Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

rectilineal figure described upon bc is equal to the similar and similarly described figures upon ba, a c

Draw the perpendicular ad; therefore, because in the right-angled triangle a bc, ad is drawn from the right angle at a perpendicular to the base bc, the triangles a bd, ad c are similar to the whole triangle a bc, and to one another (vi. 8), and because the triangle abc is similar to abd; as cb to ba, so is ba to bd (vi. 4); and because these three

a

b

d

:

straight lines are proportionals, as the first to the third, so is the figure upon the first to the similar and similarly described figure upon the second (vi. 20, cor. 2): therefore as cb to bd, so is the figure upon cb to the similar and similarly described figure upon ba and inversely (v. B.). as db to bc, so is the figure upon ba to that upon bc. For the same reason, as dc to cb, so is the figure upon c a to that upon cb. Wherefore as bd and dc together to bc, so are the figures upon ba, a c to that upon bc (v. 24): but bd and dc together are equal to bc Therefore the figure described on bc is equal (v. A.) to the similar and similarly described figures on ba, a c.. Wherefore, in right-angled triangles, &c. Q. E. D.

PROPOSITION XXXII.—THEOREM.

If two triangles which have two sides of the one proportional to two sides of the other be joined at one angle, so as to have their homologous sides parallel to one another, the remaining sides shall be in a straight line.

LET a bc, dce be two triangles which have the two sides ba, ac proportional to the two cd, de, viz. ba to ac as cd to de; and let ab be parallel to dc, and a c to de, bc and ce are in a straight line. Because a b is parallel to dc, and the straight line ac meets them,

[ocr errors]

d

the alternate angles bac, a cd are equal (i. 29); for the same reason, the angle cde is equal to the angle acd; wherefore also bac is equal to cde: and because the triangles abc, dce have one angle at a equal to one at d, and the sides about these angles proportionals, viz. ba to a c as cd to d e, the triangle abc is equiangular (vi. 6) to e dce: therefore the angle a bc is equal to the angle dce: and the angle bac was proved to be equal to a cd therefore the whole angle ace is equal to the two angles abc, bac; add the common angle acb, then the angles ace, acb are equal to the angles abc, bac, acb: but abc, bac, a cb are equal to two right angles (i. 32); therefore also the angles ace, a cb are equal to two right angles. And since at the point c, in the

b

:

straight line a c, the two straight lines bc, ce, which are on the opposite sides of it, make the adjacent angles a ce, a cb equal to two right angles; therefore (i. 14) bc and ce are in a straight line. Wherefore, if two triangles, &c. Q. E. D.

PROPOSITION XXXIII.-THEOREM.

In equal circles, angles, whether at the centres or circumferences, have the same ratio which the circumferences on which they stand have to one another; r; so also have the sectors.

LET abc, def be equal circles; and at their centres the angles bgc, ehf, and the angles bac, e df, at their circumferences; as the circumference bc to the circumference ef, so is the angle bgc to the angle ehf, and the angle bac to the angle edf: and also the sector bgc to the sector ehf.

Take any number of circumferences ck, kl, each equal to bc, and any number whatever fm, mn, each equal to ef: and join gk gl hm, hn. Because the circumferences bc, ck, kl are all equal, the angles bgc, cgk, k gl are also all equal (iii. 27): therefore what multiple soever the circumference bl is of the circumference bc, the same multiple is the angle bg1 of the angle bgc: for the same reason, whatever multiple the circumference en is of the circumference ef, the same multiple is the angle ehn of the angle ehf: and if the circumference bl be equal to the circumference en, the angle bg1 is also equal (iii. 27) to the angle ehn; and if the circumference bl be greater than en, likewise the angle bgl is greater than ehn: and if less, less there being then four magnitudes, the two circumferences bc, ef, and the two angles bgc, ehf; of the circumference bc, and of the angle bgc, have been taken any equimultiples whatever, viz. the circumference bl, and the angle bgl; and of the circumference ef, and of the angle ehf, any equi

:

[blocks in formation]

multiples whatever, viz. the circumference en, and the angle ehn: and it has been proved, that if the circumference bl be greater than en, the angle bgl is greater than ehn; and if equal, equal; and if less, less: as therefore the circumference bc to the circumference ef, so (v. def. 5) is the angle bgc to the angle ehf: but as the angle bgc is to the angle ehf, so is (v. 15) the angle bac to the angle edf; for each is double of each (iii. 20); therefore, as the circumference bc is to ef

so is the angle bgc to the angle ehf, and the angle bac to the angle ed f

Also, as the circumference bc to ef, so is the sector bgc to the sector ehf. Join bc, ck, and in the circumferences bc, ck take any points X, 0, and join bx, xc, co, ok: then, because in the triangles gbc, gck the two sides bg, gc, are equal to the two cg, gk, and that they contain equal angles; the base bc is equal (i. 4) to the base ck, and the triangle gbc to the triangle gck: and because the circumference bc is equal to the circumference ck, the remaining part of the whole circumference of the circle a bc, is equal to the remaining part of the whole circumference of the same circle: wherefore the angle bxc is equal to the angle cok (iii. 27) and the segment bx c is therefore similar to the segment cok (iii. def. 11); and they are upon equal straight lines bc, ck: but similar segments of circles upon equal straight lines are equal (iii. 24) to one another therefore the segment bxc is equal to the segment cok: and the triangle bgc is equal to the triangle cgk: therefore the whole, the sector bgc is equal to the whole, the sector cgk: for the same reason, the sector k gl is equal to each of the sectors bgc, cgk: in the same manner, the sectors ehf, fhm, mhn may be proved equal to one another; therefore, what multiple soever the circumference bl is of the circumference bc, the same multiple is the sector bgl of the sector bgc for the same reason, whatever multiple the circumference en is of ef, the same multiple is the sector ehn of the sector ehf; and if the circumference bl be equal to en, the sector bgl is equal to the sector

[blocks in formation]

ehn; and if the circumference bl be greater than en, the sector bgl is greater than the sector ehn; and if less, less: since, then, there are four magnitudes, the two circumferences bc, ef, and the two sectors bgc, ehf, and of the circumference bc, and sector bg c, the circumference bl and sector bg1 are any equal multiples whatever; and of the circumference ef, and sector ehf, the circumference en, and sector e hn, are any equimultiples whatever and that it has been proved, if the circumference bl be greater than en, the sector bgl is greater than the sector ehn; and if equal, equal; and if less, less; therefore (v. def. 5), as the circumference bc is to the circumference ef, so is the sector bgc to the sector ehf. Wherefore, in equal circles, &c. Q. E. D.

:

PROPOSITION B.-THEOREM.

If an angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle.

LET abc be a triangle, and let the angle ba c be bisected by the straight line ad; the rectangle ba, a c is equal to the rectangle b d. d c, together with the square of a d

:

b

C

d

Describe the circle (iv. 5) a cb about the triangle, and produce ad to the circumference in e, and join ec. Then because the angle bad is equal to the angle cae, and the angle a bd to the angle (iii. 21) a e c, for they are in the same segment; the triangles a bd, a e c, are equiangular to one another therefore as ba to a d, so is (vi. 4) ea to a c, and consequently the rectangle ba, a c is equal (vi. 16) to the rectangle ea, ad that is (ii. 3), to the rectangle ed, da, together with the square of a d: but the rectangle ed, da is equal to the rectangle (iii. 35) bd, dc. Therefore the rectangle ba, a c is equal to the rectangle bd, dc, together with the square of a d. Wherefore, if an angle, &c. Q. E. D.

e

PROPOSITION C.-THEOREM.

If from any angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

LET abc be a triangle, and a d the perpendicular from the angle a to the base bc; the rectangle ba, a c is equal to the rectangle contained by ad, and the diameter of the circle described

about the triangle.

Describe (iv. 5) the circle acb about the triangle, and draw its diameter a e, and join ec: because the right angle bda is equal b (iii. 31) to the angle e ca in a semicircle, and the angle abd to the angle aec in the same segment (iii. 21); the triangles a bd, aec are equiangular therefore as (vi. 4) ba to a d, so is ea to a c; and consequently the rectangle ba, ac is equal (vi. 16) to the rectangle e a, a d. If therefore from an angle, &c. Q. E. D.

e

d

a

L

PROPOSITION D.-THEOREM.

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

LET abcd be any quadrilateral inscribed in a circle, and join ac, b d ; the rectangle contained by a c, bd is equal to the two rectangles contained by a b, cd, and by a d, b c.

*

Make the angle a be equal to the angle dbc: add to each of these the common angle ebd, then the angle abd is equal to the angle ebc : and the angle bda is equal (iii. 21) to the angle bce, because they are

b

C

in the same segment: therefore the triangle a bd is equiangular to the triangle bce: wherefore (vi. 4), as bc is to ce, so is bd to da; and consequently the rectangle bc, ad is equal (vi. 16) to the rectangle bd, ce again, because the angle a be is equal to the angle db c, and the angle (iii. 21) bae to the angle bd c, the triangle a be is equiangular to the triangle bcd: as therefore ba to ae, so is bd to dc; wherefore the rectangle ba, dc is equal to the rectangle bd, ae but the rectangle bc, ad has been shewn equal to the rectangle bd, ce; therefore the whole rectangle ac, bd (ii. 1) is equal to the rectangle ab, dc, together with the rectangle ad, bc. Therefore, the rectangle, &c. Q. E. D.

d

EXERCISES ON BOOK VI.

SECT. I.-PROBLEMS.

1. GIVEN the base, the vertical angle, and the ratio between the two sides, to construct the triangle.

2. Given the vertical angle, the perpendicular from it to the base, and the ratio between the segments and the base on each side of the perpendicular, to construct the triangle.

3. Given a scalene triangle, to construct an isosceles triangle equal to it, and with an equal vertical angle.

4. Given a finite straight line, to describe on it a right-angled triangle whose sides shall be continual proportionals.

5. Given an angle and a point, to draw from the point a straight line cutting the lines that contain the angle, in such a way that the parts be

*This is a Lemma of Cl. Ptolemæus, in page 9 of his μeyaλN OVVTAĞIS.

« ΠροηγούμενηΣυνέχεια »