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Book VI.

PROP. VIII. THEOR.

See N. IN a right angled triangle, if a perpendicular be drawn from the right angle to the base; the triangles on each side of it are similar to the whole triangle, and to one another.

Let ABC be a right angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicular to the base BC: The triangles ABD, ADC are similar to the whole triangle ABC, and to one another.

Because the angle BAC is equal to the angle ADE, each of them being a right angle, and that the angle at B is common to the two trian

gles ABC, ABD: the re-
maining angle ACB is equal
to the remaining angle
32. 1. BAD: Therefore the tri-
angle ABC is equiangular to
the triangle ABD, and the B

c 1 Def. 6.

4. 6. sides about their equal angles are proportionals; wherefore the triangles are similare: In the like manner it may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC: And the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right angled, &c. Q. E.D.

COR. From this it is manifest that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the segments of the base: And also, that each of the sides is a mean proportional between the base, and its segment adjacent to that side: Because in the triangles BDA, ADC, BD is to DA, as DA to DC; and in the triangles ABC,DBA,BC is to BA, as BA to BD; and in the triangles ABC, ACD, BC is to CA, as CA to CDb.

Book VI.

PROP. IX. PROB.

FROM a given straight line to cut off any part re- Sea N. quired.

Let AB be the given straight line; it is required to cut off any part from it.

From the point A draw a straight line AC, making any angle with AB; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it: join BC, and draw DE parallel to it: Then AE is the part required to be cut off.

E

Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so is a BE to EA; and by composition", CA is to AD, as BA to AE: But CA is a multiple of AD; therefore c BA is the same multiple of AE: What- B

C

A

D

C

ever part therefore AD is of AC, AE is the same part of AB: Wherefore, from the straight line AB the part required is cut off. Which was to be done.

2.6. b 18. 5.

c D. 5.

PROP. X. PROB.

To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have.

Let AB be the straight line given to be divided, and AC the divided line: it is required to divide AB similarly to AC.

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Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle and join BC, and through the points D, E, draw a DF, EG parallels to it; 31.1. and through D draw DHK parallel to AB: Therefore each of the figures FH, HB, is a parallelogram: wherefore DH is equal to FG, and HK to GB: And because HE

34. 1.

Book VI.

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 parallelogram AB is equal to A

BC, and that FE is another parallelogram, AB is to FE, 7. 5. as BC to FE:. But as AB

to FE, so is the base DB to..
1. 6. BE: and as BC to FE, so

is the base of GB to BF; ..
therefore, as DB to BE, so is

G

E

11. 5. GB to BF4. 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 GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FEd; 9. 5. Wherefore the parallelogram AB is equal to the parallelogram BC. Therefore equal parallelograms, &c. Q.E.D.

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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.

B

a

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, C

E

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7.5.

1. 6.

so is the base EA to ABC: as therefore CA to AD, so is EA to ABd; 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.

e

Book VI.

a 11. 1.

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 E to F; the rectangle contained by AB, F is equal to the rectangle contained by CD, E.

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
7. 5. F to AG; AB is b 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 propor-
11.6. tional, are equal to one another; therefore the parallel-
ogram BG is equal to the parallelogram DH: And the
parallelogram BG is con-
E

tained by the straight lines
AB, F; because AG is F
equal to F; and the pa- G
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

A

that which is contained by CD and E.

B

H

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 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 reetangle 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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