planes which pass through AB are at right angles to the plane Book XI. CK. Therefore, if a straight line, &c. Q. E. D. F two planes cutting one another be each of them I perpendicular to a third plane; their common fection fhall be perpendicular to the fame plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section of the first two; BD is perpendicular to the third plane. E B F a4.Def.11. If it be not, from the point D draw, in the plane AB, the ftraight line DE at right angles to AD, the common section of the plane AB, with the third plane; and in the plane BC draw DF at right angles to CD, the common fection of the plane BC, with the third plane. And because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common fection, DE is perpendi cular to the third plane 2. In the fame manner, it may be proved, that DF is perpendicular to the third plane. Wherefore, from the point D two straight lines ftand at right angles to the third plane, upon the fame fide of it; which is impoffible : Therefore, from the point D there cannot be any ftraight line at right angles to the third plane, except BD, the common fection of the planes AB, BC. BD, therefore, is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D. If a folid PROP. XX. THEOR. A D Cb 13.11. Fa folid angle be contained by three plane angles, any two of them are greater than the third. Let the folid angle at A be contained by the three plane angles BAC, CAD, DAB. Any two of them are greater than the third. If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the ftraight line AB, make, in the plane which paffes through BA, AC, the angle BAE equal to the a 23. 1. a angle b 4. I. D BOOK XI. angle DAB; and make AE equal to AD, and through E draw BEC, cutting AB, AC in the points B, C, and join DB, DC. And becaufe DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and the angle DAB is equal to the angle EAB; therefore the base DB is equal to the bafe BE. And becaufe BD, G 20. 1. DC are greater c than CB, and one of them BD has been proved equal to BE a part of CB; therefore the other DC is greater than the remaining part EC. And becaufe DA is equal to AE, and AC B A E C common, but the bafe DC greater than the bafe EC; therefore d 25. 1. the angle DAC is greater than the angle EAC; and, by the conftruction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not less than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other. Wherefore, if a folid angle, &c. Q. E. D. See N. E PROP. XXI. THEOR. VERY folid angle is contained by plane angles, which together are less than four right angles. Let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAB. Thefe together are less than four right angles. Let the planes in which the angles are be cut by a plane, and let their common fections with it be BC, CD, DE, EB: Then, because the folid angle at B is contained by the three plane angles CBA, ABE, EBC, any two of them are greater than the third; therefore the angles CBA, ABE are greater than the angle EBC: For the fame reafon, the two plane angles BCA, ACD are greater than BCD, and the two CDA, ADE greater than CDE; and DEA, AEB greater than DEB: Therefore all the angles at the bafes of the triangles ABC, ACD, ADE, ABE, which have their common vertex at A, are together greater than all the angles of the rectilineal figure BCDE: and because the three angles of a triangle are equal to two right angles, all the angles of the triangles ABC, ACD, ADE, AÈB are equal to twice as many right angles as there are triangles; that is, as there are fides of the figure BCDE; and all the angles angles of the figure BCDE, together with four right angles, Book XI. ✔ are alfo equal to twice as many right angles as there are fides of the figure; therefore all the angles of the PROP. XXII. THEOR. A D E [F three ifofceles triangles have their fides equal, and any two of the vertical angles greater than the third; a triangle may be made of their bafes. Let ABC, DEF, GHK be three ifofceles triangles, having their fides AB, BC, DE, EF, GH, HK all equal, and any two of the angles at B, E, H greater than the third; any two of the bafes AC, DF, GK are also greater than the third. a 4. I. If two of the angles at B, E, H be equal, and not less than the third, it is evident that two of the bafes are equal a and not lefs than the third; and therefore any two of the bafes are b 24. I. greater than the third. In any other cafe, let the angle ABC be greater than either b of the other angles at E, H, therefore AC is greater than either DF or GK. At B, in AB, make the angle ABL equal to DEF, and BL equal to one of the equal fides, and C 23. I. join AL, LC: and because the two fides AB, BL are equal to the two DE, EF, and they contain equal angles; therefore the base d AL is equal to DF: and because the angles at E, H are greater d 24. 1. than ABC, and that E is equal to ABL, the remaining angle at His greater than the remaining angle CBL and because the two fides LB, BC are equal to the two GH, HK, but the angle LBC lefs than GHK; therefore the bafe LC is lefs than GK; and it was proved, that AL is equal to DF; therefore DF, GK are greater than AL, LC: but AL, LC are greater A a b than AC; 20. 1 therefore • BOOK XI.therefore DF, GK are greater than AC: and it is evident, that ~ AC with either of the two DF, GK are greater than the other; f 22. 1. therefore a triangle may be made f, the fides of which shall be equal to AC, DF, GK. Q. E. D. See N. 24. I. T O 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 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 fides of the angles cut off AB, BC, DE, EF, GH, HK all equal to one another; and join AC, DF, GK: and let b B P the angle GHK not be less than either of the other two; therea 4. or fore GK is not lefs than either AC or DF. At the point B make the angle CBP equal to DEF, and make BP equal to one of the equal fides, and join CP: and because the two fides CB, BP are equal to the two DE, EF, and the angle CBP is 4. 1. equal to the angle DEF; the base CP is equal to DF. If AB b 23. I. greater e e 24. I. greater than GK. But, if AB and BP be not in a ftraight line, Book XI. join AP: and if the point B be without the triangle ACP; because the two angles ABC, DEF are greater than GHK; and CBP is equal to DEF; therefore the two ABC, CBP, that is, the angle ABP is greater than GHK: and the two fides AB, BP are equal to the two GH, HK; therefore the base AP is than the bafe GK. But if the point B be within the triangle ACP, the three angles ABC, CBP, ABP are equal f to four f 2. Cor. right angles and the three angles ABC, CBP, GHK are lefs than four right angles; therefore the angle ABP is greater than GHK and the fides AB, BP are equal to GH, HK; therefore the base AP is greater than GK. Wherefore in every cafe AP is greater than GK. : P 15. 1, 1 F G K g 22. 11. k 5.4. 11. 3. And because every two of the angles ABC, DEF, GHK are greater than the third, a triangle may be made of the three bafes AC, DF, GK: let this be the triangle LMN, so that LM be equal to AC, MN to DF or CP, and LN to GK: and because the two fides AC, CP are equal to the two LM, MN, but the base AP greater than the base LN; therefore the angle ACP is greater h than the angle LMN. About the triangle LMN de- h 25. 1. fcribe k a circle, and find its centre X, and join XL, XM, XN: AB is greater than XL: For if AB be equal to XL, BC is equal to XM and becaufe BC, CA are equal to XM, ML, and AB equal to XL; the angle ACB is equal to LMXm: For the fame m 8. 1. reason, the angle BCP is equal to XMN; therefore the whole angle ACP is equal to the whole LMN: and it was shown to be greater than it; which is impoffible; therefore AB is not equal to LX: Neither is it lefs, for, if it be, make the triangle LOM on the same fide of LM with the triangle LXM, so that LO, OM be equal to AB, BC; therefore LO, OM are less than LX, XM; and therefore the triangle LOM falls within the triangle n 21. 1. LXM. Confequently the angle LMO is lefs than LMX: but LMO is equal to ACB; therefore ACB is less than LMX. For the fame reason, the angle BCP is less than XMN; therefore the whole angle ACP is lefs than the whole LMN; and it A a 2 is O 8. I. |