PROPOSITION 17. THEOREM. Straight lines which are cut by parallel planes are cut pro portionally. Let the st. lines AB, CD be cut by the three par1 planes GH, KL, MN at the points A, E, B, and C, F, D. Then shall AE: EB :: CF : FD. Join AC, BD, AD; and let AD meet the plane KL at the point X: Then because the two par1 planes KL, MN are cut by the plane ABD, .. the common sections EX, BD are par1. XI. 16. And because the two par1 planes GH, KL are cut by the plane DAC, .. the common sections XF, AC are par1. Now since EX is par1 to BD, a side of the ▲ ABD, .. AE : EB :: AX: XD. XI. 16. Again because XF is par1 to AC, a side of the ▲ DAC, AX: XD :: CF : FD. Hence AE: EB :: CF : FD. VI. 2. VI. 2. v. 1. Q.E.D. DEFINITION. One plane is perpendicular to another plane, when any straight line drawn in one of the planes perpendicular to their common section is also perpendicular to the other plane. [Book XI. Def. 6.] PROPOSITION 18. THEOREM. If a straight line is perpendicular to a plane, then every plane which passes through the straight line is also perpendicular to the given plane. A B X Let the st. line AB be perp. to the plane XY; and let DE be any plane passing through AB. Then shall the plane DE be perp. to the plane XY. Let the st. line CE be the common section of the planes XY, DE. From F, any point in CE, draw FG in the plane DE perp. to CE. XI. 3. I. 11. Then because AB is perp. to the plane XY, Нур. .. AB is also perp. to CE, which meets it in that plane, XI. Def. 1. But the GFB is also a rt. angle; .. GF is par to AB. And AB is perp. to the plane XY, ... GF is also perp. to the plane XY. Constr. Hyp. XI. 8. Hence it has been shewn that any st. line GF drawn in the plane DE perp. to the common section CE is also perp. to the plane XY. ... the plane DE is perp. to the plane XY. XI. Def. 6. Q.E.D. EXERCISE. Shew that two planes are perpendicular to one another when the dihedral angle [see x1. Def. 7] formed by them is a right angle. If two intersecting planes are each perpendicular to a third plane, their common section shall also be perpendicular to that plane. E. A Let each of the planes AB, BC be perp. to the plane ADC, and let BD be their common section. Then shall BD be perp. to the plane ADC. For if not, from D draw in the plane AB the st. line DE perp. to AD, the common section of the planes ADB, ADC: I. 11. and from D draw in the plane BC the st. line DF perp. to DC, the common section of the planes BDC, ADC. Then because the plane BA is perp. to the plane ADC, and DE is drawn in the plane BA perp. to AD the common section of these planes, Constr. DE is perp. to the plane ADC. XI. Def. 6. Similarly DF is perp. to the plane ADC. .. from the point D two st. lines are drawn perp. to the plane ADC; which is impossible. XI. 13. Hence DB cannot be otherwise than perp. to the plane ADC. Q.E.D. Of the three plane angles which form a trihedral angle, any two are together greater than the third. Let the trihedral angle at A be formed by the three planes BAD, DAC, BAC. Then shall any two of them, such as the ▲ BAD, DAC, be together greater than the third, the CASE I. If the of the BAD, DAC; it is evident that the the BAC. BAC. BAC is less than, or equal to, either BAD, DAC are together greater than CASE II. But if the BAC is greater than either of the < BAD, DAC; then at the point A in the plane BAC make the ▲ BAE equal to the BAD ; and cut off AE equal to AD. Through E, and in the plane BAC, draw the st. line BEC cutting AB, AC at B and C join DB, DC. Again in the BDC, since BD, DC are together greater than BC, .'. and BD BE, I. 20. Proved. B 8 E And in the A3 DAC, EAC, ... the DAC is greater than the ▲ EAC. Constr. Proved. I. 25. Constr. ... the two ▲ BAD, DAC are together greater than the ▲ BAC. PROPOSITION 21. THEOREM. Q.E.D. Every (convex) solid angle is formed by plane angles which are together less than four right angles. S E C X Let the solid angle at S be formed by the plane < ASB, BSC, CSD, DSE, ESA. Then shall the sum of these plane angles be less than four rt. angles. |