Page images
PDF
EPUB
[graphic]

* 33 Dat.

in position, be cut by the straight line GHK; the ratio of GH to HK is given.

A

CH M D

EK

In AB take a given point L, and draw LM perpendicular to CD, meeting EF in N; therefore* LM is given in position: and CD, EF, are given in position, wherefore the points M, N, are given and the point L is *29 Dat. given; therefore the straight lines LM, MN, are given in magnitude; and the ratio of LM to MN is therefore given*: but as LM to MN, so is GH to HK; wherefore the ratio of GH to HK is given.

* 1 Dat.

[blocks in formation]

PROP. XLII.

If each of the sides of a triangle be given in magnitude; the triangle is given in species.

*

Let each of the sides of the triangle ABC be given in magnitude, the triangle ABC is given in species. Make a triangle DEF, the sides of which are equal, each to each, to the given straight lines AB, BC, CA; which can be done, because any two of them must be greater than the third; and let DE be equal to AB, EF to BC, and FD to CA; and because the two sides ED, DF, are equal to the two BA, AC, each

*

A A

B

E

[graphic]

to each, and the base EF equal to the base BC; the angle EDF is equal to the angle BAC; therefore, because the angle EDF, which is equal to the angle BAC, has been found, the angle BAC is given *: in like manner the angles at B, C, are given. And because the sides AB, BC, CA, are given, their ratios to one another are given *; therefore the triangle ABC is given in species.

*

PROP. XLIII.

If each of the angles of a triangle be given in magnitude; the triangle is given in species.

Let each of the angles of the triangle ABC be given in magnitude, the triangle ABC is given in species.

*

Take a straight line DE given in position and magnitude, and at the points D, E, make the angle EDF equal to the angle BAC, and the angle DEF equal to ABC; therefore the other angles EFD, BCA are equal, and

[ocr errors][merged small][ocr errors][subsumed][subsumed]

each of the angles at the points A, B, C is given, wherefore each of those at the points D, E, F, is given : And because the straight line FD is drawn to the given point D in DE which is given in position, making the given angle EDF; therefore DF is given in position *;* 32 Dat. In like manner EF also is given in position; wherefore the point F is given: and the points D, E are given; therefore each of the straight lines DE, EF, FD is given in magnitude; wherefore the triangle DEF is given in species*; and it is similar to the triangle ABC; which therefore is given in species.

PROP. XLIV.

If one of the angles of a triangle be given, and if the sides about it have a given ratio to one another; the triangle is given in species.

Let the triangle ABC have one of its angles BAC given, and let the sides BA, AC, about it have a given. ratio to one another; the triangle ABC is given in species.

Take a straight line DE given in position and magnitude, and at the point D, in the given straight line DE, make the angle EDF equal to the given angle BAC: wherefore the angle EDF is given; and because the straight line FD is drawn to the given point D in ED, which is given in position, making the given angle EDF; therefore FD is given in position *. And because the ratio of BA to AC is given, make the ratio of ED to DF the same B with it, and join EF; and be

A

C E F

[ocr errors]

29 Dat. 42 Dat.

* 4.6.

41.

1 Def. 6.

* 32 Dat.

[ocr errors]

cause the ratio of ED to DF is given, and ED is given, therefore* DF is given in magnitude: and it is given * Dat. also in position, and the point D is given, wherefore

the point F is given *; and the points D, E are given,* 30 Dat.

[blocks in formation]

wherefore DE, EF, FD are given * in magnitude; and the triangle DEF is therefore given in species; and because the triangles ABC, DEF have one angle BAC equal to one angle EDF, and the sides about these angles proportionals; the triangles are similar, but the triangle DEF is given in species, and therefore also the triangle ABC.

[graphic]

42.

See N.

[blocks in formation]

PROP. XLV.

If the sides of a triangle have to one another given ratios; the triangle is given in species.

Let the sides of the triangle ABC have given ratios to one another, the triangle ABC is given in species.

Take a straight line D given in magnitude; and because the ratio of AB to BC is given, make the ratio of D to E the same with it; and D is given, therefore* E is given. And because the ratio of BC to CA is given, to this make the ratio of E to F the same; and E is given, and therefore* F. And because as AB to BC, so is D to E; by composition AB and BC together are to BC, as D and E to F: but as BC to CA, so is E to F; therefore, ex æquali

*

*

as

*

AB and BC are to CA,
so are D and E to F, and B
AB and BC are greater
than CA; therefore D
and E are greater than
F. In the same manner,
any two of the three D, E, F are greater than the
third. Make the triangle GHK, whose sides are
equal to D, E, F, so that GH be equal to D, HK to
E, and KG to F; and because D, E, F are, each of
them, given, therefore GH, HK, KG are each of them
given in magnitude; therefore the triangle GHK is
given in species: but as AB to BC, so is (D to E,
that is) GH to HK; and as BC to CA, so is (E to F,
that is) HK to KG; therefore, ex æquali, as AB to
AC, so is GH to GK. Wherefore the triangle
ABC is equiangular and similar to the triangle GHK;
and the triangle GHK is given in species: therefore
also the triangle ABC is given in species.

*

COR. If a triangle be required to be made, the sides of which shall have the same ratios which three given straight lines D, E, F have to one another; it is necessary that every two of them be greater than the third.

PROP. XLVI.

If the sides of a right angled triangle about one of the acute angles have a given ratio to one another; the triangle is given in species.

Let the sides AB, BC about the acute angle ABC of the triangle ABC, which has a right angle at A, have a given ratio to one another; the triangle ABC is given in species.

43.

Take a straight line DE given in position and magnitude; and because the ratio of AB to BC is given, make as AB to BC, so DE to EF; and because DE has a given ratio to EF, and DE is given, therefore * * 2 Dat. EF is given; and because as AB to BC, so is DE to

*

EF; and AB is less than BC, therefore DE is less** 19. 1. than EF. From the point D draw DG at right angles

to DE, and from the centre E, at the distance EF, describe a circle

which shall meet DG in

* A. 5.

F

[blocks in formation]

*

6 Def.

* 42 Dat.

cumference of the circle is given in position; and the straight line DG is given in position, because it is * 32 Dat. drawn to the given point D in DE given in position, in a given angle; therefore* the point G is given, and * 28 Dat. the points D, E are given, wherefore DE, EG, GD are given * in magnitude, and the triangle DEG in * 29 Dat. species*. And because the triangles ABC, DEG, have the angle BAC equal to the angle EDG, and the sides about the angles ABC, DEG proportionals, and each of the other angles BCA, EGD, less than a right angle; the triangle ABC is equiangular* and similar to *7. 6. the triangle DEG; but DEG is given in species; therefore the triangle ABC is given in species: and in the same manner, the triangle made by drawing a straight line from E to the other point in which the circle meets DG, is given in species.

B B

[blocks in formation]

PROP. XLVII.

[graphic]

If a triangle have one of its angles which is not a right angle given, and if the sides about another angle have a given ratio to one another; the triangle is given in species.

Let the triangle ABC have one of its angles ABC a given, but not a right angle, and let the sides BA, AC, about another angle BAC have a given ratio to one another; the triangle ABC is given in species.

First, let the given ratio be the ratio of equality, that is, let the sides BA, AC, and consequently the angles ABC, ACB, be equal; and because the angle ABC is given, the angle ACB, and also the remaining angle BAC is given; therefore the triangle ABC is given in species; and it is evident that in this case the given angle ABC must be acute.

[ocr errors]

*

*

A

Next, let the given ratio be the ratio of a less to a greater, that is, let the side AB adjacent to the given angle be less than the side AC: take a straight line DE given in position and magnitude, and make the angle DEF equal to the given angle ABC: therefore EF is given in position; and because the ratio of BA to AC is given, as BA to AC, so make ED to DG; and because the ratio of ED to DG is given, and ED is given, the straight line DG is given *, and BA is less than AC, therefore ED is less than DG. From the centre D, at the distance DG, describe the circle GF meeting EF in F, and join DF; and because the circle is given in position, as also

[graphic]

G

the straight line EF, the point F is given *; and the points D, E, are given; wherefore the straight lines DE, EF, FD are given in magnitude, and the triangle DEF in species *. And because BA is less than AC,

the angle ACB is less
therefore ACB is less

than the angle ABC, and than a right angle. In the

« PreviousContinue »