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Take two straight lines DE, DF containing any angle EDF; and Book VI: upon these make DG equal

D to A, GE equal to B, and

Α. DH equal to C; and having

B joined GH, draw EF parallel' to it through the point

C

a. 31.1 E. and because GH is paral

G

H lel to EF one of the sides of the triangle DEF, DG is to GE, as DH to HF b. but E

b. 2.6

F 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.

PROP. XIII.

. PROB. То O find a mean proportional between two given

Itraight lines.

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

Place AB, BC in a straight line, and upon AC describe the fe-
micircle ADC, and from the point
B draw a BD at right angles to
AC, and join AD, DC.

Because the angle ADC in a few micircle is a right angle b, and be

b. 31.38 Cause in the right angled triangle ADC, DB is drawn froin the right angle perpendicular to the base, DB A

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

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Look VI.

PROP. XIV. THEOR.
EQUAL parallelograms which have one angle of

the one equal to one angle of the other, have their fides about the equal angles reciprocally proportional, and parallelograms that have one angle of the one equal to one angle of the other, and their fides about the equal angles reciprocally proportional, are equal to one another.

3.14 1.

b. 7.5

C. 1. 6.

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 alfo 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.

Complete the parallelogram FE; and because the parallelogram
AB is equal to BC, and that

A T
FE is another parallelogram,
AB is to FE, as BC to FE .
but as AB to FE, fo is the base

E
DB to BE“; and as BC to FE, D B
fo is the base GB to BF; there-
fore as DB to BE, fo is GB to
BF a. Wherefore the sides of

G C 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, fo GB to BF; the parallelogram AB is equal to the parallelogram BC.

Because as DB is to BE, fo GB to BF; and as DB to BE, so is the parallelogram AB to the parallelogram FE; and as GB to BF, fo is parallelogram BC to parallelogram FE; therefore as AB to FR, fo BC to FE d. wherefore the parallelogram AB is equal o to the parallelogram BC. Therefore equal parallelograms, &c. Q. E. D.

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Book VI.

PRO P. XV.

THEOR.

EQUAL, triangles which have one angle of the one

equal to one angle of the other, have their fides about the equal angles reciprocally proportional, and triangles which have one angle in the one equal to one angle in the other, and their fides about the equal angles reciprocally proportional, are equal to one another.

E C.1.6.

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 line ; and join BD. Because the

2. 14.2

B triangle ABC is equal to the triangle ADE, and that ABD is another triangle; therefore as the triangle CAB is to the triangle BAD fo is triangle EAD to triacgle DAB b. but as tri

b.. angle CAB to triangle BAD, fo is the base CA to AD; and as triangle EAD to triangle DAB, fo is the base EA to AB; as therefore CA to AD, fo is LA to AB4, wherefore d. zz. si 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 as CA to AD, fo is EA to AB; and as CA to AD; so is triangle BAC to triangle BAD; and as EA to AB, so triangle EAD to triangle BAD"; therefore o as triangle BAC to triangle BAD, fo 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- 6.9.5 angle ADE. Therefore equal triangles, &c. Q. E. D.

Book V1.

IF

PRO P. XVI. THEOR.
F 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.

2. II. I.

b. 7. 5.

C: 14.6.

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

From the points A, C draw * 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), fo 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 fides of the parallelograms BG, DH abont the equal angles are reciprocally proportional ; but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another o ; therefore the parallelogram BG is equal to the parallelogram DH. and the parallelogram BG is contained by the straight lines F

H

F.
AB, F, because AG is equal
to F; and the parallelogram G
DH is contained by CD and
E, because CH is equal to E.
therefore the rectangle con-
tained by the straight lines

A B C D
AB, F is equal to that which
is contained by CD and E.

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 fame 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 ; tberefore the parallelogram BG is equal to the parallelogram DH; and they are equiangular. but the tides

about the equal angles of equal parallelograms are reciprocally pro- Book VI.
portional wherefore as AB to CD, so is CH to AG; and CH
is equal to E, and AG to F. as therefore AB is to CD, so E to F. C. 14. .
Wherefore if four, &c. Q. E. D.

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IF
F 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 ex-
tremes be equal to the square of the mean, the three
straight lines are proportionals.

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b. 16.

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, fo B to C, and that B is equal to D; A is to B, as D to C. but if four straight lines a. 7. 5, be proportionals, the rectangle contained by the A extremes is equal to that

B
which is contained by the
means, therefore the rec-

C
tangle contained by A, C
is equal to that contained

C

D
by B, D. but the rectan-
gle contained by B, D is

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

And if the rectangle contained by A, C be equal to the square
of B; A is to B, as B to 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 b. therefore A is to B, as D to C; but B is equal to Di

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