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Book VI.

See N.

a 23. I.

b 32 I.

C 4 6.

dir. 5.

€ 9.5. f 5. I.

g 13. I.

IF

PROP. VII. THEOR.

[F two triangles have one angle of the one equal to one angle of the other, and the fides about two other angles proportionals, then, if each of the remaining angles be either lefs, or not lefs, than a right angle; or if one of them be a right angle: The triangles shall be equiangular, and have thofe angles equal about which the fides are proportionals.

Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAČ to the angle EDF, and the fides about two other angles ABC, DEF proportionals, fo that AB is to BC, as DE to EF; and, in the firft cafe, let each of the remaining angles at C, F be less than a right angle. The triangle ABC is equiangular to the triangle DEF, viz. the angle ABC is equal to the angle DEF, and the remaining angle at C to the remaining angle at F.

B

A

D

G

CE

F

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 ftraight line AB, make the angle ABG equal to the angle DEF: And because the angle at A is equal to the angle at D, and the angle ABG to the angle DLF; the remaining angle AGB is equal to the remaining angle DFE: Therefore the triangle ABG is equiangular to the triangle DEF; wherefore as AB is to BG, fo is DE to EF; but as DE to EF, fo, by hypothefis, is AB to BC; therefore as AB to BC, to is AB to BG; and because AB has the fame ratio to each of the lines BC, BG; BC is equal to BG, and therefore the angle BGC is equal to the angle BCG f: But the angle BCG is, by hypothefis, lefs than a right angle; therefore alfo the angle BGC is lefs than a right angle, and 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 the hypothefis, it is lefs than a right angle; which is abfurd. There

fore

fore the angles ABC, DEF are not unequal, that is, they are Book VI. equal: And 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 alfo in this cafe equiangular to the triangle DEF.

The fame conftruction 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 the angle B at C is not less than a right angle; therefore the angle

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BGC is not lefs than a right angle: Wherefore two angles of the triangle BGC are together not less than two right angles, which is impoffible; and therefore the triangle ABC may be h 17. 1. proved to be equiangular to the triangle DEF, as in the first

cafe.

Laftly, Let one of the angles at C, F, viz. the angle at C, be a right angle; in this cafe likewife the triangle ABC is equiangular to the triangle DEF.

For, if they be not equiangular, make, at the point B of the ftraight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the firft cafe, that BG is e B qual to BC: But the angle' BCG is a right angle, therefore the angle BGC is alfo a right angle; whence two of the angles of the triangle BGC are together not less than two right angles, which is impoffible: Therefore the triangle ABC is equiangular to the

A

C

A

G

i s. I.

E

C

G

triangle DEF. Wherefore, if two triangles, &c. Q. E. D.

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Book VI.

PROP. VIII.

THEOR.

See N.

a 32. I.

IN

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.

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 two triangles ABC,

ABD; the remaining angle ACB is equal to the remaining angle BAD: Therefore the triangle ABC is equiangular to the triangle ABD, and the fides about their equal angles are prob 4. 6. portionals; wherefore the tri- B c 1. Def. 6. angles are fimilar: In the like

D C

manner it may be demonftrated, that the triangle ADC is e quiangular and fimilar to the triangle ABC: And the triangles ABD, ADC, being both equiangular and fimilar to ABC, are equiangular and fimilar to each other. Therefore, in a right angle, &c. Q. E. D.

COR. From this it is manifeft, that the perpendicular drawn from the right angle of a right angled triangle to the base, is a mean proportional between the fegments of the bafe: And alfo 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 b; and in the triangles ABC, ACD, BC is to CA, as CA to CD b.

PROP.

Book VI.

PROP. IX. PROB.

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

FROM

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

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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 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 part required to be cut off.

A

D

Because ED is parallel to one of the fides E of the triangle ABC, viz. to BC, as CD is to DA, fo is BE to EA; and, by compofition b, CA is to AD, as BA to AE: But CA is a multiple of AD; therefore BA is the fame multiple of AE: Whatever part therefore AD is of AC, AE is the fame part of AB: Wherefore, from the straight line AB the part required is cut off. Which was to be done.

B

a 2. 6.

b 18. 5.

c D. 5.

Tdi

PROP. X. PROB.

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 ftraight 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 so as to contain any angle, and join BC, and through the points D, E, draw DF, EG parallels to it; and through Da 31. 1. draw DHK parallel to AB: Therefore each of the figures FH,

HB, is a parallelogram; wherefore DH is equal to FG, and b 34. 1.

L 4

HK

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€ 2. 6.

F

Book VI. HK to GB: And because HE is pa-
rallel to KC, one of the fides of the
triangle DKC, as CE to ED, so 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, G
because FD is parallel to EG, one of
the fides of the triangle AGE, as ED B
to DA, fo is GF to FA: But it has
been proved that CE is to ED, as

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

a 31. I.

b 2. 6.

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Let AB, AC be the two given ftraight 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 B
joined BC, through D, draw DE parallel to

it a.

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 D

E

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

T

PROP. XII. PROB.

O 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

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