b 34. 1. c 2. 6.

b

C

Book VI. equal to FG, and HK to GB:
And because HE is parallel to KC,
one of the sides of the triangle
DKC, as CE to ED, so is KH to
HD: But KH is equal to BG, and
HD to GF: therefore, as CE to G
ED, so is BG to GF: Again, be-
cause FD is parallel to EG, one of
the sides of the triangle AGE, as
ED to DA, so is GF to FA: But

a 31. 1.

b 2. 6.

B

it has been proved, that CE is to ED, as BG to GF; and as ED to DA, so is 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; it is required to find a third proportional to AB, AC.

E;

A

C

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

Because BC is parallel to DE, a side of
the triangle ADE, AB is to BD, as AC
to CE: But BD is equal to AC; as there- D
fore AB to AC, so is AC to CE. Where-

E

fore, 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.

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 a to it through the point E: and because GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH to HFb:

but DG is equal to A, GE E

to B, and DH to C; there

fore, 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 Which was to be done.

found.

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

D

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

a 11. 1.

Because the angle ADC in a semicircle is a right angle ", and because in the right angled triangle ADC, DB is drawn from the right angle

perpendicular to the base, DB is a mean proportional between

b 31. 3.

AB, BC the segments of the base: Therefore, between the c Cor. 8.6. two given straight lines AB, BC, a mean proportional DB

is found.

Which was to be done.

Book VI.

a 14. 1.

b 7. 5.

c 1. 6.

d 11. 5.

e 9. 5.

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 straight line. The sides of the parallelograms AB, BC about the equal angles, are reciprocally proportional: that is, DB is to BE, as GB to BF.

A

Complete the parallelogram FE; and because the parallelogram AB is equal to BC, and FE is another parallelogram, AB is to FE, as BC to FE: But as AB to FE, so is the base DB to BE; and as BC to FE, so is the base GB to BF; therefore, as DB to BE, so is GB to BFd. Wherefore the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

F

E

D

B

G C

But, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so is 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 GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so is BC to FE : Wherefore, the parallelogram AB is equal to the parallelogram BC. Therefore," equal parallelograms," &c. Q. E. D.

е

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 these triangles are reciprocally proportional; that is, CA is to AD, as EA to AB.

B

D

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; and join BD. Because the triangle ABC is a 14. 1. equal to the triangle ADE, and ABD is another triangle: therefore as the triangle CAB is to the triangle BAD, so is triangle EAD to triangle DABb: But as triangle CAB to triangle BAD, so is the base CA to AD: and as triangle EAD to triangle DA B, so is the base EA to AB; as therefore CA to AD, so is EA to AB4; wherefore, the sides of the tri

E

angles 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.

b 7. 5.

c l. 6.

d 11. 5.

Having joined BD as before; because, as CA to AD, so is EA to AB: and since as CA to AD, so is triangle BAC to triangle BAD; and as EA to AB, so is triangle EAD to triangle BAD: therefore, 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 triangle ADE. e 9. 5. Therefore, "equal triangles," &c. Q. E. D.

e

L

Book VI.

a 11. 1.

b 7. 5.

PROP. XVI. THEOR.

If 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.

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

ს

From the points A, C, draw a 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 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 proportional, are equal to one c 14. 6. 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 parallelogram DH is contained by CD and E; because CH is equal to E: Therefore the rectangle contained by the straight lines AB, F is equal to that which is contained by CD and E.

F

G

A

H

BC

D

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 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 :