PROPOSITION VII. It is always possible to draw a straight line perpendicular to a given straight line from a given point in the same. -- - - - Let BC be the given straight line, and A the given point in it. From AB, AC cut off equal parts AD, AE. With D and E as centres, describe equal os intersecting in F. Join AF. AF shall be 1 to BC. :: DA is = EA, DF= EF, and AF common to the two As DAF, EAF, ... DAF is = 1 EAF, (Def.) PROPOSITION VIII. From a given point in a given straight line there cannot be drawn more than one straight line perpendicular to it. F16 B let AF and AG be each of them I to BC. .:25 FAB, FAC are equal to one another, and also z s GAB, GAC are equal to one another, which is impossible. .: from a given point in a given straight line there cannot be drawn more than one straight line I to it. CoR. All right angles are equal to one another. - K and let the ABC be applied to the 2 GHK, then will BA fall along HG. For if it fell in any other direction, as HX, then from H there would be drawn two straight lines HG, HX, 1 to HK, which is impossible; .: BA must fall along HG, .. ABC is = 2 GHK. C. G. 2 and .......... . :) PROPOSITION IX. The angles which one straight line makes with another upon one side of it are either two right angles or are together equal to two right angles. Let the straight line AB make with CD upon one side of it the ZS ABC, ABD. These shall either be two right 2S; or shall together be = two right 2 s. If the LABC is = the . ABD; then each of them is a right 2. But if not, through B draw BEI to CD; then the - ABC is = the s CBE and EBA together; .. the ZS ABC and ABD are together = the z s CBE, EBA and ABD; . of which CBE is a right 2, and EBA, ABD make up a right 2 ; .: LS ABC, ABD are together = two right 2 s. Cor. If a number of straight lines BA, BC, BF, BE meeting in the point B form angles ABC, CBF, FBG, GBA which fill up the space about B, then these is are together = 4 right 2 s. For produce CB to X. Then all the 4s ABC, CBF, FBG, GBA are equal to the zs on one side of CBX together with the us on the other side of CBX, and are , ; = four right 2 s. PROPOSITION X. If two straight lines, drawn from one extremity of a straight line on opposite sides of it, make the adjacent angles together equal to two right angles, these two straight lines shall be in one and the same straight line. o Let BC, BD drawn from B one extremity of AB make the adjacent LS ABC, ABD together = two right angles. Then shall BD be in the same straight line with BC. For if not, if possible, let BE be in the same straight line with BC; .:. LS ABC, ABE are together = two right angles; (1. 9) and are .:. = LS ABC, ABD; (hyp.) .: LABE is = 1 ABD, which is impossible. .:, no other straight line than BD is in the same straight line with BC; .. BD is in the same straight line with BC. same PROPOSITION XI. It is always possible to draw a perpendicular to a given straight line of an unlimited length from a given point without it. Let A be any point without the straight line BC of unlimited length. Then from A a straight line can be drawn I to BC. With centre A describe a O cutting BC in D and E. With centres D and E describe equal os intersecting in F. Join AF, cutting BC in G. Join AD, AE, FD, FE. Then ::: DA, AF, FD are respectively = EA, AF, FE; .. .As DAF, EAF are equal in all respects, and ...DAF is = . EAF. (1. 5) Again, ::: DA, AG and the included - DAG are respectively = EA, AG, and 2 EAG; .:: the as DAG, EAG are equal in all respects; ... LDGA is = . EGA; (1. 1) .. AG is 1 to BC. |