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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 b d, adc are similar to the whole triangle a b c, and to one another (vi. 8), and because the triangle a bc is similar to abd; as cb to ba, so is ba to bd (vi. 4); and because these three

straight lines are proportionals, as the first a

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 d b to b d

b c, so is the figure upon b a to that upon bc. For the same reason, as dc to cb,

so is the figure upon c a to that upon c b. 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 bcTherefore the figure described on b c 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 pa

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

the alternate angles ba c, acd are a

equal (i. 29); for the same reason, the angle cde is equal to the angle acd;

wherefore also bac is equal to cde: d

and because the triangles a b c, dce have one angle at a equal to one at d, and the sides about these angles proportionals, viz. b a to a c as cd to d e, the

triangle a b c is equiangular (vi. 6) to b

dce: therefore the angle a b c is equal

to the angle dce: and the angle bac was proved to be equal to a cd: therefore the whole angle a ce is equal to the two angles a b c, bac; add the common angle a cb, then the angles a ce, a cb are equal to the angles a b c, bac, acb: but a b c, bac, a c b are equal to two right angles (i. 32); therefore also the angles ace, a c b are equal to two right angles. And since at the point c, in the straight line a c, the two straight lines bc, ce, which are on the opposite sides of it, make the adjacent angles ace, a cb equal to two right angles ; therefore (i. 14) b c 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; so also have the sectors. LET a b c d e f 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 b c, 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 bgl of the angle bgc: for the same reason, whatever multiple the circumference en is of the circumference e f, the same multiple is the angle ehn of the angle ehf: and if the circumference bl be equal to the circumference en, the angle bgl is also equal (iii. 27) to the angle ehn; and if the circumference bl be greater than en, likewise the angle bg 1 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 b c, 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

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multiples whatever, viz. the circumference en, and the angle ehn: and it has been proved, that if the circumference bl be greater than e n, the angle bgl is greater than ehn; and if equal, equal; and if less, less : as therefore the circumference b c to the circumference e f, 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,

80 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 e hf. Join bc, ck, and in the circumferences b c, ck take any points x, 0, and join bx, xc, Cook: 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 gсk: and because the circumference bc is equal to the circumference ck, the remaining part of the whole circumference of the circle a b c, 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 bxc 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 kgl is equal to each of the sectors bgc, cgk: in the same manner, the sectors e h f, 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

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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 b c, ef, and the two sectors bgc, ehf, and of the circumference bc, and sector bg c, the circumference b1 and sector bg) are any equal multiples whatever ; and of the circumference e f, 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 b c is to the circumference ef, so is the sector bg c to the sector e h f. 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 a b c be a triangle, and let the angle bac be bisected by the straight line a d; the rectangle ba, a c is equal to the rectangle b d. đc, together with the square of a d.

Describe the circle (iv. 5) a cb about the triangle, and produce a d to the circumference in e, and join ec. Then because the angle bad is equal to the angle ca e, and the angle a bd to the angle (iii. 21) a e c, for they are in the same segment; the triangles a b d, a e c, are equiangular to one

b another : therefore as b a to a d, so is (vi. 4)

d ea to a C, and consequently the rectangle ba, ac is equal (vi. 16) to the rectangle ea, ad that is (ii. 3), to the rectangle ed, da, together with the square of ad: but the rectangle ed, da is equal to the rectangle (iii. 35) bd,

e dc. Therefore the rectangle ba, a c is equal to the rectangle bd, dc, together with the square of ad. Wherefore, if an angle, &c. Q. E. D.

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 a bc 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 ac b about the triangle, and draw its diameter a e, and join ec: because the right angle bda is equal b

C (iii. 31) to the angle e ca in a semicircle, and

d the angle a bd to the angle a ec in the same segment (iii. 21); the triangles a b d, a ec are equiangular : therefore as (vi. 4) ba to ad, so is e a to ac; and consequently the rectangle ba, a c is equal (vi. 16) to the rectangle e a, a d. If therefore from an angle, &c. Q. E. D.

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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 a b c d be any quadrilateral inscribed in a circle, and join a c, bd; 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 e bd, then the angle a b d is equal to the angle ebc: and the angle b da is equal (iii. 21) to the angle bce, because they are

in the same segment: therefore the triangle b

a bd is equiangular to the triangle bce: wherefore (vi. 4), as b c is to ce, so is bd to da; and consequently the rectangle bc, ad is equal (vi. 16) to the rectangle b d, ce: again, because the angle a be is equal to the angle dbc, and the angle (iii. 21) ba e 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; whered

fore the rectangle ba, dc is equal to the a

rectangle bd, ae : but the rectangle b c,

ad has been shewn equal to the rectangle bd, ce; therefore the whole rectangle ac, bd ii. 1) is equal to the rectangle a b, dc, together with the rectangle ad, bc. Therefore, the rectangle, &c. Q. E. 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 ueyann Ouvrages.

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