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F

BOOK VI. is parallel to KC, one of the sides
of the triangle DKC, as CE to ED,
2. 6. so is KH to HD: But KH is
equal to BG, and HD to GF;
therefore, as CE to ED, so is BG
to GF: Again, because FD is pa- G
rallel to EG, one of the sides of
the triangle AGE, as ED to DA, B
so is GF to FA: But it has been

D

HE

K

proved that CE is to ED, as BG to GF; and as ED to DA, so GF to FA: Therefore the given straight line AB is divided similarly to AC. Which was to be done.

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; A it is required to find a third proportional

to AB, AC.

Produce AB, AC, to the points D, E; and make BD equal,to AC; and having joined BC, through D, draw DE parallel a 31. 1. to it a.

B

Because BC is parallel to DE, a side of 12. 6. the triangle ADE, AB is to BD, as AC D

E

to CE: But BD is equal to AC; as therefore AB to AC, so is AC to CE. Wherefore, to the two 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 Book VI.

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but DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done.

b 2.6.

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 draw a BD at right angles to AC, and join AD, DC.

Because the angle ADC in a semicircle is a right angleb, and because in the right

angled triangle ADC, BD is A

D

B

11. 1.

b 31. 3.

drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the base: Therefore between the two given straight lines Cor. 8. 6. AB, BC, a mean proportional DB is found. Which was

to be done.

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BOOK VI.

a

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 that have one 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.

Let AB, BC be equal parallelograms, which have the angles at B equal, and let the sides DB, BE be placed in the same straight line; wherefore also FB, BG are in one 14. 1. straight line a. The sides of the parallelograms AB, BC about the equal angles, are reciprocally proportional; that is, DB is to BE, as GB to BF.

Complete the parallelogram FE; and because the rallelogram AB is equal to A

pa

b

F

BC, and that FE is another

parallelogram, AB is to FE,

7.5.

as BC to FEb: But as AB

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to FE, so is the base DB to

B

G

C

1. 6. BE: and as BC to FE, so
is the base of GB to BF;
therefore, as DB to BE, so is

11. 5. GB to BFd. 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, as DB to BE, so is GB to BF: and as DB to BE, so is the parallelogram AB to the parallelogram FE; and as CB to BF, so is the parellelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FEd: 9. 5. Wherefore the parallelogram AB is equale to the parallelogram BC. Therefore equal parallelograms, &c. Q. E.D.

BOOK VI.

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 angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE; the sides about the equal angles of the triangles are reciprocally proportional; 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 linea; and join BD. Because the triangle * 14. 1. ABC is equal to the triangle

:

ADE, and that ABD is another triangle therefore as the triangle CAB, is to the triangle BAD, so is triangle AED to triangle DABb: But as triangle CAB to triangle BAD, so is the base CA to AD; and as triangle EAD to triangle DAB,

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so is the base EA to AB: as therefore CA to AD, so is EA to AB; wherefore the sides of the triangles ABC, 11. 5. 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, as CA to AD, so is EA to AB; and as CA to AD, so is triangle ABC to triangle BAD; and as EA to AB, so is triangle EAD to triangle BAD; therefore d as triangle BAC to triangle BAD, so is triangle EAD to triangle BAD; that is, the triangles BAC, EAD have the same ratio to the triangle BAD: Wherefore the triangle ABC is equal to the tri- 9. 5. angle ADE. Therefore equal triangles, &c. Q. E. D.

BOOK VI.

11. 1.

PROP. XVI. THEOR.

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

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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 drawa 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, as AB to CD, so is E to F; and that E. is equal to CH, and 7. 5. F to AG; AB is to CD as CH to AG. 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 propor14. 6. tional, are equal to one another; therefore the parallelogram BG is equal to the parallelogram DH: And the parallelogram BG is con- E

tained by the straight lines
AB, F; because AG is
equal to F; and the pa-
rallelogram DH is con-
tained by CD and E; be-
cause CH is equal to E:
Therefore the rectangle
contained by the straight
lines AB, F is equal to

F

G

A

that which is contained by CD and E.

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"And if the rectangle contained by the straight lines, AB, F be equal to that which, is contained by CD, E; these four lines are proportional, 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 that 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

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