For conceive a plane passing through B C to revolve about that line till it pass through the point E. Then because the A D points E and C are in that plane, the line E C is in it; E) and for the same reason the line E B is in it; and BC is in it, by hypothesis. Hence the lines A B, CE, and B C are all in one plane. BI Cor. Any two straight lines which meet each other are in one plane; and any three points whatever are in one plane. THEOREM XC. If two planes, as A E, C F cut one another, the line of their common section is a straight line. For let B and D, any two points in the line of their common section, be joined by the straight line BD; then B because the points B and D are both in the plane E A E, the whole line B D is in that plane; and C F for the same reason B D is in the plane C F. 'A D The straight line B D is therefore common to both planes ; and it is therefore the line of their common section. THEOREM XCI. If a straight line, as A B, stand at right angles to each of two other straight lines, as A C, AG, at A, their point of meeting ; A B is also at right angles to the plane passing through A C, A G. For through A, draw any line A E, in the plane passing through AG, A C; and from any point E, in that line, draw EF В. parallel to AG; make F C equal to AF; join C E, and produce till it meet A G in D. Then because F E is parallel to A D, C F is to F A as CE is to ED (Theo. 78.); but C F is equal to FA; therefore C E A is equal to E D. Now the square of BD is equal to the squares of B A and AD (Theo. 44.), and the square G of B C is equal to the squares of B A and A C. Hence the squares of B D and B C are together equal to the square of A C, the square of A D, and twice the square of A B. But because C D is bisected in E, the squares of B D and B C are together equal to double the squares of B E and ED (Theo. 50.); and for a like reason, the squares of A D and A C are together equal to double the squares of E D and E A. Hence double the squares of B E and E D are together equal to double the squares of A E, A B, and E D; or by omitting double the square of E D from each, double the square of E B is equal to double the square of B A and double the square of AE; or the square of B E is equal to the sum of the squares of BA and A E; and hence B A E is a right angle. The line B A therefore being at right angles to any straight line which it meets with с THEOREM XCII. If a straight line, as A B, stand at right angles to each of three straight lines, as B C, B D, B E, at B, their point of meeting, these three straight lines are all in one plane. If not, let B D and D Е, if possible, be in one plane, and BC above it ; and let the plane passing through A B and A B C meet the plane in which В D and B E are, in the straight line B F. Then because A B is at right с angles to B D and B E, it is also at right angles to В! F B F, which it meets in the plane of B D and B E D (Theo. 91.). A B is also by hypothesis at right E angles to B C, and therefore the angles ABC, ABF, in the same planes, are equal to each other, which is impossible. B C is therefore not above the plane of B E and B D, and consequently the three lines B C, BD, and B E, are all in one plane. THEOREM XCIII. If two straight lines, as A B, CD, be at right angles to the same plane, as B DE, A B is parallel to C D. In the plane B D E draw D E at right angles to B D, take any point E in it, and join E A, E B, and A D. Then the A C square of E A is equal to the squares of E B and BA; and the square of E B is equal to the squares of E D and D B (Theo. 44.) Hence the square of E A is B! D equal to the squares of E D, D B, and B A, and as the squares of B D and B A are equal to the square of E DA, the square of E A is equal to the sum of the squares of E D and D A; and consequently E D A is a right angle. E D being therefore at right angles to BD, AD, and C D, these three lines are all in one plane. But A D and B D are in the same plane with A B. Hence A B and C D are in one plane ; and as each of the angles A BD, CD B is a right angle, A B is parallel to CD, (Cor. 1. Theo. 19.) THEOREM XCIV. If A B, one of two parallel straight lines, be at right angles to a plane, as E F, the other C D is also at right angles to the same plane. For if C D be not perpendicular to the plane E F, let D G be perpendicular to it. Then as A B and D G are per AGIC pendicular to the same plane, they are parallel to each other (Theo. 93.). Hence DC and D G drawn E down through the same point are parallel to the D same straight line, which is impossible. Therefore F no line drawn through D, except DC, is at right angles to the plane E F; and D C is consequently at right angles to it. 'B THEOREM XCV. If A B, C (), any two lines, be each parallel to another line E F, and not in the same plane with it, A B and C O are parallel to eath other. From any point G in E F, let G H be drawn, in the plane passing through A B and E F, and at right angles to E F; A H B and in the plane passing through E F and C O let G K be drawn also at right angles to EF. Then E G F as E F is perpendicular to G H and GK, it is perpendicular to G HK, the plane passing through K them. (Theo. 91.) And as E F is parallel to A B, A B is also at right angles to the plane GHK, (Theo. 94.) For the same reason CO is at right angles to the plane G HK, and hence A B and C O being at right angles to the same plane, are parallel to each other. (Theo.93.) THEOREM XCVI. If two straight lines, as A B, BC, which meet one another, be parallel to two other straight lines, as D E, E F, that meet one another, but are not in the same plane with A B, B C; the angle A B C is equal to the angle D EF. Take BA, BC, ED, EF, all equal to one another, and join AD, C F, BE, A C, and D F. Then as A B is equal B and parallel to DE, A D is equal and parallel to B E. (Theo. 30.) For the same reason C F is equal and A С parallel to B E. Hence A D and C F being equal and parallel to the same straight line are equal and paral E lel to each other, (Theo. 95.); and consequently AC DL F and D F are equal and parallel. Therefore the triangles A B C, D E F, having A B and B C equal respectively to DE and E F, and A C equal to DF, have also the angles A B C and DEF equal. (Theo. 5.) THEOREM XCVII. If any straight line, A B, be at right angles to a plane, C K, any plane, DE, passing through A B, will be at right angles to the plane C K. For let C E be the common section of the planes DE, and from any point F in C E, draw F G in the plane D G A H DE, at right angles to CE. Then as A B is perpendicular to the plane C K, it is perpendicular to K C E meeting it in that plane; therefore A B F is a right angle; and as G F B is also a right angle, C F B E G F is parallel to A B. (Cor. 1. Theo. 19.) Hence as A B is at right angles to the plane CK, F G is at right angles to the same plane, (Theo. 94.) ; and consequently the plane D E is at right angles to the plane CK (Def. 2. p. 53); and thus may any other plane passing through A B be shewn to be at right angles to the CK; TheoREM XCVIII. If each of two planes AB, BC be perpendicular to a third plane ADC, B D, the common section of the first two planes will also be perpendicular to the third plane. For from D in the plane A D C draw D E perpendicular to A D, and D F to DC. Then as D E is at right angles to DA, the common section of the planes A B and B В AD C, and those planes A D C are at right angles to each other, therefore E D is at right angles to D D B. For the same reason D F is at right angles to D B; and hence as B D is at right angles to both D F and D Е, it is at right angles to the A plane A D C in which these lines are. (Theo. 91.) F E THEOREM XCIX. A If a solid angle, as A, be contained by three plane angles, as B AC, CAD, D A B, any two of these plane angles are greater than the third. If the angles BAC, CAD, DAB, be all equal, it is evident that any two of them are greater than the third. If D they are not all equal, let B AC be that angle which is not less than either of the others, and is greater than one of them, as D AB; and in the place passing through A B, AC, draw the line B Е С A E, making the angle B A E equal to the angle BAD; make A E equal to AD, and through E draw any straight line B E C, cutting A B and A C in B and C, and join B D C D. Then as D A is equal to A E, the angle B A D to the angle B A E, and the side B A common to the two triangles B A D and B A E, their bases BD and B E are equal, (Theo. 1.); and consequently E C is the difference of B C and B D, and it is less than D C the other side of the triangle B D E. (Theo. 16.) Now D A and A E are equal, and A C common to the triangles DAC and E AC, but D C the base of the triangle D A C is greater than E C the base of the triangle E AC; therefore the angle D AC is greater than the angle E A C. (Theo. 14.) But D A B and E A В are equal hy construction, whence D A B and DA C together are greater than B A E and E A C together, or than B AC. And as B A C is not less than either of the angles B-AD, DAC; therefore B A C with either of them is greater than the other. THEOREM C. If A be a solid angle contained by any number of plane angles B AC, CAD, D A B, E A B, these plane ungles together are less than four right angles. Let the planes which contain the solid angle at A be cut by another plane, and let the section of them by that plane A E С D greater than C D E; and AED, and A E B greater than D EB. Hence the angles at the bases of the triangles, which have their common vertex at A, are greater than the interior angles of the rectilineal figure BCDE. But all the angles of the triangles BAC, CAD, DAE, BAE, are equal to twice as many right angles as there are triangles, or as there are sides in the figure B C D E; and the interior angles of that figure, together with four right angles, are also equal to twice as many right angles as the figure has sides. (Theo. 25.) Therefore all the angles of the triangles are equal to all the angles of the figure, together with four right angles. And as all the angles at the base of the triangles are greater than all the interior angles of the figure, the remaining angles of the triangles, or those which contain the solid angle A, are less than four right angles. GEOMETRICAL PROBLEMS. PROBLEM I. m. To bisect a given straight line A B. From the points A and B, as centres, with any distance greater than half A B, describe arcs cutting each other in n and Join n m, and the point C where it cuts A B will be the middle of the line required. For join An, A m, B n, and Bm; then because An and A m are respectively equal to B n and B m, and nm is common to both the triangles, A m n, and B nm, the angles A n m and Bnm are equal. And hence as A n and B n are equal, n C common to both triangles, and the angle An C equal to the angle B n C, the remaining sides, A C and B C, are equal; or A B is bisected in C. C |