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Book XI. 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 be
cause the two sides LB, BC are equal to the two DE, EF, and
that the angle DEF is greater than the angle LBC, the base DF d 24. 1. is greater d than the base LC: and it has been proved that GK
is equal to AL; therefore DF and GK are greater than AL and e 20. 1. LC: but AL and LC are greater e 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, f 22. 1. therefore, a triangle may be made f, 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 toge: ther 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 AB, BC, Book XI. DE; EF, GH, HK, all equal to one another; and join AC, DF, GK: then a triangle may be made a of three straight lines equal a 22. 11.
to AC, DF, GK. Let this be the triangle LMNb, so that AC 6 22. 1. be equal to LM, DF to MN, and GK to LN; and about the triangle LMN describe c a circle, and find its centre X, which will c 5. 4. 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 d 8 1. reason, the angle DEF is equal to the angle MXN, and the angle GHK to the angle NXL ; therefore the three angles ABC,
R DEF, GHK are equal to the three angles LXM, MXN, NXL: but the three angles LXM, MXN, NXL
L are equal to four right angles :
e 2. Cor. therefore also the three angles
15. 1. ABC, DEF, GHK are equal to four right angles : but, by the hypothesis, they are less than four right an
X gles; which is absurd; therefore AB is not equal to LX: but nei- M
N ther 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 the triangle L.OM, 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
Book XI. to the angle ABCd: and AB, that is, LO, by the hypothesis, is
w less than LX; wherefore LO, OM fall within the triangle LXM; d 8.1. for, if they fell upon its sides, or without it, they would be equal
to, or greater than LX, XMf: therefore the angle LOM, that
is, the angle ABC, is greater than f 21. 1. the angle LXM*: in the same man
Therefore the three
Next, let the centre X of the circle fall in one of the sides of
EF are equal to DF, which is im- M † 20. 1. possible t: wherefore AB is not
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 Book XI. angle MXN is equal to the angle DEF: and it has been prov. ed that it is greater than DEF, which is absurd. Therefure AB d 8. l; is not equal to LX. Nor yet is it less; for 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 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 8 than the angle ACB at the base. For the same rea- g 32. 1. son, because the angle GHK, or CBP, is greater than the angle
R LXN, the angle XLN is greater than the angle BCP. Therefore the whole angle MLN is greater than the whole angle ACP. And
L 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 h than the base AP. M
N h 24.1. And MN is equal to DF; therefore also DF is greater than AP.
X 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 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. a 12. 11. From the point X erect a XR at right angles to the plane of
the circle LMN. And because it has been proved in all the cases
cause RX is perpendicular to the b 3. defe plane of the circle LMN, it is b per11.
pendicular to each of the straight
therefore the square of AB is equal to the squares of LX, XR. c 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 to the base AC; the angle LRM is equal d 8.1. d 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.