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Book XI. angle ABL equal to the angle GHK, and make BL equal to one of the ftraight 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 bafe AL is equal to the bale GK: And becaufe 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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And because the two fides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle LBC, the base DF 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 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 ftraight lines AC, DF, GK are greater than the third; and, therefore, a triangle may be made f, the fides of which fhall be equal to AC, DF, GK. Q. E. D.

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PRO P. XXIII. PRO B.

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

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

lines equal to AC, DF, GK. Let this be the triangle LMN b, b 22 1. fo that AC be equal to LM, DF to MN, and GK to LN; and about the triangle LMN defcribe a circle, and find its centres c 5. 4. X, which will either be within the triangle, or in one of its fides, or without it.

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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 lefs than LX; firft, let it be equal: Then because AB is equal to LX, and that AB is alfo 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 bafe LM; wherefore the angle ABC is equal to the angle LXM 4. For the fame reason, the angle DEF is equal to the d 8. r. angle MXN, and the angle GHK to the angle 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 angles ABC, DEF, GHK are equal to four right angles: But, by the hypothefis, they are less M than four right angles, which is abfurd; therefore AB is not equal to LX: But neither can AB be

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lefs than LX: For, if poffible, let it be lefs, and upon the ftraight line LM, on the fide of it on which is the centre X, describe the triangle LOM, the fides LO, OM of which are equal to AB, BC; and because the base LM is equal to the

bafe

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Book XI. bafe AC, the angle LOM is equal to the angle ABC &; And AB, that is, LO, by the hypothefis, is lefs than LX; wherefore LO, OM fall within the triangle LXM; for, if they fell upon its fides, 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 the angle LXM f: In the fame manner it may be proved that the angle DEF is greater than the angle MXN, and the angle GHK greater than the angle NXL. Therefore the three angles ABC, M DEF, GHK are greater than the three angles LXM, MXN, NXL; that is, than four right angles: But

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the fame angles ABC, DEF, GHK are lefs than four right angles; which is abfurd: 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.

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Next, Let the centre X of the circle fall in one of the fides of the triangle, viz. in MN, and join XL: In this cafe alfo AB is greater than LX. If not, AB is either equal to LX, or lefs 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: But,by the conftruction, Mi MN is equal to DF; therefore DE, EF are equal to DF, which is impoffible +: Wherefore AB is not equal to LX; nor is it lefs; for then, much more, an abfurdity would, follow: Therefore AB is greater than LX.

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But, let the centre X of the circle fall without the triangle LMN, and join LX, MX, NX. In this cafe likewife 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 fame manner, as in the fift cafe, 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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EF are equal to MX, XN, and the bafe DF to the bafe Book XI. MN, the angle MXN is equal to the angle DEF: And it has been proved, that it is greater than DEF, which is abfurd. Therefore AB is not equal to LX. Nor yet is it lefs; 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 to

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HK, and join CP, AP. And becaufe CB is equal to GH; CB, BP are equal to GH, HK, each to each, and they contain equal angles; wherefore the bafe CP is equal to the bafe GK, that is, to LN. And in the ifofceles triangles ABC, MXL, because the angle ABC is greater than the angle MXL, therefore the angle MLX at the bafe is greater g than the angle g 32, 1. ACB at the bafe. For the fame reafon, because the angle GHK, or CBP, is greater than the angle LXN, the angle XLN is greater than the angle CBP. Therefore the whole angle MLX 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 angle ACP, the bafe MN is greater hthan the bafe M AP. And MN is equal to DF; therefore alfo DF is greater than AP. Again, because DE, EF are equal to AB BP, but the bafe DF greater than the base AP, the angle DEF is greater than the angle

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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 alfo lefs than these which is impoffible. Therefore AB is not less than

LX;

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Book XI. LX; and it has been proved that it is not equal to it; therea fore AB is greater than LX.

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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 cafes, that AB is greater than LX, find a fquare equal to the excefs of the fquare of AB above the fquare of LX, and make RX equal to its fide, and join RL, RM, RN. Because RX is perpendicular to the plane of the circle LMN, it b 3. def.11, is b perpendicular to each of the ftraight lines LX, MX, NX. And because LX is equal to MX, and XR common, and at right angles M to each of them, the base RL is equal to the bafe RM. For the fame reafon, RN is equal to each of the two RL, RM. Therefore the three straight lines RL, RM, RN are all equal. And because the fquare of XR is equal to the excefs of the fquare of AB above the fquare of LX; therefore the fquare of AB is equal to the fquares of LX, XR. But the fquare of RL is equal to the fame fquares, because LXR is a right angle. Therefore the square of AB is equal to the fquare of RL, and the ftraight line AB to RL. But each of the ftraight 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 ftraight lines RL, RM, RN. And because RL, RM, are equal to AB, BC, and the bafe LM to the bafe AC; the angle LRM is equal to the angle ABC. For the fame reason, the angle MRN is equal to the angle DEF, and NRL to GHK. Therefore there is made a folid 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, Which was to be done.

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