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 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, and ... ▲ DAF is = LEAF (1.5) Again, ·.· DA, AG and the included DAG are re spectively = EA, AG, and ▲ EAG; ..the As DAG, EAG are equal in all respects; ..4 DGA is: = LEGA; .. AG is 1 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. A R C 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 QFAQ. Join FR. Then. AQR is a right 4, .. 4 FQR is a right 4. (1.9) Hence in the As AQR, FQR, AQ, QR and ▲ AQR are respectively equal to FQ, QR, and FQR; LARQ = L FRQ. .. if ARQ were a right angle, = (I. I) the SARQ, FRQ would be together two right angles, 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 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. D Let the side BC of ▲ ABC be produced to D. Then shall the Bisect AC in E; join BE and produce BE to F, making EF- BE. Join CF. Then, ·. AE, EB are respectively = CE, EF, But the ACD is > the ▲ ECF; .. ▲ ACD is > the EAB; i.e. ▲ CAB. Similarly it may be shewn, if AC be produced to G, that BCG is the ABC. But ACD is = LBCG; (1.6) .. LACD is the ABC. Hence it follows that Any two angles of a triangle are together less than two right angles. D Let ABC be 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. (I. 9) PROPOSITION XIV. The greater side of every triangle has the greater angle opposite to it. Let ABC be a ▲ having the side AC greater than the side AB. Then shall the ABC be greater than the ACB. And ⚫. the side CD of ▲ BCD has been produced to A, (I. 13) .. LADB is greater than ▲ DCB; i.e. LACB. |