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Book VI.

PROP. VI. THEOR.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have those angles equal which are opposite to the homologous sides.

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 proportionals; 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, make a the a 23. 1. angle FDĠ equal to either of the angles BAC, EDF; and the angle DFG equal to the angle ACB: Wherefore

A the remaining angle at B is equal to the remaining angle at G,b and consequently the

D

G

b 32. 1. triangle ABCis equiangular to the triangle DGF; and therefore as BA to AC, so is c GD to DF; But, by hy

c 4. 6. pothesis, as BA to AC, so is ED to DF; and therefore

B CE as ED to DF, so is d GD to

dll. 5. DF; wherefore ED is equale to DG; and DF is common e 9. 5. to the two triangles EDF, GDF: Therefore, the two sides ED, DF are equal to the two sides GD, DF, and the angle EDF is also equal to the angle GDF; wherefore, the base EF is equal to the base FG', and the triangle EDF to the f 4. 1. triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides : Therefore, 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: And the angle BAC is equal to the angle EDF8; wherefore also the remaining angle at B is equal to the remaining angle at E. Therefore, the triangle ABC is equiangular to the triangle DEF. Wherefore, * if two tri" angles," &c. Q. E. D.

g Hyp.

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.

A

a 23. 1.

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 the
point B in the straight line
AB, make the angle ABG
equal to the angle a DEF:

D
And because the angle at A
is equal to the angle at D,

G
and the angle ABG to the
angle DEF; the remaining

B СЕ F
angle AGB is equal to the
remaining angle DFE: Therefore the triangle ABG is equi-
angular to the triangle DEF; wherefore as AB is to BG, so
is DE to EF; but as DE to EF, so, by hypothesis, 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, BG; BC is
equale

to BG, and therefore the angle BGC is equal to the angle BCGf: But the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less 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 hypothesis, it is less than a

b 32. 1.

c 4. 6.

d 11. 5.

e 9. 5. f 5. 1.

g 13, 1,

СЕ

right angle; which is absurd. Therefore the angles ABC, Book VI. 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 may be proved in like manner, that BC

A is equal to BG, and the

D
angle at C equal to the an-
gle BGC: But the angle
at C is not less than a right

G
angle; therefore the angle
BĞC is not less than a right
B

F angle: Wherefore, two angles of the triangle BGC are together not less than two right angles, which is impossible"; and therefore the triangle ABC h 17. 1. may be proved to 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

A of the straight line AB, the angle ABG equal to the angle DEF; then, it may be

G proved, as in the first case, that BG is equal to BC: But

С

D the angle BCG is a right an

A gle; therefore the angle

i 5. 1, BGC is also a right angle; whence two of the angles of the triangle BGC are toge

E F ther not less than two right B

с angles, which is impossible: h

G Therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, “ if two tri

angles,&c. Q. E. D.

Book VI.

PROP. VIII.

THEOR.

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 the angle at B common to the two triangles ABC, ABD; the remaining angle ACB is equal

А a 32. 1.

to the remaining angle BAD a :
Therefore, the triangle ABC is
equiangular to the triangle

ABD, and the sides about their b 4. 6. equal angles are proportionals b;

wherefore the triangles are sicl. Def. 6. milar c: In like

B

C it

manner, may be demonstrated, that the triangle ADC is equiangular and similar to the triangle ABC: And the triangles ABD, ADC, 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 CD b.

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 AB.

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 same 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. Because ED is parallel to one of the

E D sides of the triangle ABC, viz. to BC, as CD is to DA, so is a BE to EA; and, by composition, CA is to AD, as BA to AE: But CA is a multiple of AD; thereforec BA is the same multiple of AE: Whatever part therefore AD is of ACB

С 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. 5.

c D, 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 BC; a 31. 1. and through D draw DHK parallel to AB : Therefore each of the figures FH, HB is a parallelogram; wherefore DH is

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