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419 the ratio of the parallelogram AC to EG is given, and that AC is equal to BL; therefore the ratio of BL to EG is given : b 35. 1. And because BL is equiangular to EG, and by the hypothefis, the ratio of BC to FG.is given ; therefore the ratio of KB to c 65. dat. EF is given, and the ratio of KB to BA is given ; the ratio therea

A K DL fore d of AB to EF is given.

d g. dat. The ratio of AB to EF may be found thus: Take the straight line

B В

C
MN given in position and magni E

M
H

N tude; and make the angle NMO

F equal to the given angle BAK,

G

P and the angle MNO equal to the given angle EFG or AKB: And 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 aequali, as AB to EF, so is MO to OP.

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1
F 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 the ratio 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 K given in magnitude, and because the ratio of AB to EF is given, make the ratio of K to the A D E Η same with it, therefore L is given * : And because the ratio

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

K Therefore M is given; and L

F G K is given, wherefore b the M

bi, dat. ratio of K to M is given : But the parallelogram AC is to the parallelogram EG, as the straight line K to the straight line M, Dd2

as

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as is demonstrated in the 23d prop. of b. 6. Elem. therefore the ratio of AC to EG is given.

From this it is plain how the ratio of two equiangular parallelograms may be found when the ratios of their Gides are given.

70.

PRO P. LXVIII.

Sce N.

IF,

the fides of two parallelograms which have unequal,

but given angles, have given ratios to one another; the parallelograms shall have a given ratio to one another.

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Let two parallelograms ABCD, EFGH which have the given unequal angles ABC, EFG have the ratios of their Gides, viz. of AB to EF, and of BC to FG, given; the ratio of the parallelogram AC to EG is given.

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

gram KBCL: And because each of the angles BAK, BKA is a 43. dat. given, the triangle ABK is given in fpecies : Therefore che

ratio of AB to BK is given ; and the ratio of AB to EF is gi. b g. dat.

ven, wherefore the ratio of BK to EF is given : And the
ratio of BC to FG is given K A L D E H
and the angle KBC is equal

to the angle EFG; therec 67. dat. fore the ratio of the parallelogram KC to EG is gi

B d 35. I. ven : But KC is equal

MON
AC; therefore the ratio of
AC to EG is given.

O

F G The ratio of the parallelogram AC to EG may be found thus : Take the straight line MN given in position and magni. tude, and make the angle MNO equal to the given angle KAB, and the angle NMO equal to the given angle AKB or FEH: And because the ratio of AB to EF is given, make the ratio of NO to P the same ; also make the ratio of P to the fame with the given ratio of BC to FG, the parallelogram AC is to EG, as MO to Q.

Because the angle KAB is equal to the angle MNO, and the angle AKB equal to the angle NMO; the triangle AKB is equiangular to NMO : Therefore as KB to BA, fo is MO 00 ON ; and as BA to EF, fo is NO to P; wherefore, ex aequali

, as KB to EF, so is MO to P: And BC is to FG, as P

d

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pleto

to

dat.

to Q, and the parallelograms KC, EG are equiangular ; therefore, as was shown in prop. 67. the parallelogram KC, that is, AC, is to EG, as MO to .

COR. 1. If two triangles ABC, DEF have two equal angles, 71. or two unequal, but given angles ABC, DEF, and if the ratios of the fides about these angles, viz.

A G D H the ratios of AB to DE, and of BC to EF be given; the triangles shall have a given ratio to one another.

Complete the parallelograms BG, EH; the ratio of BG to EH is gi- B C E F ven'; and therefore the triangles which are the halves o of a 67. or 68. them bave a given ratio to one another.

b 34. 1. COR. 2. If the bases BC, EF of two triangles ABC, DEF have a given ratio to one another, and if also the straight lines AG, 72. DH which are drawn to the bases from the opposite angles, either in equal angles, or unequal, but given angles AĞC, DHF have a given ratio to one K A

L D another; the triangles shall have a given ratio to one another.

Draw BK, EL parallel to AG, DH, and complete the paralle- B G C E H F lograms KC, LF. And because the angles AGC, DHF, or their equals, the angles KBC, LEF are either equal, or unequal, but given; and that the ratio of AG to DH, that is, of KB to LE, is given, as also the ratio of BC to EF; therefore the ra- a 67. or 68. tio of the parallelogram KC to LF is given ; wherefore also the ratio of the triangle ABC to DEF is given

$ 41. 1. 215.5.

CIS. S.

.

dat.

b

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IF a parallelogram which has a given angle be applied to

one side of a rectilineal figure given in species ; if the figure have a given ratio to the parallelogranı, the parallelogram is given in species.

Let ABCD be a rectilineal figure given in species, and to one side of it AB, let the parallelogram ABEF having the given angle ABE be applied ; if the figure ABCD has a given ratio to the parallelogram BF, the parallelogram BF is given in species.

Through the point A draw AG parallel to BC, and through the point C draw CG parallel to AB, and produce GA, CB to

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a 3. def. · the points H, K; because the angle ABC is given", and the

ratio of B to BC is given, the figure ABCD being given in species; therefore, the parallelogram BG is given in species. And because upon the same straight line AB the two rectilineal

figures BD, BG given in species are described, the ratio of b 53. dat. BD 10 BG is given b; and, by hypothesis, the ratio of 09. dat.

BD to the parallelogram BF is given ; wherefore the ratio of d 35. I.

BF, that is, of the parallelogram BH, to BG is given, and therefore the ratio of the straight line KB to BC is given ; and the ratio of - BC to BA is given, wherefore the ratio of KB to BA is given : And because the angle ABC is given, the adjacent angle ABK is given ; and the angle ABE is given, therefore the remaining angle KBE is given. The angle EKB is also given, because it is equal to the angle ABK ; therefore the triangle BKE is given in species, and consequently the 12• tio of EB to BK is given ; and the ratio of KB to BA is given, wherefore the ratio of EB

D to BA is given; and the

N angle ABE is given, there

G

C fore the parallelogram

0 BF is given in species.

M
A

B
A parallelogram fimi-
lar to BF may be found
thus : Take a straight line
LM given in position and H F KE plole
magnitude; and because the angles ABK, ABE are given,
make the angle NLM equal to ABK, and the angle NLO
equal to ABE. And because the ratio of BF to BD is given,
make the ratio of LM to P the same with it; and because the
ratio of the figure BD to BG is given, find this ratio by the
53d dat. and make the ratio of P to Q the same. Also, because
the ratio of CB to BA is given, make the ratio of Q to R the
same; and take LN equal to R; through the point M draw OM
parallel to LN, and complete the parallelogram NLOS; then
this is similar to the parallelogram BF.

Because the angle ABK is equal to NLM, and the angle ABE to NLO the angle KBE is equal to MLO; and the angles BKE, LMO are equal, because the angle ABK is equal to NLM; therefore, the triangles BKE, LMO are equiangular to one another ; wherefore as BE to BK, fo is LO to LM ; and because as the figure BF to BD, fo is the straight line LM to P; and as BD to BG, fo is P to Q; ex aequali, as BF, that is d BH, to BG, fo is LM to Q: But BH is too

BG,

.

BG, as KB to BC; as therefore KB to BC, fo is LM to Q; and because BE is to BK as LO to LM; and as BK to BC, fo is LM to Q: And as BC to BA, so Q was made to R; therefore, ex aequali, as BE to BA, fo is LÒ to R, that is to LN; and the angles ABE, NLO are equal; therefore the parallelogram BF is fimilar to LS.

PROP.

LXX.

62. 78.

IF two straight lines have a given ratio to one another, See N.

and upon one of them be described a rectilineal figure given in species, and upon the other a parallelogram having a given angle; if the figure have a given ratio to the parallelogram, the parallelogram is given in species,

b

9.

Let the two straight lines AB, CD have a given ratio to one another, and upon AB let the figure AEB given in fpecies be defcribed, and upon CD the parallelogram DF having the given angle FCD; if the ratio of AEB to DF be given, the parallelogram DF is given in species.

Upon the straight line AB, conceive the parallelogram AG to be described similar, and similarly placed to FD; and because the ratio of AB to CD is given, and upon them are described the similar rectilineal figures AG,

E. FD; the ratio of AG to FD is gi. ven; and the ratio of FD to AEB

B В

a 54. dat. is given; therefore the ratio of

dat. AEB to AG is given ; and the angle

GC ABG is given, because it is equal to the angle FCD ; because therefore M the parallelogram AG which has a

N

1 1 given angle ABG is applied to a lide AB of the figure AEB given in fpe

H KL cies, and the ratio of AEB to AG is given, the parallelogram AG is given in species ; but FD is similar to AG; therefore c 69, date FD is given in spccies.

A parallelogram similar to FD may be found thus: Take a straight line H given in magnitude; and because the ratio of the figure AEB to FD is given, make the ratio of H to K the same with it: Also, because the ratio of the straight line CD to. AB is given, find by the 54th dat. the ratio which the figure FD described upon CD has to the figure AG described upon AB similar to FD; and make the ratio of K to L the same with this ratio : And because the ratios of H to K, and of K

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