the diameter AC, and its segment AD: then

B ABXAF=ACXAD. Join BC.

Because ABC is an angle in a semicircle (a); it is equal to the

DI F E right angle ADF; and the angles DAF, BXC are either identical or

FD vertical, and therefore equal (b): therefore the triangles ABC, ADF are equiangular (c); and BA : AC :: DA: AF. But the rectangle contained by the means is equal to that contained by the extremes (d); therefore DAXAC=BAXÀF. Wherefore, if a chord, &c.

Q. E. D. Recite (a) p. 31, 3; (6) p. 15, 1; (c) p. 4, 6;

(d) p. 15, 6.

G

H Th. The perpendiculars drawn from the three angles of a triangle to the opposite sides, intersect each other in the same point.

Given the triangle ABC; perpendiculars BD, CE intersecting each other in F, and AF joined; which last produced to G will fall perpendicular to BC. Join ED.

Since AF subtends a right angle on either side of it, it is the diameter of a circle passing through the points A, E, F, D (a). Now the vertical angles EFB, DFC are equal (b); the right angles BEF, CDF are equal (c), and the triangles EFB,

B DFC are equiangular (d); therefore FB : FE:: FC : FD, and alternately FB : FC : : FE : FD (e). Therefore also, the triangles EFD, BFC are equiangular: for their vertical angles are contained by proportional sides, as proved. Wherefore the angles FCB, EDF are equal : but EDF equals EAF as angles in the same segment (f); therefore EAF equals FCB: the vertical angles EFA, CFG are also equal (b); and the third angles AEF, CGF are equal (g). But AEF is a right angle;

Wherefore, CGF is a right angle; and AG, passing through the point F, is perpendicular to BC.

Q. E. D Recite (a) p. 31, 3; (b) p. 15, 1; (c) ax. 10, 1;

(d) p. 4 and 6,6; (e) p. 16, 5; (f) p. 21, 3:

(8) p. 32, 1. Cor. The triangles ADE, ABC are similar: for the triangles ABD, ACE, having right angles at D and E, and the angle A common, are equiangular, and have BA to AD as CA to AE; and alternately BA to CA as AD to AE. Therefore the two triangles BAC, DAE have the angle A common; and the sides about it proportionals; therefore they are equiangular (6, 6), and similar: and the rectangles BAXAF CAXÀD are equal.

END OF THE SIX BOOKS.

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PLANE TRIGONOMETRY teaches the measurement of distance and elevation by means of angles. The principles of it are all contained in the preceding six Books.

B

Lemma 1. An angle at the centre of a circle is to four right angles, as the arc on which it stands is to the whole circumference.

Given ABC, a central angle, and AC the arc which measures it (a); the angle ABC is to four right angles, as the arc AC is to the whole circumference ACF.

GA Produce AB to meet the circle in E; and at right angles to AE draw DBF.

Then, because ABC, ABD are central angles ; and AC, AD their arcs ;-ABC : ABD :: AC: AD (b): moreover ABC : 4ABD :: AC : 4AD (C). But ABD is a right angle, and AD is the arc which measures it: therefore the angle ABC is to 4 right angles as the arc AC is to the whole circumference.

Recite (a) def. 1, trig. ; (6) p. 33 of b. 6; (c) def. 3, and p. 4 of b. 5.

Cor. As the arcs AC and GH measure the same angle; AC is to the circumference ACE, as GH is to the circumference GHK: therefore equal angles at the centres of different circles stand on arcs which have the same ratio to their circumferences.

Definitions. 1. The measure of an angle is the arc of the circle on which it stands : thus, the arc AC is the measure of the angle ABC.

2. Degrees, minutes, seconds, &c., are the terms of circle measure.

Example. Every circle great or small is divided into 360 degrees, each degree into 60 minutes, each minute into 60 seconds. Hence an angle and its arc contain the same number of degrees, &c.

3. The complement of an angle is its defect from a right angle. 4. The supplement of an angle is its defect from a semicircle. 5. The sine of an angle is a straight line drawn in the bosom

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