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Book VI.

See N.

2. 32. I.

IN

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N a right angled triangle, if a perpendicular be drawn from the right angle to the bafe; the triangles on each fide of it are fimilar 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 fimilar to the whole triangle ABC, and to one another.

A

Because the angle BAC is equal to the angle ADB, each of them being a right angle, and that the angle at B is common to the twe triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BADa. therefore the triangle ABC is equiangular to the triangle ABD, and the fides about their equal angles are proportionals, wherefore the c. 1. Def. 6. triangles are fimilar c. in the like

b. 4. 6.

B

D C

manner it may be demonftrated that the triangle ADC is fimilar to the triangle ABC.

Alfo the triangles ABD, ADC are fimilar to one another.

Because the right angle BDA is equal to the right angle ADC, and also the angle BAD to the angle at C, as has been proved; the remaining angle at B is equal to the remaining angle DAC. therefore the triangle ABD is equiangular and fimilar to the triangle ADC. Therefore in a right angled, &c. Q. E. D.

c

COR. From this it is manifeft that the perpendicular drawn from the right angle of a right angled triangle to the bafe, is a inean proportional between the fegments of the base. and also that each of the fides is a mean proportional between the base, and its fegment adjacent to that fide. 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 CD,

PROP. IX. PROB.

Book VI.

ROM a given ftraight line to cut off any part re- See N. quired.

FRO

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 which is the fame 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 fame
of AB that AD is of AC; that is, AE E
part
is the part required to be cut off.

Because ED is parallel to one of the fides
of the triangle ABC, viz. to BC, as CD is to
DA, fo is a BE to EA; and, by compofition,
CA is to AD, as BA to AE. but CA is a B
multiple of AD, therefore BA is the fame'

D

multiple of AE. whatever part therefore AD is of AC, AE is the fame part of AB. wherefore from the ftraight line AB the part required is cut off. Which was to be done.

PROP. X. PROB.

O divide a given ftraight line fimilarly to a given divided straight line, that is, into parts that fhall have the fame 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

fimilarly to AC.

Let AC be divided in the points D, E; and let AB, AC be placed fo as to contain any angle, and join BC, and F

G

through the points D, E draw a DF,
EG parallels to it; and through D draw
DHK parallel to AB. therefore each of B
the figures FH, HB is a parallelogram;

2. 2. 6.

b. 18. 5.

c. D. 5.

A

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b. 34. I. C. 2. 6.

c

A

Book VI. wherefore DH is equal to FG, and HK to GB. and because HE is parallel to KC one of the fides of the triangle DKC, as CE to ED, fo is © KH to HD. but KH is equal to BG, and HD to GF; therefore as CE to ED, fo is BG to GF. again, because FD is parallel to EG one of the fides of the tri- G angle AGE, as ED to DA, fo is GF to FA. but it has been proved that CE is B

to ED, as BG to GF; as therefore CE

F

H

K

to ED, fo is BG to GF; and as ED to DA, fo GF to FA. therefore the given ftraight line AB is divided fimilarly to AC. Which was to be done.

a. 31. I.

b. 2. 6.

PROP. XI. PROB.

find a third proportional to two given straight

lines.

Let AB, AC be the two given straight lines, and let them be placed fo as to contain any angle; 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, thro' D draw DE parallel to it 2.

b

'B

A

Because BC is parallel to DE a fide of the
triangle ADE, AB is to BD, as AC to CE.
but BD is equal to AC, as therefore AB to.
AC, fo is AC to CE. Wherefore to the two D

E

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

PROP. XII. PROB.

10 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 Book VI.

upon these make DG equal to A, GE equal to B, and DH equal to C; and having joined GH, draw EF paral lel a to it through the point E. and because GH is parallel to EF one of the fides of the triangle DEF, DG is to GE, as DH to HF 6. but

DG is equal to A, GE to B,

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and DH to C; therefore as A is to B, fo is C to HF. Wherefore to the three given ftraight lines A, B, C a fourth proportional HF is found. Which was to be done.

To

PROP. XIII. PROB.

find a mean proportional between two given
ftraight lines.

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

Place AB, BC in a ftraight line, and upon AC defcribe the fe

micircle ADC, and from the point

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

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

angle perpendicular to the bafe, DB A

is a mean proportional between

b. 2. 6.

D

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B

AB, BC the feginents of the bafe, therefore between the two c. Cor. 8. 6.

given straight lines AB, BC, a mean proportional DB is found.

Which was to be done.

Book VI.

a. 14. I.

b. 7.5.

C. I. 6.

d. 11. 5.

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QUAL 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.

Let AB, BC be equal parallelograms which have the angles at B equal, and let the fides DB, BE be placed in the same straight line; wherefore also FB, BG are in one straight line 2. the fides 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 FE is another parallelogram, AB is to FE, as BC to FE ь. but as AB to FE, fo is the base DB to BE; and as BC to FE, fo is the bafe GB to BF; there-fore as DB to BE, fo is GB to

BF d. Wherefore the fides of

the parallelograms AB, BC a

A

F

E

D

B

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e. 9. 5.

bout their equal angles are reciprocally proportional.

But let the fides 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, fo is the parallelogram AB to the parallelogram FE; and as GB to BF, fo is parallelogram BC to parallelogram FE; therefore as AB to FE, fo BC to FE. wherefore the parallelogram AB is equal e to the 'parallelogram BC. Therefore equal parallelograms, &c. Q. E. D.

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