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angle ABL equal to the angle GHK, and make BL equal to one of the straight lines AB, BC, 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; and because B

E

H

AA

A


F C

K L the two sides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle L.BC, the base DF is greater (24. 1.) 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 are greater (20. 1.) 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, therefore, a triangle may be made (22. 1.) the sides of which shall be equal to AC, DF, GK. Q. E. D.

PROP. XXIII. PROB.

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.

. See Note.

From the straight lines containing the angles, cut off AB, BC, DE, EF, GH, HK, all equal to one another; and join AC, DF, GK; then a triangle may be made (22. 11.) of three straight lines B

H

E

с

G
A

K
D

F
equal to AC, DF, GK. Let this be the triangle LMN (22. 1.)
so that AC be equal to LM, DF to MN, and GK to LN; and
about the triangle LMN describe (5. 4.) a 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, ÅB 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 LXM (8. 1.). For the same reason, the angle DEF is equal to the angle MXN, and the angle GHK to the

R angle NXL; therefore the three angles ABC, DEF, GHK are equal to the three angles LXM, MİN,

L NXL: but the three angles LXM, MXN, NXL are equal to four right angles (2. Cor. 15. i.): therefore also the three angles ABC, DEF,

X GHK, are equal to four right angles; but by the hypothesis, they M 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, the side of it on which is the centre X, describe the triangle LOM, the sides LO, OM of which are equal to AB, BC; and because the base LM is equal to the base AC, the angle LOM is equal to the angle ABC (8. 1.): and AB, that

US

is, LO, by the hypothesis, is less than LX; wherefore LO, OM
fall within the triangle LXM; for if they fell upon its sides,
or without it, they would be equal to, or greater than LX, XM
(21. 1.): therefore the angle LOM, that is, the angle ABC, is
greater than the angle LXM (21. 1.):

R in the same manner it may

be proved
that the angle DEF is greater than

L
the angle MXN, and the angle GHK
greater than the angle NXL. There-
fore the three angles ABC, DEF,
GHK are greater than the three angles
LXM, MXN, NXL; that is than

x
four right angles ; but the same an-

N gles ABC, DEF, GHK are less than

M
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, and

R
join XL: in this case also AB is great-
er than LX. If not, AB is either
equal to LX, or less than it; first, let

L
it be equal to XL: therefore AB and
BC, that is, DE and EF, are equal to
MX and XL, that is, to MN: but,
by the construction, MN is equal to M
DF; therefore DE, EF are equal to

X
DF, which is impossible (20. 1.):
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.

N

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,

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LLI

EF are equal to MX, XN, and the baseaDF to the base MN, the angle MXN is equal (8. 1.) 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 ČB make the angle CBP equal to the angle GHK, and make BP equal 10 HK, and join CP, AP. B

II

P

E

C

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D 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 (32. i.) than the angle ACB at the base.

For the same reason, because the angle GHK, or

R CBP, is greater than the angle LXN, the angle XLN is greater than the angle CBP. Therefore the whole angle MLN is greater than the whole

I angle ACP. And because ML, LN are equal to AC, CP, each to each, but the angle MLN is greater than the angle ACP, the base MN is greater M

N (24. 1.) than the base AP. And

X 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 (25. 1.) than the angle ABP. And 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 than these ;

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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.

From the point X erect (12. 11.) 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 AB above the

R
square of LX, and make RX equal to
its side; and join RL, RM, RN. Be-
cause RX is perpendicular to the plane
of the circle LMN, it is (3. def. 11.)
perpendicular to each of the straight
lines LX, MX, NX. And because
LX is equal to MX, and XR common,
and at right angles to each of them,

M

N the base RL is equal to the base RM. For the same reason, RN is equal

X 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; therefore the square of AB is equal to the squares of LX, XR. But the square of RL is equal (47. 1.) 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 R... 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 to the base AC; the angle LRM is equal (8. 1.) to the angle ABC. For the same reason, the angle MŘN 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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