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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 given a, and therefore the ratio of D to F;

a Hyp: and because as A to B, so is C to D; and as B to E so D to F; therefore, ex æquali, as A 10 E, so is A B E C to F; and E is the straight line to which B has a CDF 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.

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

Let E be the straight line to which B has 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 gi A B E ven: and because X is to B, as () to D; and as B F C D to E, so F to C; therefore, ex æquali in proportione perturbata 2, A is to E, as F to D; that is, A is to E

# 23.5 to which B has a given ratio, as F, to which C has a given ratio, is to D.

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IF 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 of AC abore the a 12. 2. squares of AB, BC, that is a, the double of the rectangle con

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

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

ven ; and the angle ADB is given ; wherefore the triangle b) 43. dat. ABD is given bin species; and therefore the ratio of AD to c1.6. DB is given : and as AD to DB. so is the rectangle AD,

BC to the rectangle DB, BC ; wherefore the ratio of the rect. angle AD, BC to the rectangle DB, BC is given, as also the ratio of twice the rectangle DB, BC to the A E rectangle AD, BC: but the ratio of the rect

angle ad, BC to the triangle ABC is gid 41. 1. ven, because it is double d of the triangle; therefore the ratio of twice the rectangle

I e 9. dat. DB, BC to the triangle ABC is given ;

and twice the rectangle DB, BC is the ex D B с
cess a of the square of AC above the squares of AB, BC; there-
fore 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 ihe angle ABC ; produce GF, and draw EH perpendicular 10 FG ; then the ratio of the excess of the square of AC above the squares of AB, BC to the triangle A BC, is the same with the ratio of quadruple the straight line HF 10 HIE.

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

angle ADB to EHF, each being a right angle ; the triangle f4. 6.

ADB is equiangular to EHF; therefore f as BD to DA, so gCor.4.5. FH to HE; and as quadruple of BD 10 DA, so is & quadru

ple of FH to HE: but as twice BD is to DA, so is ctrice

the rectangle DB, BC to ihe rectangle AD, BC ; and as DA h Cor. 5. to the half of it, so is be the rectangle AD, BC 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 quadrupl: of FH to HE, so is twice the rectangle DB, BC lo the triangle ABC.

PROP. LXXV.

65.

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, the double of the a 13. * 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,

A 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 triangle ABC; therefore b the B

D с

b 9. dan ratio of twice the rectangle CB, 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.

IF from the vertex A of an iscosceles 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 Al which will be perpendicular a to

a 8. 1. BC; wherefore the square of AB is equal bto the squares of AE, EB; but the square

b 47.1. of EB is equal to the rectangle BD, DC

c 5. 1. together with the square of DE; there. Ú

B DE с fore the square of B is cqual to the

b 47. 1. 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.

IF 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 10gether, that is, of BD, is greater than

D the square of BC by the rectangle DC, CE; and this rectangle has a given

А

F ratio to the triangle ABC: because the angle B.AC is given, the adjacent

с angle CAD is given ; and each of the

angles ADC, DCA is given, for each a 5. & 32. of them is the half a of the given angle G BAC; therefore the triangle ADC is

E b 43. dat. given b in species ; and Al is drawn

from its vertex to the base in a given

angle ; wherefore the ratio of AF to the base CD is given ~; and c 50. dat. as CD to AF, so is d the rectangle DC, CE to the rectangle AF, d 1. 6.

CE; and the ratio of the rectangle AF, CE te its half e the trie 41 1. ingle ACE is given ; therefore the ratio of the reclangle DC,

CE to the triangle ACE, that is f, to the triangle ABC, is given &: 8 9. dat. and the rectangle DC, 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.

The ratio which the rectangle DC, CE has to the triangle ABC is found thuse take the straight line GH given in posi

B

1.

tion 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 GL perpendicular to it; then the ratio of HK to the half of GL 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 another, these triangles are similar ; and because GL, AF are perpendicular to the bases HK, DC, as HK to GL, so ish (DC to AF, and so is) the rectangle DC, CE to the rect

54.6. angle AF, CE; but as GL to its half, so is the rectangle, AF,

h

22.5. CE to its half, which is the triangle ACE, or the triangle ABC ; therefore, ex æquali, HK is to the half of the straight line GL, as the rectangle DC, CE is to the triangle ABC.

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 triangle. This is demonstrated the same way as the preceding proposition, by help of the second case of the lemma.

PROP. LXXVII.

L.

IF the perpendicular drawn from a given angle of a See Note. 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 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 evident a that if any a 5.& 32.

1.
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one of its angles be given, the rest are also given ; and 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 straight line EF given in position and magnitude, and upon it describe

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