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 4. (1.9) Hence in the As AQR, FQR, AQ, QR and AQR are respectively equal to FQ, QR, and FQR; .. LARQ= L FRQ. (I. 1) .. if ARQ were a right angle, the S ARQ, 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 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. B Let the side BC of ▲ ABC be produced to D. Then shall the ACD be> either of the S BAC, ABC. Bisect AC in E; join BE and produce BE to F, making EF BE. H Join CF. Then, AE, EB are respectively CE, EF, = Similarly it may be shewn, if AC be produced to G, that BCG is the ABC. But ACD is = L BCG; (1.6) .. LACD is > the ABC. Hence it follows that Any two angles of a triangle are together less than two right angles. B 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. (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 = LADB. (1. 2) .. LADB is greater than 4 DCB; ... also ABD is greater than ▲ DCB. (I. 13) Much more then is ABC greater than DCB, i.e. LACB. PROPOSITION XV. The greater angle of every triangle has the greater side opposite to it. B Let ABC be a ▲ having the ABC > the ACB, then shall the side AC be > the side AB. For if not, AC must either be = AB or < AB. AB, for, then would ABC be = LACB, (1. 2) And AC is not < AB, for then would ▲ ABC be < LACB, but it is not. (I. 14) .. AC is > AB. |