1 Let a plane pass through BA, AC, and let DAE be the common section of it with the given plane: . AB, AC, DAE, are in one plane. Then CA is to the plane, and meets DAE, . CAE is. a right; for the same reason B C V. A E but one perpendicular to that plane; for if there could be two they would be ||, which is absurd. PROPOSITION XV. Planes to which the same straight line is perpendicular are parallel to one another. Let A B be to each of the planes CD, EF; these planes are to each other. If not, they will meet if produced: let them meet; their common section is a straight line GII, in which take any point K, and join AK, BK. Then AB is EF it is to BK If two straight lines meeting one another be parallel to two other straight lines which meet one another, but are not in the same plane with the first two; the plane which passes through these is parallel to the plane through the others. Let AB, BC, meeting in B, be || to DE, EF, meeting in E, but which are not in the same plane with AB, BC: the planes through AB, BC, and DE, EF, are . From B, draw BG to plane through ED, DF, through G, where BG meets the plane; draw GH || to ED, and GK || to EF. to the GBA right, . GB is to BA. For the same reason GB is BC, and .. GB is to plane through DE, EF. .. BG is at once to the plane through AB, BC, and the hence these planes are parallel. plane through AB, BC; and it is plane through DE, EF; PROPOSITION XVII. If two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes, AB, CD, be cut by EFHG, and let EF, HG, be their common sections: EF shall be parallel to GH. If not, EF, GH, will meet if produced either on the side of FH or EG. D First let them be pro duced on the side of FH, and meet in the point K. Hence, EFK is in the plane AB, every point of EFK is in that plane, .. K is a point in the plane For the same reason, K is a point in the plane CD; .. the planes AB, CD, meet if produced, which is contrary to the hypothesis. AB. .. EF and GH do not meet when produced on the side of FH. In the same manner it may be shown that they do not meet when produced on the side EF: wherefore EF and HG are parallel. PROPOSITION XVIII. If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let A B, CD, be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D; then AE If a straight line be perpendicular to a plane, every plane passing through it shall be perpendicular to that plane. Let AB be to plane CK; every plane through AB shall be to CK. Let plane DE pass through AB; and let CE be the common section of D E and CK. any point F in CE, draw FG in DE, Take to CE; then AB is to CK, ABF is a right, but GFB is a right, AB and FG are ||, and AB is to the plane cK, .. FG is to the same plane. But one plane is to another plane, H when the straight lines drawn in one of the planes, to their common section, are also to the other plane. And any line FG, in the plane DE, drawn to the line CE, is also to CK. And the same may be proved of .. the plane DE is to CK. PROPOSITION XX. If two planes which intersect be each perpendicular to a third plane, their common section shall be perpendicular to the same plane. Let the two planes AB, BC, be each to the third plane, and let BD be the common section of AB, BC; BD shall be to the third plane. If it be not, from D in AB draw DE AD, the common section of AB with the third plane; in вc draw DF CD, the common section of BC with the same plane. Then AB is to AD, .. DE is to plane ADC, and DE is to ADC; similarly, DF is to ADC; .. from the same point D, two LS, DE, DF, to the same plane ADC, have been c drawn, which is impossible; .. the only line that can be drawn from D to ADC is BD, the common section of AB and BC. PROPOSITION XXI. If a solid angle be contained by three plane angles, any two of them are greater than the third. Let the solid A, be contained by the three planes, BAC, CAD, DAB; any two of them shall be greater than the third. If the three angles be all, the proposition is evident. If not, let BAC be not less than either of the other two, and > DAB. At the E point A, in the plane through BA, AC, make BAE= Ĺ DAB. Make AEAD; through E draw BEC, cutting AB, AC, in B and c, and join DB, DC. DAB EAB, And DA AE, and AB is common, and ..DB= EB; but. BD, DC, are > BC and BD BE, .. DC is > EC; and DA AE and AC is common, also DC DAC is > < EAC; and DAB BAE,.. EC, .. / > BAC. Also, BAC is not less than either of the other two; . L BAC+ either of the other two, must be > than the remaining angle. PROPOSITION XXII. Every solid angle is contained by plane angles, which together are less than four right angles. First, let the solid A be contained by the three planes, BAC, CAD, DAB; take in each of the lines AB, AC, AD, any points B, C, D; join BC, CD, DB. Then the CBA, ABD, DBC, make the solid B, . CBA + ABD are > DBC; similarly BCAACD, are > DCB; and CDA+ADB, are > BDC. ..s CBA, ABD, BCA, ACD, CDA, S ADB, are > DBC, BCD, BDC; but DBC + DCB+ BDC two rights. CBA + ABD + BCA + ACD + CDA + 2 two rights; and AS ABC, ACD, ADB, are the three s of each of the two rights, .. the nines of these As, viz., the s CBA, BAC, ACB, ACD, ADC, DAC, ADB, ABD, BAD, are six rights; and of these the s CBA, ABD, BCA, ACD, ADC, ADB, are > two rights. .. the three remaining, BAC, BAD, DAC, are four rights. Next, let the solid at A be contained by any number of |