278 BOOK XI [XI. 4 For let a straight line EF be set up at right angles to the two straight lines AB, CD, which cut one another at the point E, from E; I say that EF is also at right angles to the plane through AB, CD. For let AE, EB, CE, ED be cut off equal to one another, E D CHB and let any straight line GEH be drawn across through E, at random; let AD, CB be joined, and further let FA, FG, FD, FC, FH, FB be joined from the point F taken at random <on EF>. Now, since the two straight lines AE, ED are equal to the two straight lines CE, EB, and contain equal angles, [1. 15] therefore the base AD is equal to the base CB, and the triangle AED will be equal to the triangle CEB; [1.4] so that the angle DAE is also equal to the angle EBC. But the angle AEG is also equal to the angle BEH ;[1. 15] therefore AGE, BEH are two triangles which have two angles equal to two angles respectively, and one side equal to one side, namely that adjacent to the equal angles, that is to say, AE to EB; therefore they will also have the remaining sides equal to the remaining sides. Therefore GE is equal to EH, and AG to BH. while FE is common and at right angles, [1. 26] therefore the base FA is equal to the base FB. [1. 4] For the same reason FC is also equal to FD. And, since AD is equal to CB, and FA is also equal to FB, the two sides FA, AD are equal to the two sides FB, BC respectively; and the base FD was proved equal to the base FC; therefore the angle FAD is also equal to the angle FBC. [1.8] Now since, again, GE was proved equal to EH, and EF is common, the two sides GE, EF are equal to the two sides HE, E and the base FG is equal to the base FH; therefore the angle GEF is equal to the angle HEF. Therefore each of the angles GEF, HEF is right. Therefore FE is at right angles to GH drawn at ran through E. Similarly we can prove that FE will also make angles with all the straight lines which meet it and are in plane of reference. [XI. D But a straight line is at right angles to a plane who makes right angles with all the straight lines which me and are in that same plane; therefore FE is at right angles to the plane of reference. But the plane of reference is the plane through the stra lines AB, CD. Therefore FE is at right angles to the plane thro AB, CD. Therefore etc. Q. E. D. The steps to be successively proved in order to establish this propo by Euclid's method are (1) triangles AED, BEC equal in all respects, (4) likewise triangles CEF, DEF, (5) triangles FAD, FBC equal in all respects, so that the angles FAG, FBH are equal, (6) triangles FAG, FBH equal in all respects, so that FG is equal to FH, [by [by [by (2), (3), (5) an 280 BOOK XI [XI. 4 (7) triangles FEG, FEH equal in all respects, [by (2), (6) and 1. 8] In consequence of the length of the above proof others have been suggested, and the proof which now finds most general acceptance is that of Cauchy, which is as follows. Let AB be perpendicular to two straight lines BC, BD in the plane MN at their point of intersection B. In the plane MN draw BE, any straight line through B. Join CD, and let CD meet BE in E. Produce AB to F so that BF is equal to AB. Since BC is perpendicular to AF at its middle point B, AC is equal to CF Similarly AD is equal to DF Since in the triangles ACD, FCD the two sides AC, CD are respectively equal to the two sides FC, CD, and the third sides AD, FD are also equal, M/ D the angles ACD, FCD are equal. N [1.8] The triangles ACE, FCE thus have two sides and the included angle equal, whence EA is equal to EF. [1.4] [1.8] The triangles ABE, FBE have now all their sides equal respectively; therefore the angles ABE, FBE are equal, and AB is perpendicular to BE. And BE is in any straight line through B in the plane MN. Legendre's proof is not so easy, but it is interesting. We are first required to draw through any point E within the angle CBD a straight line CD bisected at E. To do this we draw EK parallel to DB meeting BC in K, and then mark off KC equal to BK. CE is then joined and produced to D; and CD is the straight line required. Now, joining AC, AE, AD in the figure above, we have, since CD is bisected at E, (1) in the triangle ACD, ACAD2 = 2AE2 + 2ED2, BC2 BD2 = 2BE2 + 2ED2. and also (2) in the triangle BCD, B D Subtracting, and remembering that the triangles ABC, ABD are rightangled, so that XI. 4, 5] PROPOSITIONS 4, 5 281 whence [1. 48] the angle ABE is a right angle, and AB is perpendicular to BE. It follows of course from this proposition that the perpendicular AB is the shortest distance from A to the plane MN. And it can readily be proved that, If from a point without a plane oblique straight lines be drawn to the plane, (1) those meeting the plane at equal distances from the foot of the perpendicular are equal, and (2) of two straight lines meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the greater. Lastly, it is easily seen that From a point outside a plane only one perpendicular can be drawn to that plane. For, if possible, let there be two perpendiculars. Then a plane can be drawn through them, and this will cut the original plane in a straight line. This straight line and the two perpendiculars will form a plane triangle which has two right angles: which is impossible. PROPOSITION 5. If a straight line be set up at right angles to three straight lines which meet one another, at their common point of section, the three straight lines are in one plane. For let a straight line AB be set up at right angles to the three straight lines BC, BD, BE, at their point of meeting at B; I say that BC, BD, BE are in one plane. For suppose they are not, but, if possible, let BD, BE be in the plane of reference and BC in one more elevated; let the plane through AB, BC be produced; F C E it will thus make, as common section in the plane of reference, a straight line. Let it make BF. [XI. 3] Therefore the three straight lines AB, BC, BF are in one plane, namely that drawn through AB, BC. Now, since AB is at right angles to each of the straight lines BD, BE, therefore AB is also at right angles to the plane through BD, BE. [XI. 4] XI. De But BF which is in the plane of reference meets it; therefore the angle ABF is right. But, by hypothesis, the angle ABC is also right; therefore the angle ABF is equal to the angle ABC. And they are in one plane: which is impossible. Therefore the straight line BC is not in a more eleva plane; therefore the three straight lines BC, BD, BE are in plane. Therefore, if a straight line be set up at right angle three straight lines, at their point of meeting, the three stra lines are in one plane. Q. E. D. It follows that, if a right angle be turned about one of the straight containing it the other will describe a plane. At any point in a straight line it is possible to draw only one plane is at right angles to the straight line. A One such plane can be found by taking any two planes through the straight line, drawing perpendiculars to the straight line in the respective planes, e.g. BO, CO in the planes AOB, AOC, each perpendicular to AO, and then drawing a plane (BOC) through the perpendiculars. If there were another plane through perpendicular to AO, it must meet the plane through AO and some perpendicular to it as OC in a straight line OC' different from OC. Then, by XI. 4, AOC' is a right angle, and in the same plane with the right angle AOC: which is impossible. Next, one plane and only one can be drawn through a point outside a line at right angles to that line. line. Let P be the given point, AB the given straight In the plane through P and AB, draw PO perpendicular to AB, and through O draw another straight line OQ at right angles to AB. Then the plane through OP, OQ is perpendicular to AB. If there were another plane through P perpendicular to AB, either |