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gram ; wherefore DH is equal (34. 1.) to
FG, and HK to GB : and because HE is
parallel to KC, one of the sides of the tri-
angle DKC, CE: ED :: (9. 6.) KH : HD:
But KH=BG, and HD=GF ; therefore F D
CE: ED: BG : GF : Again, because
FD is parallel to EG, one of the sides of

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H Н ES
thé triangle AGE, ED: DA : GF : FA:
But it has been proved that CE: ED : :
BG : GF; therefore the given straight line

B K
AB is divided similarly to AC.
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PROP. XI. PROB. To find a third proportional to two given straight lines. Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle ; it is required to find a third proportional to AB,

А. AC.

Produce AB, AC to the points D, E; and make BD equal to AC ; and having B

C с joined BC, through D draw DE parallel to it (31. 1.).

Because BC is parallel to DE, a side of the triangle ADE, AB : (2. 6.) BD :: AC: CE; but BD=AC ; therefore AB : AC:: AC: CE. Wherefore to the two

D given straight lines AB, AC a third proportional, CE is found. Which was to be done.

PROP. XII. PROB. To find a fourth proportional to three given straight lines. Let A, B, C be the three given straight lines; it is required to find a fourth proportional to A, B, C.

Take two straight lines DE, DF, containing any-angle EDF ; and upon these make DG equal to A, GE equal to B, and DH equal to C; and having joined GH, draw EF parallel (31. 1.) to it through the

A

B

1

H

point E.

And because GH is parallel to EF, one of the sides of the triangle DEF, DG : GE :: DH : HF (2. 6.); but DG=A, GE=B, and DH=C; and therefore A : B::C: HF. Wherefore to the three given straight lines, A, B, C, a fourth proportional HF is found. Which was to be done.

PROP. XIII. PROB.

To find a mean proportional between two given straight lines.

Let AB, BC be the two given straight lines; it is required to find a mean proportional between them.

Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B

D (11. 1.) draw BD at right angles to AC, and' join AD, DC.

Because the angle ADC in a semicircle is a right angle (31. 3.) and because in the right angled triangle ADC, DB is drawn from the right angle, perpendicular to

B.

С the base, DB is a mean proportional between AB, BC, the segments of the base (Cor.8.6.) ; therefore between the two given straight lines AB, BC, a mean proportional DB is found. Which was to be done.

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PROP. XIV. THEOR.

Equal parallelograms which have one angle of the one equal to one angle

of the other, have their sides about the equal angles reciprocally proportional : And parallelograms which have une angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

1

Let AB, BC be equal paralle. A

IT lograms, which have the angles at B equal, and let the sides DB, BE be placed in the same straight

D line.; wherefore also FB, BG D

B are in one straight line (14. 1.): the sides of the parallelograms AB, BC, about the equal angles, are reciprocally proportional ; that is, DB is to BE, as GB to

c BF.

Complete the parallelogram FE ; and because the parallelograms 4B, BC are equal, and FE is another parallelogram,

AB : FE :: BC: FE (7.5.) ;

G

but because the parallelograms AB, FE have the same altitude,

AB : FE :: DB : BE (1. 6.), also,
BC: FE :: GB : BF (1.6.); therefore

DB : BE:: GB : BF (11. 5.). Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

But, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF ; the parallelogram AB is equal to the parallelogram BC.

Because, DB : BE :: GB : BF, and DB : BE :: ÅB : FE, and GB : BF :: BC : EF, therefore, AB : FE :: BC : FE (11. 5.) : Wherefore the parallelogram AB is equal (9. 5.) to the parallelogram BC. Therefore equal parallelograms, &c. Q. E. D.

PROP. XV. THEOR.

Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: And triangles which have one angle in the one equal to one anglë in the other, and their sides about the equal angles reciprocally pro: portional, are equal to one another. Let ABC, ADE be equal

B triangles, which have the angle BAC equal to the angle DAE ; the sides about the equal angles of the triangles are reciprocally proportion

с

ΑΙ al ; that is, CA is to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD be in one straight line ; wherefore also EA and AB are in one straight line (14.

E 1.) ; join BD. Because the triangle ABC is equal to the triangle ADE, and ABD is another triangle ; therefore, triangle CAB : triangle BAD :: triangle EAD : triangle BAD ; but CAB : BAD :: CA : AD, and EAD : BAD :: EA : AB ; therefore CA : AD :: EA : AB (11. 5.), wherefore the sides of the triangles ABC, ADE about the equal angles are reciprocally proportional.

But let the sides of the triangles ABC, ADE, about the equal angles be reciprocally proportional, viz. CA to AD, as EA to AB; the triangle ABC is equal to the triangle ADE.

Having joined BD as before ; because CA : AD :: EA : AB ; and since CA : AD :: triangle ABC : triangle BAD (1. 6.); and also EA : AB :: triangle EAD : triangle BAD (11. 5.); therefore, triangle ABC : triangle BAD : : triangle EAD : triangle BAD; that is,

.

the triangles ABC, EAD have the same ratio to the triangle BAD : wherefore the triangle ABC is equal (9. 5.) to the triangle EAD. Therefore equal triangles, &c. Q. E. D.

PROP. XVI. THEOR.

I four straight lines be proportionals, the rectangle contained by the

extremnes is equal to the rectangle contained by the means : And if the rectangle contained by the extremes be equal to the rectangle contained by means, the four straight lines are proportionals.

Let the four straight lines, AB, CD, E, F be proportionals, viz. as AB to CD, so E to F ; the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

From the points A, C draw (11. 1.) AG, CH at right angles to AB, CD ; and make AG equal to F, and CH equal to E, and complete the parallelograms BG, DH. Because AB : CD ::E : F.; and since E=CH, and F=AG, AB : CD (7. 5.):: CH : AĢ: therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional ; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another (14.6.); therefore the parallelogram BG is equal to the parallelogram DH : and the parallelogram BG is Econtained by the straight lines AB,

H F; because AG is equal to F; and F the parallelogram DH is contained by CD and E, because CH is equal

G to E: therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E. And if the rectangle contained A

B

с by the straight lines AB, F be equal to that which is contained by CD, E; these four lines are proportionals, viz. AB is to CD, as E to F.

The same construction being made, because the rectangle contained by the straight lines AB, F is equal to that which is contained by CD, E, and the rectangle BG is contained by AB, F, because AG is equal to F; and the rectangle DH, by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram DH and they are equiangular : but the sides about the equal angles of equal parallelograms are reciprocally proportional (14. 6.): where. fore AB : CD :: CH : AG ; but CH=E, and AG=F, therefore AB : CD :: E:F. Wherefore, if four, &c Q. E. D.

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D

PROP. XVII. THEOR

If three straight lines be proportionals, the rectangle contained by the

extremes is equal to the square of the mean : And if the rectangle contained by the extremes be equal to the square of the mean, the three straight lines are proportionals.

Let the three straight lines A, B, C be proportionals, viz. as A to B, so B to C ; the rectangle contained by A, C is equal to the square of B.

Take D equal to B : and because as A to B, so B to C, and that B is equal to D; A is (7. 5.) to B, as D to C: but if four straight lines be proportionals, the rectangle contained by the extremes is equal to that which is contained by the means (16. 6.): A therefore the rectangle A.C = the rectan

B gle B.D; but the rectangle B.D is equal to the

D square

of B, because B=D; therefore the rectangle A.C is equal to the square of B. C

And if the rectangle contained by A, C be equal to the square of B; A:B::B:C.

The same construction being made, because the rectangle contained by A, C is equal to the square of B, and the square of B is equal to the rectangle contained by B, D, because B is equal to D; therefore the rectangle contained by A, C is equal to that contained by B, D; but if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals (16.6.): therefore A:B ::D:C, but B=D; wherefore A : B ::B:C: Therefore, if three straight lines, &c. Q. E. D.

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PROP. XVIII. PROB.

Upon a given straight line to describe a rectilineal figure similar, and

similarly situated to a given rectilineal figure. Let AB be the given straight line, and CDEF the given rectilineal figure of four sides; it is required upon the given straight line AB to describe a rectilineal figure similar, and similarly situated to CDEF.

Join DF, and at the points A, B in the straight line AB, make (23. 1.) the angle BAG equal to the angle at 'C, and the angle ABG equal to the angle CDF; therefore the remaining angle CFĎ is equal to the remaining angle AGB (32. 1.): wherefore the triangle FCD is equiangular to the triangle GAB : Again, at the points G, B in the straight line GB make (23. 1.) the angle BGH equal to the angle DFE, and the angle GBH equal to FDE ; therefore the remaining angle FED is equal to the remaining angle GHB, and the triangle FDE equiangular to the triangle GBH: then, because the angle AGB is equal to the angle CFD, and BGH to DFE, the whole angle AGH is equal to the

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