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the rectangle AB, BC to AE, BC that is c, to the parallelogram c 35. 1, AC is given.

And it is evident how the ratio of the rectangle to the parallelogram may be found by making the angle FGH equal to the given angle ABC, and drawing, from any point F in one of its sides, FK perpendicular to the other GH; for GF is to FK, as BA to AE, that is, as the rectangle AB, BC, to the parallelo

gram AC.

COR. And if a triangle ABC has a given angle ABC, the rect.

66. angle AB, BC contained by the sides about that angle, shall have a given ratio to the triangle ABC.

Complete the parallelogram ABCD ; therefore, by this proposition, the rectangle AB, BC has a given ratio to the parallelogram AC; and AC has a given ratio to its half the triangled d 41. 1. ABC ; therefore the rectangle AB, BC has a givene ratio to the e 9. dat. triangle ABC.

And the ratio of the rectangle to the triangle is found thus : make the triangle FGK, as was shown in the proposition; the ratio of GF to the half of the perpendicular FK is the same with the ratio of the rectangle AB, BC to the triangle ABC. Because, as was shown, GF is to FK, as AB, BC to the parallelogram AC; and FK is to its half, as AC is to its half, which is the triangle ABC; therefore, ex æquali, GF is to the half of FK, as AB, BC rectangle is to the triangle ABC.

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IF two parallelograms be equiangular, as a side of the first to a side of the second, so is the other side of the second to the straight line to which the other side of the first has the same ratio which the first parallelogram has to the second. And consequently, if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that Straight line is given; and if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second is given.

Let AC, DF be two equiangular parallelograms, as BC, a side of the first, is to EF, a side of the second, so is DE, the other side of the second, to the straight line to which AB, the

a 14. 6.

other side of the first has the same ratio which AC has to DF.

Produce the straight line AB, and make as BC to EF, so DE to BG, and complete the parallelo A gram BGHC ; therefore, because BC or GH, is to EF, as DE to BG, the sides about the equal angles BGH, DEF are B reciprocally proportional ; wherefore a the parallelogram BH is equal to DF; G

'H and AB is to BG, as the parallelogram

D
AC is to BH , that is, to DF; as there-
fore BC is to EF, so is DE to BG, which

E
is the straight line to which AB has the

F same ratio that AC has to DF.

And if the ratio of the parallelogram AC to DF be given, then the ratio of the straight line AB to BG is given ; and if the ratio of AB to the straight line BG be given, the ratio of the parallelogram AC to DF is given.

74. 73.

PROP. LXIV.

See Note. IF two parallelograms have unequal but given an

gles, and if as a side of the first to a side of the second, so the other side of the second be made to a certain straight line; if the ratio of the first parallelogram to the second be given, the ratio of the other side of the first to that straight line shall be given. And if the ratio of the other side of the first to that straight line be given, the ratio of the first parallelogram to the second shall be given.

Let ABCD, EFGH be two parallelograms which have the unequal, but given, angles ABC, EFG; and as BC to FG, so make EF to the straight line M. If the ratio the parallelogram AC to EG be given, the ratio of AB to M is given.

At the point B of the straight line BC make the angle CBK equal to the angle EFG, and complete the parallelogram

KBCL. And because the ratio of AC to EG is given, and that a 35. 1

AC is equal a to the parallelogram KC, therefore the ratio of

KC to EG is given ; and KC, EG are equiangular; thereb 63. dat. fore as BC to FG, so is b EF to the straight line to which KB

has a given ratio, viz. the same which the parallelogram KC has to EG ; but as BC to FG, so is EF to the straight line M; therefore KB has a given ratio to M; and the ratio

of AB to BK is given, because the triangle ABK is given in speciesc; therefore the ratio of AB to M is given d.

c 43. dat And if the ratio of AB to M be given, the ratio of the pa-d 9. dat. rallelogram AC to EG is given : for since the ratio of KB to BA is given, as also the ratio of AB to M, KA L D the ratio of KB to M is given d; and because the parallelograms KC, EG are equiangular, as BC to FG, so is b EF to B

Cb 63. dats the straight line to which KB has the same ratio which, the parallelogram KC has to EG ; but as BC to FG, so is EF to M; therefore KB is to M, as the parallel

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G ogram KC is to EG; and the ratio of KB to M is given, therefore the ratio of the parallelogram KC, that is, of AC to EG, is given.

Cor. And if two triangles ABC, EFG, have two equal angles, 75. . or two unequal, but given angles ABC, EFG, and if as BC a side of the first to FG a side of the second, so the other side of the se. cond EF be made to a straight line M; if the ratio of the triangles be given, the ratio of the other side of the first to the straight line M is given.

Complete the parallelograms ABCD, EFGH; and because the ratio of the triangle ABC to the triangle EFG is given, the ratio of the parallelogram AC to EG is given e, because the parallelo.e 15.5. grams are double f of the triangles; and because BC is to FG, as f 41. 1. EF to M, the ratio of AB to M is given by the 63d dat. if the angles ABC, EFG are equal ; but if they be unequal, but given angles, the ratio of AB to M is given by this proposition.

And if the ratio of AB to M be given, the ratio of the parallelogram AC to EG is given by the same proposition ; and there. fore the ratio of the triangle ABC to EFG is given.

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If two equiangular parallelograms have a given ratio to one another, and if one side has to one side a given ratio ; the other side shall also have to the other side a given ratio.

Let the two equiangular parallelograms AB, CD have a given ratio to one another, and let the side EB have a given ratio to the side FD ; the other side AE has also a given ratio to the other side CF.

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Because the two equiangular parallelograms AB, CD have a

given ratio to one another; as EB, a side of the first, is to FD, a 63. dat. a side of the second, so is a FC, the other side of the second, to

the straight line to which AE, the other side of the first, lias
the same given ratio which th first parallelogram AB has
to the other CD. Let this straight line be EG; therefore the
ratio of AE to EG is given;
and EB is to FD, as FC to
EG, therefore the ratio of FC A
to EG is given, because the
ratio of EB to FD is given ; E
and because thie ratio of AE
to EG, as also the ratio of FCG

to EG is given; the ratio of H K L b 9. dat. AE to CF is given b.

The ratio of AE to CF may be found thus : take a straight line H given in magnitude ; and because the ratio of the parallelogram AB to CD is given, make the ratio of H 10 K the same with it. And because the ratio of FD to LB is given, make the ratio of K to L the same : the ratio of AE 10 CF is the same with the ratio of H to L. Make as EB to FD, so FC to EG, therefore, by inversion, as FD to EB, so is LG to FC ; and as AE to EG, so is a (the parallelogram AB to CD, and so is) H to K; but as EG to FC, so is (FD to EB, and so is) K to L; therefore, ex æquali, as AE to FC, so is H to L.

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IF two parallelograms have unequal, but given angles, and a given ratio to one another; if one side has to one side a given ratio, the other side has also a given ratio to the other side.

Let the two parallelograms ABCD, EFGH which have the given unequal angles ABC, EFG, have a given ratio to one another, and let the ratio of BC to FG be given ; the ratio also of AB to EF is given.

At the point B of the straight line BC make the angle CBK equal to the given angle LFG, and complete the parallelo

gram BKLC; and because each of the angles BAK, AKB is & 43. dat.

given, the triangle ABK is given a in species ; therefore the ratio of AB to BK is given ; and because, by the hypothesis,

the ratio of the parallelogram AC 10 EG is given, and that AC is equal to BL; therefore the ratio of BL to EG is given : and b 35. 1. because BL is equiangular to EG, and, by the hypothesis, the ratio of BC to FG is given ; therefore c the ratio of KB to EF is c 65. dat. given, and the ratio of KB to BA is A K D L given; the ratio therefore d of AB to

d 9. dat: EF is given. The ratio of AB to EF may be found

C thus : take the straight line MN given E

M
R

N
in position and magnitude ; and make
the angle NMO equal to the given F G

P angle BAK, and the angle MNO equal to the given angle EFG or AKB : and

O because the parallelogram BL is equiangular to EG, and has a given ratio to it, and that the ratio of BC to FG is given ; find by the 65th dat. the ratio of KB to EF; and make the ratio of NO to OP the same with it: then the ratio of AB to EF is the same with the ratio of MO to OP: for since the triangle ABK is equiangular to MON, as AB to BK, so is MO to ON: and as KB to EF, so is NO to OP; therefore, ex æquali, as AB to EF, so is MO to OP.

70.

PROP. LXVII.

IF the sides of two equiangular parallelograms have See N. given ratios to one another; the parallelograms shall have a given ratio to one another.

Let ABCD, EFGH be two equiangular parallelograms, and let he atio of AB to EF, as also the ratio of BC to FG, be given; the ratio of the parallelogram AC to EG is given.

Take a straight line given in magnitude, and because the ratio of AB to EF is given, make A

DE H the ratio of K to L the same with it; therefore L is given a : and because the ratio of BC to FG B

a 2. dat, is given, make the ratio of L to K M the same : therefore M is L

F

G given a : and K is given, where. M foreb the ratio of K to Mis given : but the parallelogram AC is b 1. datu to the parallelogram EG, as the straight line K to the straight

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