greater than all the angles of the polygon : and because all the Book XI. angles of the triangles are together equal to twice as many right angles as there are triangles b; that is, as there are sides in the b 32. 1. polygon BCDEF: and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are sides in the polygonc; therefore all c 1. Cor. the angles of the triangles are equal to all the angles of the lygon together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved. Wherefore the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right angles. Therefore, every solid angle, &c. Q. E. D. PROP. XXII. THEOR. IF every two of three plane angles be greater than See N, the third, and if the straight lines which contain them be all equal; a triangle may be made of the straight lines that join the extremities of those equal straight lines. Let ABC, DEF, GHK be three plane angles, whereof every two are greater than the third, and are contained by the equal straight lines AB, BC, DE, EF, GH, HK; is their extremities be joined by the straight lines AC, DF, GK, a triangle may be made of three straight lines equal to AC, DF, GK; that is, every two of them are together greater than the third. If the angles at B, E, H are equal, AC, DF, GK are also equal“, and any two of thein greater than the third : but if a 4. La the angles are not all equal, let the angle ABC be Dot less than either of the two at E, H; therefore the straight line AC is not less than either of the other two DF, GKb; and it is b 4. or plain that AC, together with either of the other two, must be 24. 1. greater than the third : also DF with GK are greater than AC: for, at the poiot B in the straight line AB make the c 23. 1. с 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 12. 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. 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 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. a From the straight lines containing the angles, cut off AB, BC, Book XI. DE; EF, GH, HẤ, 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. АДА K A 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 cither 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: 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 angles; 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, ban 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 R ner it may be proved that the angle DEF is greater than the angle MXN, and the angle GHK greater than the L angle NXL. Therefore the three X N Next, let the centre X of the circle fall in one of the sides of R L N + 20.1. possible +: wherefore AB is not X 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 d to the angle DEF: and it has been prov. m ed that it is greater than DEF, which is absurd. Therefure AB d 8. d; 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 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 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. 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 x k 25.1 |