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Book XI. line AB, make the angle ABL equal to the angle GHK,

and make BL equal to one of the straight lines AB, BC, * 23. 1. DE, EF, GH, HK, and join AL, LC; Then, because AB,

BL, are equal to GH, HK, and the angle ABL to the angle GHK, the base AL is equal to the base GK: And because the angles at E, H, are greater than the angle ABC, of which the angle at H is equal to ABL, therefore the remaining angle at E is greater than the angle LBC:

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D
F

K
L
And because the two sides LB, BC are equal to the two

DE, EF, and that the angle DEF is greater than the angle à 24. 1. LBC, the base DF is greaterd than the base LC: And it

has been proved that GK is equal to AL; therefore DF

and GK are greater than AL and LC: But AL and LC 20. 1. are greater than AC; much more then are DF and GK

greater than AC. Wherefore every two of these straight

lines AC, DF, GK are greater than the third; and, there* 22.1. fore, a triangle may be made f, the sides of which shall be

equal to AC, DF, GK Q. E. D.

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See N. To make a solid angle which shall be contained by

three given plane angles, any two of them being greater than the third, and all three together less than four right angles.

Let the three given plane angles be ABC, DEF, GHK, any two of which are greater than the third, and all of them together less than four right angles. It is required to make a solid angle contained by three plane angles equal to ABC, DEF, GHK, each to each.

From the straight lines containing the angles, cut off Boox XI. AB, BC, DE, EF, GH, HK, all equal to one another; and join AC, DF, GK: Then a triangle may be made of • 29. 11.

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A

three straight lines equal to AC, DF, GK. Let this be the triangle LMNb, so that AC be equal to LM, DF to MN, 22. 1. and GK to LN; and about the triangle LMN describe a 15.4. circle, and find its centre X, which will either be within the triangle, or in one of its sides, or without it.

First, Let the centre X be within the triangle, and join LX, MX, NX: AB is greater than LX: If not, AB must either be equal to, or less than LX: first, let it be equal : Then because AB is equal to LX, and that AB is also equal to BC, and LX to XM, AB and BC are equal to LX and XM, each to each ; and the base AC is, by construction, equal to the base LM; wherefore the angle ABC is equal to the angle LXMd. For the same reason, the angle DEF.8. 1. is equal to the angle MXN, and

R the angle GHK to theangle NXL: Therefore the three angles ABC, DEF, GHK are equal to the three angles LXM, MXN, NXL: But the three angles LXM, MXN, NXL are equal to four right anglese; therefore also the three

e 2 Cor. angles ABC, DEF, GHK are equal to four right angles : But, M

N by the hypothesis, they are less than four right angles, which is absurd; therefore AB is not equal to LX. But neither can AB be less than LX: For, if possible, let it be less; and upon the straight line LM, on the side of it on which is the centre X, describe thie triangle LOM, the sides LO, OM of which are equal to AB, BC; and because the base

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15. 1.

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Boox XI. LM is equal to the base AC, the angle LOM is equal to

the angle ABCd: And AB, that is, LO, by the hypothesis,
is less than LX; wherefore LO, OM fall within the trian-
gle LXM; for, if they fell upon its sides, or without it, they
would be equal to or greater than
LX, XM: Therefore the angle
LOM, that is, the angle ABC, is

L 21. 1. greater than the angle LXMf: In

the same mannerit may be proved
that the angle DEF is greater than
the angle MXN, and the angle
GHK greater than the angle

X
NXL. Therefore the three angles
ABC, DEF, GHK are greater M

N
than the three angles LXM,
MXN, NXL; that is, than four
right angles : But the same angles ABC, DEF, GHK are
less than four right angles; which is absurd : Therefore
AB is not less than LX, and it has been proved that it is
not equal to LX; wherefore AB is greater than LX.

Next, Let the centre X of the circle fall in one of the sides of the triangle, viz. in MN,

RA

R
and join XL: In this case also,
AB is greater than LX. If not,
AB is either equal to XL, or L
less than it. First, let it be equal
to XL; Therefore AB and BC,
that is, DE and EF, are equal
to MX and XL, that is, to MN:M

NN
But, by the construction, MN

X is equal to DF; therefore DE,

ud EF, are equal to DF, which is + 20. 1. impossible +: Wherefore AB is

not equal to LX; nor is it less; for then, much more, an absurdity would follow : Therefore AB is greater than LX.

But, let the centre X of the circle fall without the triangle LMN, and join LX, MX, NX. In this case likewise AB is greater than LX: If not, it is either equal to, or less than LX: First, let it be equal; it may be proved, in the same manner as in the first case, that the angle ABC is equal to the angle MXL, and GHK to LXN; therefore the whole angle MXN is equal to the two angles ABC, GHK: But ABC and GHK are together greater than the angle DEF: therefore also the angle MXN is greater than DEF. And because DE, EF are equal to MX, XN, and the base DF to the base MN, the angle Boor XI. MXN is equal d to the angle DEF: And it has been proved, that it is greater than DEF, which is absurd. Therefore AB is not equal to LX. Nor yet is it less; for then, as has been proved in the first case, the angle ABC is greater than the angle MXL, and the angle GHK greater than the angle LXN. At the point B, in the straight line CB, make the angle CBP equal to the angle GHK, and make BP equal

B

H

CA
A С

K

F to HK, and join CP, AP. And because CB is equal to GH; CB, BP are equal to GH, HK, each to each, and they contain equal angles; wherefore the base CP is equal to the base GK, that is, to LN. And in the isosceles triangles ABC, MXL, because the angle ABC is greater than the angle MXL, therefore the angle MLX at the base is greater than the angle ACB at the base. For the same = 32. 1. Teason, because the angle GAK

R or CBP, is greater than the angle LXN, the angle XLN is greater than the angle BCP. Therefore the whole angle MLN is greater than the whole angle ACP. And because ML, LN, are equal to AC, CP, each to each, but the angle MLN is greater than the M

N angle ACP, the base MN is greater than the base AP. And

X

h 24. 1. MN is equal to DF; therefore also DF is greater than AP. Again, because DE, EF are equal to AB, BP, but the base DF greater than the base AP, the angle DEF is greater than the angle ABP. And * 25. 1. ABP is equal to the two angles ABC, CBP, that is, to the two angles ABC, GHK; therefore the angle DEF, is greater than the two angles ABC, GHK; but it is also less

Book XI. than these, which is impossible. Therefore AB is not less

than LX; and it has been proved that it is not equal to it;

therefore AB is greater than LX. 2 12. 11. From the point X erecta XR at right angles to the plane

of the circle LMN. And because it has been proved in all
the cases, that AB is greater than LX, find a square equal
to the excess of the

square
of

R
AB above the square of LX, and
make RX equal to its side, and
join RL, RM, RN. Because

RX is perpendicular to the bs Def. 11. plane of the circle LMN, it is b

perpendicular to each of the
straight lines LX, MX, NX.
And because LX is equal to M

N
MX, and XR common, and at
right angles to each of them, the

X base RL is equal to the base

ME RM. For the same reason, RN is equal to each of the two RL, RM. Therefore the three straight lines RL, RM, RN, are all equal. And because the square of XR is equal to the excess of the square of AB above the square of LX; there

fore the square of AB is equal to the squares of LX, XR. • 47. 1. But the square of RL is equal to the same squares, because

LXR is a right angle. Therefore the square of AB is equal to the square of RL, and the straight line AB to RL, But each of the straight lines BC, DE, EF, GH, HK, is equal to AB, and each of the two RM, RN, is equal to RL. Wherefore AB, BC, DE, EF, GH, HK, are each of them equal to each of the straight lines RL, RM, RN. And

because RL, RM, are equal to AB, BC, and the base LM * 8. I. to the base AC; the angle LRM is equald to the angle

ABC. For the same reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore there is made a solid angle at R, which is contained by three plane angles LRM, MRN, NRL, which are equal to the three given plane angles ABC, DEF, GHK, each to each. Which was to be done.

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