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Book VI.

PROP. VIII.

THEOR.

See N.

IN
N a right angled triangle, if a perpendicular be drawn

from the right angle to the base; the triangles on each fide of it are similar to the whole triangle, and to one another.

1

3. 32. I.

Let ABC be a right angles 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.

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 re-
maining angle ACB is equal to

А
the remaining angle BAD. there-
fore the triangle ABC is equian-
gular to the triangle ABD, and

the fides about their equal angles
b. 4. 6.
are proportionals, wherefore the

B

D. c Co 1. Def. 6.

triangles are similar c. in the like
manner it may be demonstrated that the triangle ADC is similar
to the triangle ABC.

Also the triangles ABD, ADC are similar 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 similar to the triangle ADC. 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 inean 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 fide. because in the triangles BDA, ADC, BD is to DA, as DA to DC b; and in the triangles ABC, DBA, BC is to BA, as BA to BD 0; and in the triangles ABC, ACD, BC is to CA, as CA to CD).

Book VI.

PROP. LX.

PROB.

Fequired.

ROM a given straight line to cut off any part re. See 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 which is the fame multiple of AD that AB is of the part A which is to be cut off from it; join BC, and draw DE parallel to it. then AE is the same part of AB that AD is of AC; that is, AE E D is the part required to be cut off.

Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, fo is a BE to EA; and, by composition", CA is to AD, as BA to AE. but CA is a aB C multiple of AD, therefore CBA is the same

c. D. 50 multiple of AE. whatever 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.

a. 2. 6.

b. 18. So

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O 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 di-
vided line; it is required to divide AB
{imilarly to AC.

A
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

H through the points D, E draw a DF, G

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

К.

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a. 31. I.

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Book VI. wherefore DH is equal to FG, and HK to GB. and because HE

is parallel to KC one of the sides of the b. 34. I.

triangle DKC, as CE to ED, so is .KH A
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-

F

H
rallel to EG one of the sides of the tri- G
angle AGE, as ED to DA, fo is GF to
FA. but it has been proved that .CE is B

K
to ED, as BG to GF; as therefore CE
to ED, fo is BG to GF; and as ED to DA, fo GF to FA. there-
fore the given straight line AB is divided similarly to AC. Which
was to be done.

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To lines.

10 find a third proportional to two given straight

lines.

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a. 31. I.

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.

А.
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 a.

B C
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 therefore AB to
AC, fo is AC to CE. Wherefore to the two D E
given straight lines AB, AC a third proportional CE is found.
Which was to be done.

b. 2. 6.

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To find

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 ftraight lines DE, DF containing any angle EDF; and Book VI. upon these make DG equal

D to A, GE equal to B, and

A DH equal to C; and having

B. joined GH, draw EF parala lel a to it through the point

C... 4. 31. 2: E. and because GH is parallel to EF one of the sides of

G H the triangle DEF, DG is to GE, as DH to HF 6. but E

b. 1.

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.

To

To find a mean proportional between two giveti

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 defcribe the sea micircle ADC, and from the point B draw · BD at right angles to

D

å. ti. t. AC, and join AD, DC.

Because the angle ADC in a femicircle is a right angle", and be

b. 3. 3. cause in the right angled triangle ADC, DB is drawn from the right angle perpendicular to the base, DB

A B is a mean proportional between AB, BC the feginents of the base c, therefore between the two c. Cor. 8. 8. given straight lines AB, BC, a mean proportional DB is found. Which was to be done.

Book VI.

PROP. XIV.

THEOR.

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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.

b. 7. 5.

C, I, 6.

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 2. 14. 1. line; wherefore also FB, BG are in one 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 BC, and that
FE is another parallelogram,

А. F
AB is to FE, as BC to FE 6.
but as AB to FE, fo is the base

E
DB to BE C; and as BC to FE, D B
so is the base GB to BF; there-

fore as DB to BE, so is GB to d. 11. 5. BF d. Wherefore the sides of

G
the parallelograms AB, BC a-
bout 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 is to BE, so GB to BF; and as DB to BE, fo is the parallelogram AB to the parallelogram FE; and as GB to BF, so is parallelogram BC to parallelogram FE; therefore as AB to FE, fo BC to FE d. wherefore the parallelogram AB is equal e to the parallelogram BC. Therefore equal parallelograms, &c. R. E. D.

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