B ز D A. therefore the three straight lines EC, CB, BE, are in one plane ; but the whole of the lines DC, AB, and BC, produced, are in the same plane with the parts of them EC, EB, BC (I. 1.) Therefore AB, CD, CB, are all in one plane. COROLLARY 1.-It is manifest that any two straight lines which cut one another are in one plane. CoR. 2.-Only one plane can pass through three points, or through a straight line and a point; and these conditions therefore are sufficient to determine a plane. PROPOSITION III. THEOREM. If two planes cut one another, their common section is a straight line. Let two planes AB, BC, cut one another, and let B and D be two points in the line of their common section. From B to D draw the straight line BD; and because the points B and D are in the plane AB, the straight line BD is in that plane (Pl. Ge. I. Def. 8); for the same reason, it is in the plane CB; the straight line BD is therefore common to the planes AB and BC, or it is the common section of these planes. PROPOSITION IV. THEOREM. If a straight line stand at right angles to each of two straight lines in the point of their intersection, it will also be at right angles to the plane in which these lines are. Let PO be perpendicular to the lines AB, CD, at their point of intersection 0, it is perpendicular to their plane. For, draw through O any straight line EF in their plane. In OD take any point G, and make GD=OG, and through G draw GF parallel to OB, to meet OF in F; join DF, and produce DF to meet AB in B, and join PD, PF, and PB. Because OG=GD, therefore (Pl. Ge. VI. 2.) BF=FD; and because, in triangle DOB, the side DB is bisected in F, therefore (Pl. Ge. II. A.) DO2 + OBP = 2 0F2 +2 FB3. P B с E A And for a similar reason, in triangle DPR, DP2 + PB = 2 PFP + 2 FB?. But since POD is given a right angle, therefore (Pl. Ge. I. 47) PD = PO2 + OD?; and for a similar reason PBP = PO? + OB2. Therefore PD2 + PB2 = 2 PO? +OD+ OB=2 PO2 +2 OF2 +2 FB?, for it was shown that ODP+OB=2 OF2 + 2 FB?. But it was also proved that PD2 + PBP = 2 PF2 + 2 FB?; and therefore 2 PF2 +2 FB2=2 PO2 +2 OF2 +2 FB? ; or, taking 2 FB2 from both, 2 PF2 = : 2 PO2 + 2 OF, or PF2 = PO2 + OF2. Therefore (Pl. Ge. I. 48) POF is a right angle. In a similar manner it may be shown that PO is perpendicular to any other line through O in the plane ACBD; therefore it is perpendicular to that plane (I. Def. 1.) Cor. 1.-If a plane be horizontal in any two directions, it is so in every direction. Cor. 2.-The perpendicular PO is less than any oblique line as PB, and therefore it measures the shortest distance from the point P to the plane. PROPOSITION V. THEOREM. If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point, these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B, the point where they meet; BC, BD, BE, are in one and the same plane. If not, let, if it be possible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which with the plane, in which BD and BE are, shall be a straight line (I. 3); let this be BF; therefore the three straight lines AB, BC, BF, are all in one plane, namely, that which passes through AB, BC; and be- B cause AB stands at right angles to each of the straight lines BD, BE, it is also at right angles to the plane passing through them (I. 4); and therefore makes right angles (I. Def. 1) with every straight line meeting it in that plane; but BF, which is in that plane, meets it ; therefore the angle ABF is a right angle; but the D B в angle ABC, by the hypothesis, is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane ; which is impossible : therefore the straight line BC is not above the plane in which are BD and BE. Wherefore the three straight lines BC, BD, BE, are in one and the same plane. PROPOSITION VI. THEOREM. If two straight lines be at right angles to the same plane, they shall be parallel to one another. o-Let the straight lines AB, CD, be at right angles to the same plane; AB is parallel to CD. Let them meet the plane in the points B, D, and draw the straight line BD, to which draw DE at right angles, in the same plane; and make DE equal to AB, and join BE, AE, AD. Then, because AB A is perpendicular to the plane, it shall make right (I. Def. 1) angles with every straight line which meets it, and is in that plane; but BD, BE, which are in that plane, do each of them meet AB. Therefore each of the angles ABD, ABE, is a right angle. For the same reason, each of the angles CDB, CDE, is a right angle; and because AB is equal to DE, and BD common, the two sides AB, BD, are equal to the two ED, DB; and they contain right angles; therefore the base AD is equal (Pl. Ge. I. 4) to the base BE. Again, because AB is equal to DE, and BE to AD, and the base AE common to the triangles ABE, EDA, the angle ABE is equal (Pl. Ge. I. 8) to the angle EDA ; but ABE is a right angle; therefore ÉDA is also a right angle, and ED perpendicular to DA; but it is also perpendicular to each of the two BD, DC; wherefore ED is at right angles to each of the three straight lines BD, DA, ĐČ, in the point in which they meet. Therefore these three straight lines are all in the same plane (I. 5); but AB is in the plane in which are BD, DA, because any three straight lines which meet one another are in one plane (I.2). Therefore AB, BD, DC, are in one plane; and each of the angles ABD, BDC, is a right angle; therefore AB is parallel to CD." A B D PROPOSITION VII. THEOREM. If two straight lines be parallel, and one of them is at right angles to a plane, the other also shall be at right angles to the same plane. Let AB, CD, be two parallel straight lines, and let one of them AB be at right angles to a plane; the other CD is at right angles to the clo same plane. For, if CD be not perpendicular to the plane to which AB is perpendicular, let DG be perpendicular to it. Then (1.6) DG is parallel to AB; DG and DÒ therefore are both parallel to AB, and are drawn through the same point D; which is impossible. PROPOSITION VIII. THEOREM. Two straight lines which are each of them parallel to the same straight line, though not both in the same plane with it, are parallel to one another. Let AB, CD, be each of them parallel to EF, and not in the same plane with it; AB shall be parallel CD. In EF take any point G, from which draw, in the plane passing through EF, AB, the straight line GH at right angles to EF; and in the plane passing a В. through EF, CD, draw GK at right angles to the same EF. And because EF is perpendicular both to GH and T D GK, EF is perpendicular to the plane HGK passing through them; and EF is parallel to AB; therefore AB is at right angles (I. 7) to the plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD, are each of them at right angles to the plane HGK. But if two straight lines are at right angles to the same plane, they are parallel (I. 6) to one another. Therefore AB is parallel to CD. PROPOSITION IX. THEOREM. If two straight lines meeting one another be parallel to two others that meet one another, though not in the same H E 1 K plane with the first two, the first two and the other two shall contain equal angles. Let the two straight lines AB, BC, which meet one another, be parallel to the two straight lines DE, EF, that meet one another, and are not in the same plane with AB, BC. The angle ABC is equal to the angle DEF. Take BA, BC, ED, EF, all equal to one another; and join AD, CF, BE, A AC, DF; because BA is equal and parallel to ED, therefore AD is (Pl. Ge. I. 33) both equal and parallel to BE; for the same reason, CF is equal and parallel to BE; therefore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the same straight line, though not in the same plane with it, are parallel (I. 8) to one another ; therefore AD is parallel to CF; and it is equal to it, and AC, DF, join them towards the same parts ; and therefore AC is equal and parallel to DF; and because AB, BC, are equal to DE, EF, and the base AC to the base DF, the angle ABC is equal (Pl. Ge. I. 8) to the angle DEF. PROPOSITION X. THEOREM. To draw a straight line perpendicular to a plane, from a given point above it. Let A be the given point above the plane BH; it is required to draw from the point A a straight line perpendicular to the plane BH. In the plane draw any straight line BC, and from the point A draw AD perpendicular to BC. If then AD be also perpendicular to the plane BH, the thing required is already done; but if it be not, from the point D draw, in the plane BH, the straight line DE at right angles to BC; and from the point A draw AF perpendicular to DE; and through F draw GH parallel to BC; and because BC is at right angles to ED and DA, BC is at right angles (I. 4) to the plane passing through ED, DA; and GH is parallel to BC; but if two straight lines be parallel, one of which is at right angles to a plane, the other E F B D |