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book VI.

PROP. VII. THEOR.

See N. If two triangles have one angle of the one equal to

one angle of the other, and the sides about two other angles proportionals, then, if each of the remaining angles be either less, or not less, than a right angle; or if one of them be a right angle: The triangles shall be equiangular, and have those angles equal about which the sides are proportionals.

Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz. the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF; and, in the first case, 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.

For if the angles ABC, DEF be not equal, one of them
is greater than the other: Let
ABC be the greater, and at

A .
the point B, in the straight
i line AB, make the angle
2 23. 1. ABG equal to the anglea

DEF: And because the an-
gle at A is equal to the angle B

at D, and the angle ABG to 6 32. 1. the angle DEF; the remaining angle AGB is equalb to

the remaining angle DFE: Therefore the triangle ABG is 64. 6. equiangular to the triangle DEF; whereforec as AB is to

BG, so is DE to EF; but as DE to EF, so, by hypothesis, à 11. 5. is AB to BC; therefore as AB to BC, so is AB to BG:

and because AB has the same ratio to each of the lines BC, .e9. 5. BG; BC is equale to BG, and therefore the angle BGC is . f5. 1. equal to the angle BCGf: But the angle BCG is, by hy

pothesis, less than a right angle; therefore also the angle

BGC is less than a right angle, and the adjacent angle ? 13. 1. AGB must be greater than a right angles. 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,

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111. 5. is A because Aguale to By? But the herefore adjacent an was

5. BG, because ÁBh crefore

by the hypothesis, it is less than a right angle; which is Book VI. absurd. Therefore the angles ABC, DEF are not unequal, that is, they are 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 also in this case equiangular to the triangle DEF.

The same construction being made, it

D) may be proved in like manner that BC is cqual to BG, and the Re angle at C equal to the angle BGC: But the angle at C is not less than a right angle; therefore the angle BGC is not less than a right angle : Wherefore two angles of the triangle BGC, are together not less than two right angles, which is impossibleh; and therefore the triangle ABC may be proved to h 17. 1. be equiangular to the triangle DEF, as in the first case.

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

For if they be not equiangular, make, at the point B of the straight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC: But the angle BCG is a right angle, thereforei the angle BGC is also a right angle ; whence two of the angles of the triangle BGC are 1 together not less than two right angles, which is impossibleh: Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if the two triangles, &c. Q. E. D.

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7:5.1.

Book VI.

Ecox VI.

PROP. VIII. THEOR.

Because the wholese BC: Pht A let

See N. 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 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 similar 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 * 32. 1. BADa: Therefore the tri

angle ABC is equiangular to

the triangle ABD, and the B 64. 6. sides about their equal angles are proportionalsb; wherec1 Def. 6. fore the triangles are similarc: In the like manner it may

be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC. And the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other. 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 mean 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 side : 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 b; and in the triangles ABC, ACD, BC is to CA, as CA to CDb.

Book VI.

PROP. IX. PROB..

From 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 AČ the same multiple of AD, that AB is A of the part which is to be cut off from it: join BC, and draw DE parallel to it: Then I AE is the part required to be cut off. EL

Because ED is parallel to one of the sides of the triangle ABC, viz. to BC, as CD is to DA, so isa BE to EA; and by

* 2. 6. composition b, CA is to AD, as BA to AE: But CA is a multiple of AD; therefore c

c5 Def. 5. BA is the same multiple of AE: What. B ever 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.

Þ 18. 5.

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.

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, and join BC, and through the points D, E, draw a DF, EG parallels to it;- 31. and through D draw DHK parallel to AB: Therefore each of the figures FH, HB, is a parallelogram: wherefore DH is equal b to FG, and HK to GB: And because HE34. .

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Book VI. is parallel to KC, one of the sides

n of the triangle DKC, as CE to ED,
62. 6. so isc KH to HD: But KH is

equal to BG, and HD to GF; FH
therefore, as CE to ED, so is BG
to GF: Again, because FD is pa- G --
rallel to EG, one of the sides of
the triangle AGE, as ED to DA, B-
so is GF to FA: But it has been
proved that CE is to ED, as BG to GF; and as ED to

DA, so GF to FA: Therefore the given straight line AB · is divided similarly to’AC. Which was to be done.

PROP. XI. PROB.

To find a third, proportional to two given straight

lines.

Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; A it is required to find a third proportional | to AB, AC.

Produce AB, AC, to the points D, E; PL and make BD equal to AC; and having

joined BC, through D, draw DE parallel * 31. 1. to it a. . Because BC is parallel to DE, a side of 1. 2. 6. the triangle ADE, AB is b to BD, as AC D

E to CE: But BD is equal to AC; as therefore AB to AC, so is AC to CE. Wherefore, to the two given straight lines AB, AC, a third proportional CE is found. Which was to be done. .

PROP. XII. PROB..

To 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.

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