D H THEOREM Xviii. (117.) In any pyramid, a section parallel to the base is similar to the base ; and these two planes are to each other as the squares of their distances from the vertex. Let ABCD be a pyramid, and EFG a section parallel to the base BCD, also AIH a line perpendicular to the two planes at H and I; then will BD, EG, be two similar planes, and the plane BD will be to the plane EG, as AH? to AI?. For, join CH, FI. Then, because a plane cutting two parallel planes, makes parallel sections (th. 108), therefore the plane ABC, meeting the two parallel planes BD, EG, makes the sections BC, EF, parallel. In like manner, the plane ACD makes the sections CD, FG, parallel. Again, because two pairs of parallel lines make equal angles (th. 107), the two EF, FG, which are parallel to BC, CD, make the angle EFG equal the angle BCD. And in like manner it is shown, that each angle in the plane EG is equal to each angle in the plane BD, and consequently those two planes are equiangular. Again, the three lines AB, AC, AD, making with the parallels BC, EF, and CD, FG, equal angles (th. 14), and the angles at A being common, the two triangles ABC, AEF, are equiangular, as also the two triangles ACD, AFG, and have therefore their like sides proportional, namely, AC:AF::BC:EF::CD:FG. And in like manner it may be shown, that all the lines in the plane FG, are proportional to all the corresponding lines in the base BD. Hence these two planes, having their angles equal and their sides proportional, are similar, by def. 68. But, similar planes being to each other as the squares of their like sides, the plane BD : EG :: BC? : EF?, or :: AC? : AF?, by what is shown above. Also, the two triangles AHC, AIF, having the angles H and I right ones (th. 98), and the angle A common, are equiangular, and have therefore their like sides proportional, namely, AC : AF :: AH : AI, or AC? : AF2 :: AH? : AI? Consequently, the two planes BD, EG, which are as the former squares AC?, AF, will be also as the latter squares AH”, AI, that is, BD : EG :: AH? : AI?. THEOREM xix. (118.) In a cone, any section parallel to the base is a circle ; and this section is to the base, as the squares of their distances from the vertex. LET ABCD be a cone, and GHI a section parallel to the base BCD; then will GHI be a circle, and BCD, GHI, will be to each other, as the squares of their distances from the vertex. For, draw ALF perpendicular to the two parallel planes, and let the planes ACE, ADE, pass through the axis of the cone AKE, meeting the section in the three points H, I, K. Then, since the section GHI is parallel to the base BCD, and the planes CK, DK, meet them, HK is parallel to CE, and IK to DE (th. 108). And because the triangles formed B FE D by these lines are equiangular, KH : EC :: AK : AE :: KI : ED. But EC is equal to ED, being radii of the same circle ; therefore KI is also equal to KH. And the same may be shown of any other lines drawn from the point K to the perimeter of the section GHI, which is therefore a circle (def. 44). Again, by similar triangles, AL : AF :: AK : AE, or :: KI : ED, hence AL? : AF2 :: KI: ED2; but KI? : ED2 :: circle GHI : circle BCD (cor. th. 93); therefore AL? : AF? :: circle GHI : circle BCD. a THEOREM 'xx. (121.) a D B Let the sphere AEBF be cut by the plane ADB ; then will the section ADB be a circle. If the section pass through the centre of the sphere, then will the distance from the centre to every point in the periphery of that section be equal to the radius of the sphere, and consequently such section is a circle. Such, in truth, is the circle EAFB in the figure. Draw the chord AB, or diameter of the section ADB; perpendicular to which, or the said section, draw the axis of the sphere ECGF, through the centre C, which will bisect the chord AB in the point G (th. 41). Also, join CA, CB; and draw CD, GD, to any point D in the perimeter of the section ADB. Then, because CG is perpendicular to the plane ADB, it is perpendicular both to GA and GD (def. 89). Su that CGA, CGD are two right-angled triangles, having the perpendicular CG common, and the two hypothenuses CA, CD, equal, being both radii of the sphere ; therefore the third sides GA, GD, are also equal (cor. 2, th. 34). In like manner it is shown, that any other line, drawn from the centre G to the circumference of the section ADB, is equal to GA or GB; consequently that section is a circle. a THEOREM XXI. If two spheres intersect one another, the common section is a circle. LET A, B, be the centres : draw any two planes through AB cutting the spheres in the circles KCNG, MCLG, and KDNH, MHLD. Join CG meeting AB in E; and join AC, CB, AG, GB, AD, DB, AH, HB, DE, EH. Then, since CA = AD, and CB= M B * The section through the centre, having the same centre and diameter as the sphere, is called a great circle of the sphere; the other plane sections being called less circles of the sphere. Again, the angles CAE, DAE, being equal, the sides CA and DA being also equal, and AE common; hence the angle DEA = CEA. But the line joining AB bisects CG at right angles in E; hence also the angle DEA is a right angle. In the same manner NEH is a right angle, and EH = EG. · Hence the lines DE, EH, being in the same plane, and making the angles AED, BEH, right angles, they are in the same straight line at right angles to AB. Also, since the two lines CG, DH, are perpendicular to AB, the plane in which they are is also perpendicular to AB. Hence all the intersections, C, G, &c. are in a plane perpendicular to AB, and are, therefore, in a circle. Again, since CE EG, DE=CE, and EH = EG, the point E is the centre of the circle CDGH. Theorems on the foregoing subjects for the student to demonstrate. : 1. If two planes be parallel to the same plane, or to the same straight lines, they are parallel to one another. 2. If a plane and a straight line be parallel to the same plane, or to the same straight line, they are parallel to one another. 3. Two parallel straight lines make equal angles with the same or with parallel planes: and two parallel planes make equal angles with the same or with parallel straight lines. 4. When two parallel planes are cut by a third plane, or by a straight line, the exterior angle is equal to the interior opposite, the alternate angles are equal, and the two interior angles are together equal to two right angles : and show whether generally the converse of either of these three conditions takes place, in the case of the planes being cut by a plane and by a line, the planes would be parallel. 5. When two straight lines are not parallel, the planes which are drawn perpendicular to them will intersect one another. 6. When a straight line and a plane intersect one another, every straight line perpendicular to the plane will intersect every plane which is drawn perpendicular to the straight line. 7. If a plane bisect a dihedral angle, and from any point in it perpendiculars be drawn to the planes which contain the dihedral angle, these perpendiculars will be equal, and the plane passing through them will be perpendicular to the line in which all the planes meet. 8. If the perpendiculars from a point to two planes forming a dihedral angle be equal, the plane passing through this point and the line in which the planes containing the dihedral angle intersect, will bisect the dihedral angle. 9. If a plane bisect a line at right angles, lines drawn from any point in the plane to the extremities of the bisected line, will be equal. 10. If from the extremities of a line, two equal lines be drawn to meet in a point, then a plane drawn through this point perpendicular to the first line will bisect this line. 11. If through the middle of a given line a plane be drawn at right angles to that line, it shall pass through the vertex of every isosceles triangle having that line for a base. 12. If on a given line as base, any number of equal triangles be constituted, having their equal sides terminated in the same extremity of the line, their vertices will all lie in the circumference, or one cube whose plane is perpendicular to the common base *. SECTION II.-OF SOLID ANGLES. A THEOREM XXII. (123.) If a solid angle be contained by three plane angles, any two of them together are greater thun the third ; and the difference of any two of them is less than the third. Let the solid angle at S be contained by the three plane angles ASB, BSC, CSA: then any two of them, as ASC, CSB, shall be greater than the third BSA. In the plane ASB make the angle BSD equal to BSC, and draw any line ADB from A in the same plane, meeting SB in B. Take SC = SD, and join AC, BC. Because the two sides BS, SD, are equal to the two BS, SC, and the angle BSD to the angle BSC, the bases BD, BC, are equal. But AC + CB are greater than BA (th. 10); and hence taking the equals BD, BC, from these, AC is greater than AD. But the sides AS, SD, being equal to AS, SC, and the base AC greater than AD, the angle ASC is greater than ASD. Hence adding to the unequals ASC, ASD, the equals DSB, BSC, the angles ASC and SCB are together greater than ASB. In the same way may the other part of the eorem be proved by means of th. 11. The great Verbal analogy, and analogy in the forms of enunciation, often leads to error. similarity in the general form of the enunciation of several theorems respecting lines in one plane, and of the lines and planes variously disposed in space, has often caused propositions respecting the latter to be tacitly assumed by the inexperienced student as truths, merely in consequence of the analogy in the form of expression, but which a little attention to the circumstances involved in the hypothesis would have shown at once to be fallacious. A few such are annexed : and the student should be required to distinctly demonstrate why they are false, and under what limited circumstances they are true. 1. When two planes intersect one another, straight lines which are perpendicular to them will also intersect. 2. When two planes intersect one another, any planes which are perpendicular to these will also intersect. 3. When two straight lines intersect, any straight lines perpendicular to these will also intersect. 4. When a straight line meets a plane, every straight line perpendicular to the given straight line will intersect every plane perpendicular to the given plane. 5. Two straight lines equally inclined to the same plane are parallel to one another. 6. Two planes equally inclined to the same straight line are parallel to one another. 7. Two planes equally inclined to the same plane are parallel to one another. 8. Two straight lines which make equal angles with the same straight line are parallel to one another. 9. Two straight lines parallel to the same plane are parallel to one another. 11. When two straight lines are cut proportionally by three planes, these three planes are parallel. A B THEOREM XXIII. (124.) Every solid is contained by plane angles, which are together less than four right angles. Let the solid angle at S be contained by the plane angles ASB, BSC, CSD, DSE, ESA: then the sum of these angles is always less than four right angles. Let the planes which contain the several plane angles be cut by a plane ABCDE, these letters representing its intersections with the sides of the several plane angles, or with the edges of the solid angle. Because the solid angle at A is contained by the three plane angles SAB, BAE, EAS, any two of which are greater than the third, the angles SAB + EAS are greater than EAB. For a similar reason, the two plane angles at B, C, D, E, which are the bases of the triangles having the common vertex S, are severally greater than the third angle at the same point, which is one of the triangles of the polygon ABCDE. Hence all the angles at the bases of the triangles SAB, SBA, &c. are together greater than all the angles EAB, &c. of the polygon ABCDE. And because all the angles at the bases of the triangles, viz. SAB, SBA, &c. together with the plane angles at S, are equal to all the angles of the polygonal base, together with four right angles, (being in each case equal to twice as many right angles as there are sides AB, BC, ...), the angles at S are less than four right angles. a : M с Σ THEOREM xxiv. (125.) If each of two solid angles be contained by three plane angles equal to one another, each to each : then the planes in which the angles are have the same inclination to one another. Let the solid angles at A and B be contained by three plane angles CAD, DAE, EAC, and FBG, GBH, HBF, respectively; and let CAD = FBG, DAE = GBH, and EAC = HBF: then shall the dihedral angle formed by DAC and EAC be equal to the dihedral angle formed by GBF and HBF. In the lines CA and FB take AK = BM, and from the points K and M draw the perpendiculars KD, KL, MG, MN, in the planes of the angles whose edges are AC and FB. Then the dihedral angles made by these planes are measured by the angles DKL and GMN respectively. Join LD, NG. Then the triangles KAD, MBG, having KAD = MBG, and AKD = BMG, and the included sides equal, viz. AK BM, therefore KD = MG, and AD = BG. In like manner, in the triangles KAL, MBN, it may be proved that KL = MN, and LA = NB. Again, the triangles LAD, NBG, having LA = NB, AD = BG, and angle LAD = angle NBG, the bases are equal, viz. LD = NG. Lastly, in the triangles KLD, NMG, since the sides are equal, viz. KL = MN, KD = MG, and the bases also equal, viz. LD NG; therefore the angles included, viz. LKD and NMG are also equal. But these are the measures of the inclination of the faces of the solid angle, or the measure of the dihedral |