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mean shall have a given ratio to the other mean, as may be shown in the same manner as in the foregoing proposition.

82.

: PROP. LXXII. If four straight lines be proportionals; as the first is to the straight line to which the second has a given ratio, so is the third to a straight line to which the fourth has a given ratio.

Let A, B, C, D, be four proportional straight lines, viz. as A to B, so C to D; as A is to the straight line to which B has a given ratio, so is C to a straight line to which D has a given ratio.

Let E be the straight line to which B has a given ratio, and as B to E, so make D to F: The ratio of B to E is givena, and therefore the ratio of 1 Hyp: D to F; and because as A to B, so is C to D; and as B to E, so D to F: therefore, ex æquali, a as A to E, so is C to F; and E is the straight line C D

to which B has a given ratio, and F that to which · D has a given ratio; therefore as A is to the

straight line to which B has a given ratio, so is C to a line to which D has a given ratio.

. PROP. LXXIII.

83.

If four straight lines be proportionals; as the first See N. is to the straight line to which the second has a given ratio, so is a straight line to which the third has a given ratio to the fourth.

Let the straight line A be to B, as C to D; as A to the straight line to which B has a given ratio, so is a straight line to which C has a given ratio, to D.

Let E he the straight line to which Bhas a given ratio, and as B to E, so make F to C; because the ratio of B to E is given, the ratio of C to F is given: And because A is to B, as C to D; and as B to E,

i i BÉ so F to C; therefore, ex æquali in proportione

C D perturbato ą, A is to E, as F to D; that is, A is to E, to which B has a given ratio, as F, to which Ç has a given ratio, is to D. .

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PROP. Lxxiv. Ir a triangle has a given obtuse angle; the excess of the square of the side which subtends the obtuse angle, above the squares of the sides which contain it, shall have a given ratio to the triangle.

Let the triangle ABC have a given obtuse angle ABC; and produce the straight line CB, and from the point A

draw AD perpendicular to BC: The excess of the square • 12. 2. of AC above the squares of AB, BC, that is, the double of

the rectangle contained by DB, BC, has a given ratio to the triangle ABC.

Because the angle ABC is given, the angle ABD is also

given ; and the angle ADB is given; wherefore the triangle . 43 Dat. ABD is given in species; and therefore the ratio of AD

1.6. to DB is given: And as AD to DB, so is the rectangle

AD, BC, to the rectangle DB, BC; wherefore the ratio of
the rectangle AD, BC, to the rectangle DB, BC, is given,
as also the ratio of twice the rectangle

A E
DB, BC, to the rectangle AD, BC:
But the ratio of the rectangle AD, BC,

to the triangle ABC, is given, because
* 41. 1. it is doubled of the triangle; therefore

the ratio of twice the rectangle DB,BC, 9 Dat. to the triangle ABC, is givene: and

twice the rectangle DB, BC, is the excess a of the square of AC above the squares of AB, BC; therefore this excess has a given ratio to the triangle ABC.

And the ratio of this excess to the triangle ABC may be found thus: Take a straight line EF given in position and magnitude; and because the angle ABC is given, at the point F of the straight line EF make the angle EFG equal to the angle ABC; produce GF, and draw EH perpendicular to FG; then the ratio of the excess of the square of AC above the squares of AB, BC, to the triangle ABC, is the same with the ratio of quadruple the straiglit lice HF to HE.

Because the angle ABD is equal to the angle EFH, and

the angle ADB to EHF, each being a right angle; the tri14.6. angle ADB is equiangular to EHF; thereforef as BD to , * Cor. 4. 5. DA, so FH to HE; and as quadruple of BD to DA, so ist

quadruple of FH to HE: But as twice BD is to DA, so

is twice the rectangle DB, BC, to the rectangle AD, BC; C. 5. and as DA to the half of it, so ist the rectangle AD, BC,

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of AC aben ratio to this excessive EF givet is givene equal

63.

to its half the triangle ABC; therefore, ex æquali, as twice BD is to the half of DA, that is, as quadruple of BD is to DA, that is, as quadruple of FH to 'HE, so is twice the rectangle DB, BC, to the triangle ABC.

PROP. LXXY, If a triangle has a given acute angle, the space by which the square of the side subtending the acute angle is less than the squares of the sides which contain it, shall have a given ratio to the triangle.

Let the triangle ABC have a given acute angle. ABC, and draw AD perpendicular to BC, the space by which the square of AC is less than the squares of AB, BC, that is a, . 13. 2. the double of the rectangle contained by CB, BD, has a given ratio to the triangle ABC.

Because the angles ABD, ADB, are each of them given, the triangle ABD is given in species; and therefore the ratio of BD to DA is given : And as BD to DA, so is the rectangle CB, BD, to the rectangle CB, AD: Therefore the ratio of these rectangles is given, as also the ratio of twice the rectangle CB, BD, to the rectangle CB, AD: But the rectangle CB, AD, has a given ratio to its half the trian- B gle ABC: Thereforeb the ratio of twice the rectangle CB, 19 Dat.' BD,'to the triangle ABC is given ; and twice the rectangle CB, BD, is a the space by which the square of AC is less than the squares of AB, BC; therefore the ratio of this space to the triangle ABC is given : And the ratio may be found as in the preceding proposition.

LEMMA. ' Ir from the vertex A of an isosceles triangle ABC, any straight line AD be drawn to the base BC, the square of the · side AB is equal to the rectangle BD, DC, of the segments of the base together with the square of AD; but if AD be drawn to the base produced, the square of AD is equal to the rectangle BD DC, together with the square of AB. · CASE 1. Bisect the base BC in E, and a join AE, which will be perpendiculara

indo 86. I. to BC; wherefore the square of AB is equalb to the squares of AE, EB; but

is 47. 1. the square of EB is equal to the rect

c5. 2. angle BD, DC, together with the square DTT of DE; therefore the square of AB is

Arctangle CB, rectanglen, us also,e ratio of

* 47. 1. equal to the squares of AE, ED, that is, to b the square of

AD, together with the rectangle BD, DC; the other case is shown in the same way by 6. 2. Elem.

67.

PROP. LXXVI. Ir a triangle have a given angle, the excess of the square of the straight line which is equal to the two sides that contain the given angle, above the square of the third side, shall have a given ratio to the triangle.

Let the triangle ABC have the given angle BAC, the excess of the square of the straight line which is equal to BA, AC, together above the square of BC, shall have a given ratio to the triangle ABC.

Produce BA, and take AD equal to AC, join DC, and produce it to E, and through the point B draw BE parallel to AC: join AE, and draw AF perpendicular to DC; and because AD is equal to AC, BD is equal to BE ; and BC is drawn from the vertex B of the isosceles triangle DBE: therefore, by the Lemma, the square of BD, that is, of BA and AC together, is equal to the rectangle DC, CE, together with the square of BC: and therefore the square of BA, AC, together, that is, of BD, is greater than the square of BC by the rectangle DC, CE; and this rectangle has a given ratio to the triangle ABC, because the angle BAC is given, the adjacent angle CAD is given; and each of the

angles ADC, DCA, is given, for 5. & 32.1. each of them is the halfa of the . K .

given angle BAC; therefore the triangle ADC is given in • 43 Dat. species b; and AF is drawn from its vertex to the base in a

given angle; wherefore the ratio of AF to the base CD is 50 Dat. given c; and as CD to AF, so isd the rectangle DC, CE, to

1: 6. the rectangle AF, CE; and the ratio of the rectangle AF, • 41. 1. CE, to its halfe, the triangle ACE, is given ; therefore the

ratio of the rectangle DC, CE, to the triangle ACE, that 137. 1. isf, to the triangle ABC, is given8; and the rectangle DC, ! 9 Dat. CE, is the excess of the square of BA, AC, together, above

the square of BC :-Therefore the ratio of this excess to the triangle ABC is given. ...in vrei siis

The ratio which the rectangle DC, CE, has to the trian

“? 22. 5.

gle ABC is found thus: Take the straight line HG given in position and magnitude, and at the point G in GH make the angle HGK equal to the given angle CAD, and take · GK equal to GH, join KH, and draw ĞL perpendicular to

it: Then the ratio of HK to the half of ĠL is the same
with the ratio of the rectangle DC, CE, to the triangle
ABC: Because the angles HGK, DAC, at the vertices of
the isosceles triangles GHK, ADC, are equal to one an-
other, these triangles are similar; and because GL, AF, are
perpendicular to the bases HK, DC, as HK to GL, so ishadi
(DC to AF, and so is) the rectangle DC, CE, to the rect-
angle AF, CE; but as GL to its half, so is the rectangle
AF, CE, to its half, which is the triangle ACE, or the tri-
angle ABC; therefore, ex æquali, HK is to the half of the
straight line GL, as the rectangle DC, CE, is to the triangle
ABČ.
- Cor. And if a triangle have a given angle, the space by
which the square of the straight line, which is the difference
of the sides which contain the given angle, is less than the
square of the third side, shall have a given ratio to the trian-
gle. This is demonstrated the same way as in the preceding
proposition, by help of the second case of the Lemma.

PROP. LXXVII. If the perpendicular drawn from a given angle of See N. a triangle to the opposite side, or base, has a given ratio to the base, the triangle is given in species.

Let the triangle ABC have the given angle BAC, and i · let the perpendicular AD drawn to the base BC, have a · given ratio to it, the triangle ABC is given in species.

If ABC be an isosceles triangle, it is evidenta, that if any * 5. & 38. 1.

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E 0 H M F one of its angles be given, the rest are also given; and is therefore the triangle is given in species, without the consideration of the ratio of the perpendicular to the base, which in this case is given by Prop. 50.

But when ABC is not an isosceles triangle, take any

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