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. C Let A be any point without the straight line BC of unlimited length. Then from A a straight line can be drawn 1 to BC. With centre A describe a 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, Again, and... ▲ DAF is = LEAF. (I. 5) DA, AG and the included DAG are respectively EA, AG, and EAG; ...the as DAG, EAG are equal in all respects; .. LDGA is = ¿ EGA ; .. AG is to BC. (I. I) PROPOSITION XII. There cannot be drawn more than one perpendicular to a given straight line from a given point without it. Let A be the given point, and BC the given straight line; and let AQ be drawn from A to BC. (I. 10) Then no other straight line besides AQ can be drawn from A to BC. Draw any other straight line AR from A to BC. Produce AQ, and make the produced part QF= AQ. Join FR. Then. AQR is a right 4,.. 4 FQR is a right . (1.9) Hence in the AS AQR, FQR, AQ, QR and 4 AQR are respectively equal to FQ, QR, and FQR; .. LARQL FRQ. = (I. I) .. if ARQ were a right angle, the S ARQ, FRQ would be together two right angles, LS and .'. ARF would be a straight line; (I. 10) and thus two straight lines would inclose a space, which is impossible. .. AR is not 1 to BC. Similarly it may be proved that no other straight line besides AQ, drawn from A to BC, is 1 to it. INEQUALITIES. PROPOSITION XIII. If a side of a triangle be produced the exterior angle is greater than either of the interior and opposite angles. Let the side BC of a ABC be produced to D. Then shall the ▲ ACD be > either of the 4s BAC, ABC. Bisect AC in E; join BE and produce BE to F, making EF-BE. Join CF. Then, AE, EB are respectively CE, EF, and ▲ AEB is = ▲ CEF; (1. 6) Similarly it may be shewn, if AC be produced to G; Hence it follows that Any two angles of a triangle are together less than two right angles. B Let ABC be a A. Then shall any two of its angles, as ABC and ACB, be together two right angles. < Produce BC to D. Then ABC is the exterior ACD; (1. 13) .. Ls ABC, ACB are together < the 4s ACD, ACB, and are ... < than two right angles. (1.9) PROPOSITION XIV. The greater side of every triangle has the greater angle opposite to it. B Let ABC be a ▲ having the side AC greater than the side AB. Then shall the ABC be greater than the ACB. From AC cut off AD = AB, Then.. AB is = AD; .. ▲ ABD is = 1 ADB. (1. 2) And... the side CD of ▲ BCD has been produced to A, .. LADB is greater than ▲ DCB; (I. 13) .. also ▲ ABD is greater than ▲ DCB. Much more then is ABC greater than ▲ DCB, i.e. LACB. |