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Book VI.

a 23. 1.

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b 32. 1.

Let the triangles ABC, DEF have the angle BAC in the one equal to the angle EDF in the other, and the sides about those angles proportional, that is, BA to AC, as ED to DF; the triangles ABC, DEF are equiangular, and have the angle ABC equal to the angle DEF, and ACB to DFE.

At the points D, F, in the straight line DF, makea the
angle FDG equal to

A
either of the angles
BAC, EDF, and the
angle DFG equal

D
to the angle ACB;
wherefore the
maining angle at B
is equal to the re-
maining angle at Gb; 3

E
consequently the tri-
angle ABC is equiangular to the triangle DGF;
therefore BA : AC :: GD: DF. But, by hypothesis,

BA : AC ::ED: DF; therefore

ED: DF::GD: DFd; wherefore ED is equale to DG. The triangles EDF, GDF have the two sides ED, DF equal to the two sides ĢD, DF, and also the angle EDF equal to the angle GDF; wherefore the base EF is equal to the base FGf, and the angle DFG is equal to the angle DFE, and the angle at G to the angle at E. But the angle DFG is equal to the angle ACB; therefore the angle ACB is equal to the angle DFE; also the angle BAC is equal to the angle EDF8; wherefore the remaining angle at B is equal to the remaining angle at E. Therefore the triangle ABC is equi, angular to the triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

C 4. 6.

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d 11. 5. e 9. 5.

f 4.1.

1

g Hyp.

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PROP. VII. THEOR.

Book VI.

IF two triangles have one angle of one equal to one angle of the other, and the sides about two other angles proportional; and if each of the remaining angles be either less, or not less, than a right angle, the triangles will be equiangular, and have those angles equal about which the sides are proportional.

Let the two triangles ABC, DEF have one angle of one equal to one angle of the other, viz. the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF proportional, AB to BC as DE to EF; and first, let each of the remaining angles at C, F be less than a right angle. The triangle ABC is equiangular to the triangle DEF.

For, if the angles ABC, DEF be not equal, one of them
is greater than the other. Let ABC be the greater, and at
the point B, in the straight
line AB, make the angle

А
ABG equal to the angle
DEF. Because the angle

D
at A is equal to the angle at
D, and the angle ABG to

G the angle DEF, the remaining angle AGB is B

СЕ F equal to the remaining

b 32. 1. angle DFE ; therefore the triangle ABG is equiangular to the triangle DEF; wherefore AB : BG :: DE: EFc.

But, by hypothesis, DE: EF :: AB : BC,
therefore,

AB : BC :: AB': BGI; d 11. 5. therefore BC is equal to BG, therefore the angle BGC is e 9. 5. equal to the angle BCGf. But the angle BCG is, by hypo- f 6. 1. thesis, less than a right angle, therefore the angle BGC is

a 23. 1.

C4. 6.

g 13. 1.

Book VI. less than a right angle, therefore the adjacent angle AGB

must be greater than a right angle. But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle. But, by hypothesis, it is less than a right angle; which is absurd.' Therefore the angles ABC, DEF are not unequal, that is, they are equal. Now the angle at A is equal to the angle at D, wherefore the remaining angle at C is equal to the remaining angle at F. Therefore the triangle ABC is equiangular to the triangle DEF.

Next, let each of the angles at C, F be not less than a
right angle; the triangle ABC is also, in this case, equiangu-
lar to the triangle DEF.
The same construction

A
being made, it may be
proved, in like manner,
that BC is equal to BG,
and the angle at C equal
to the angle BGC. But

B

E the angle at C is not less than a right angle, therefore the angle BGC is not less than a right angle. Where

fore two angles of the triangle BGC are together not less h 17. 1.

than two right angles, which is impossibleh. Therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. Therefore, if two triangles, &c. Q. E. D.

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PROP. VIII. THEOR.

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 each other.

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 A RC, and to each other.

Because the angle BAC is equal to the angle ADB, each Book VI. of them being a right angle, and the angle at B common to the two triangles ABC, ABD, the remain

A ing angle ACB is equal to the remaining angle BADa;

a 32. 1. therefore the triangle ABC. is equiangular to the triangle ABD; therefore the: sides about the equal angles B

C are proportionalb; wherefore

b 4. 6. the triangles are similar. In like manner it may

be demon- c Def. 1.6. strated that the triangle ADC is equiangular and similar to the triangle ABC. The triangles ABD, ADC being equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right angled triangle, &c. Q. E. D.

Cor. 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 each of the sides is a mean proportional between the base and its segment adjacent to that side. For, in the triangles BDA, ADC, BD: DA :: DA : DCb; and in the triangles ABC, DBA, BC: BA :: BA: BDb; and in the triangles ABC, ACD, BC: CA '::CA: CDb.

PROP. IX. PROB.

FROM a given straight line to cut off any part required, that is, a part which shall be contained in it a given number of times.

Let AB be the given straight line ; it is required to cut off from AB a part which shall be contained in it a given number of times.

Book VI.

From the point A draw a straight line AC, making any
angle with AB; in AC take any point D, and take AC such
that it shall contain AD as: often as
AB is to contain the part which is to be

A
cut off from it; join BC, and draw DE
parallel to it; then AE is the part re-

E D
quired to be cut off.

Because ED is parallel to BC, one of the sides of the triangle ABC,CD : DA :: BE : EAa; therefore, by composition, CA: AD:: BA: AEB. But CA is a multiple of AD, therefore BA is the

B same multiple of AE®; therefore, whatever part AD is of AC, AE is the same of AB; wherefore, from the straight line AB the part required is cut off. Which was to be done.

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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.

a 31.1.

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

Let AC be divided in the points D, E ; and let AB, AC be placed so as to contain any angle; join BC, and through the points D, E draw a DF,

A
EG parallel to BC, and through
D draw DHK parallel to AB;
therefore each of the figures FH,

D
HB is a parallelogram ; wherefore
DH is equalb to FG, and HK to

H
GB. Because HE is parallel to KC,

с
one of the sides of the triangle
DKC, CE : ED: : KH : HD.

B K
But KH = BG, and HD=GF;

b 34. 1.

c 2. 6.

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