« ΠροηγούμενηΣυνέχεια »
parallel to CD, and through F draw KL parallel to AB. Then (Prop. vii.) the plane through AB and GH will be parallel to the plane through CD and KL. One such pair, therefore, can be drawn.
(2.) There cannot be a second pair of such parallel planes.
For if there can, each plane must also be parallel to the other line (Prop. viii. Cor. 1.); let us admit, then, that a second plane PQ can be drawn through AB which shall be parallel to CD.
Then, since CD is parallel to the two planes PQ and AGBH, it is parallel to their intersection AB (Prop. ix.) and hence, AB, CD are in one plane. But by hypothesis they are not in one plane, which is impossible. Whence only one such pair of planes can be drawn.
PROPOSITION XX. One parallelopiped, and only one, can always be constituted, which
shall have three of its non-contiguous edges coincident with three given lines ; provided that no two of these lines be in one plane, nor all three parallel to one plane.
(1.) Let AB', BC', CA' be three straight lines, no two of which are in one plane, nor all three parallel to one plane; then a parallelopiped DD' may be constituted so as to have three non-contiguous edges coincident with these lines.
For through AB', planes B'D, AD can be drawn respectively parallel to the lines BC', CA'; through BC', planes C'D, BD' parallel to the lines CA', AB', and through CA', the planes CD', A'D
D' parallel to BC', AB'.
Then the planes B'D and BD' will be parallel, C'D and CD' will be parallel, and A'D and AD' will we parallel (Prop. xix.). These three pairs of parallel planes, when limited by their intersections, constitute a parallelopiped, the non-contiguous edges of which coincide with the three lines AB', BC", CA'.
(2.) There cannot be a second parallelopiped, since there cannot be a second pair of component parallel planes drawn through either pair of lines (Prop. xix.).
PROPOSITION XXI. The six faces of a parallelopiped are parallelograms; those which are
opposite are equal : the edges are equal and parallel in fours; and the diagonals which join the four pairs of opposite angles meet in one point, and are bisected there.
(1) Let the parallelopiped have for its opposite planes AC and A'C', AB' and A'B, BC' and B'C: then these figures are parallelograms.
For since the parallel planes AC, A'C' are cut by the parallel planes AB', A'B, the four lines of section AD, BC, B'C', A'D' are all parallel (Prop. xvi. Cor.).
In the same manner it is shown that AC", DB', A'C, D'B are parallel; and A AB, CD, A'B', C'D' are parallel.
Wherefore AD being parallel to BC, and AB to CD, the figure ABCD (one of the faces of the parallelopiped) is a parallelogram: and in the same way it is proved that all the others are parallelograms.
D' (2.) The opposite pairs of parallelo- 6 grams, as AC, A'C', are equal.
For AD, BC, A'D', B'C' being parallel and intercepted between parallel planes AD', A'D, they are equal (Prop. xv.). In the same manner AB, CD, A'B', C'D' are equal; and likewise AC', B'D, BD', A'C are equal.
And since AB, BC are parallel to A'B', B'C', the angle ABC is equal to the angle A'B'C'; and parallelograms which have equal angles and the sides about those angles equal each to each, are wholly equal. Whence the parallelograms AC, A'C' are equal.
In the same way is proved the equality of the parallelograms AB' and A'B, and of AD' and A'D.
(3.) The opposite edges are equal in fours. This has been proved in the demonstration of the preceding case.
(4.) The four diagonals AA', BB’, CC', DD' intersect in one point Q, and Q is the middle of each of them.
For, if AB', A'B were drawn, the figure AB'A'B would be a parallelogram, of which the diagonals are AA' and BB'; and Q their intersection is the middle of each of them.
In the same manner if AD', A D were drawn, ADA'D' would be a parallelogram, of which the diagonals are AA', DD'; and the diagonal DD' therefore passes through Q the middle of AA', and is bisected there.
In a similar manner CC' passes through Q, and is bisected there.
Whence all the four pass through one point, and are bisected in that point.
If any number of lines, anyhow situated, be all cut by any number of
parallel planes, the planes will divide the lines proportionally. Let, for instance, the three lines AC, LE, HV, be cut by the three parallel planes MN, PQ, RS, in A, L, II, in B, F, K, and in C, E, V respectively : then they will be divided in these points, so that
AB: AC::LF:LE::HK:HV. For, draw the planes ACE, AEL, ELV, VLH. These will cui each other, the two former in the line AE, and the two latter in the line LV. They will also cut the parallel planes in parallel lines (Prop. 1.); viz., so that BD is parallel to CE, DF to AL, FG to EV,
and GK to LH. Whence the pairs of triangles ABD and ACE, etc., are similar ; consequently,
AB: AC :: AD: AE::LF:LE::LG:LV::HK:HV; where we have the enunciated property.
In the same way, the demonstration may be extended to any number of lines and any number of planes.
Cor. 1. Conversely, if any number of lines, more than two, between parallel planes be similarly divided, the sets of homologous points of division will always be situated in so many planes parallel to the original ones.
COR. 2. When there are only two such lines divided, planes through those pairs of points may be drawn parallel to the original planes; but as two points do not fix the position of a plane, there may be innumerable planes drawn through the pairs of homologous points which are neither parallel to the original planes nor to one another.
| PROPOSITION XXIII. Any section of a prism or pyramid parallel to the base is similar to
the base; and that of the prism also equal to the base. (1.) Let ABCDE be the base of a prism, and the prism itself be cut by a plane parallel thereto in abcde: then the figure abcde will be similar and equal to ABCDE.
For since the plane AB ba cuts the two parallel planes ABCDE, abcde, the lines AB, ab are parallel (Prop. 1.). Similarly BC, bc are parallel, and so on how many sides soever there be.
Whence again, since AB, BC which meet in B are parallel to ab, bc which meet in b, the angle abc is equal to ABC (Prop. vii.). In the same manner all the other angles are equal each to each in the two figures.
Again, since Aa, Bb, Cc, etc., are all parallel (Def. 9.), the figures ABba, BCcb, etc., are parallelograms, and hence ab = AB, bc = BC, etc. Consequently,
AB:BC:: ab:bc, or the sides about the equal angles ABC, abc are proportional; and so on for all the others. The figures are therefore similar.
But similar figures which have their homologous sides also equal, are altogether equal.
(2.) Let the pyramid SABCDE (S being the vertex, and ABCDE the base), be cut by a plane abcde parallel to the base: then the figure abcde will be similar to the base ABCDE.
For, as in the preceding case, ab, be, cd, de, ea are respectively parallel to AB, BC, CD, DE, EA; and the angles abc, bcd, etc., equal to the angles ABC, BCD, etc. The figure abcde is therefore equiangular with the base.
Again, the triangles aSb, ASB are similar, since ab is parallel to AB; and similarly with respect to the other pairs. Whence ab: AB::6S: BS::bc: BC, or
ab:bc:: AB:BC; that is the sides about the equal angles abc, ABC are proportional.
In the same manner it may be proved for the sides about the other equal angles. The figure abcde is hence similar to ABCDE.
Scholium. The cutting plane may be below the base, between the base and vertex, or beyond the vertex. It is analogous to the three cases of Euc. VI. 2. In the last case, however, the figure will have a reversed position, as may be seen in the figure.
If a cylinder or cone be cut by a plane parallel to the base, the section
will be a circle ; and in the cylinder this circle will be also equal to the base.
(1.) Let ABC be the base of a cylinder, and E its centre; let the cylinder be cut by a plane parallel to the base in abc: then abc is a circle.
For let Ee be the intercepted portion of the axis, and Aa, Bb, Cc, etc., the intercepted portions of the edges. All these, by the definition of the cylinder, are parallel ; and being between parallel planes, they are equal (Prop. xv.); and each equal to Ee.
Also, since the plane A ae E cuts the parallel planes in AE, ae, these lines are parallel (Prop. 1.); and hence A ae E is a parallelogram, having ae = AE.
= In like manner in all the corresponding cases, be = BE, ce = CE, etc. But AE = BE CE, etc.; and hence ae = be = ce, etc., and e is the centre of a circle, the
radius of which is equal to that of the base. The section, therefore, is a circle equal to the base.
(2.) Let S be the vertex of a cone, of which the base is ABC and axis SE; and let it be cut by a plane in abc; then abc is a circle.
For draw the planes SEA, SEB, etc., cutting ABC in AE, BE, etc., and abc in ae, be, etc.
Then the parallel planes ABC, abc, being cut by SEA, the lines AE, ae are parallel (Prop. 1.). In the same manner BE, be are parallel, and so on.
Wherefore the triangle Sae is similar to SAE, Sbe to SBE, etc.; and hence
ae: AE::eS: ES::be: BE, or
ae:eb::AE:EB. But E is the centre of the circle ABC, and hence AE EB; and therefore ae = eb. In the same manner ae = ec, etc.
Consequently, the points a, b, c, etc., being in one plane, and all equidistant from a point e in it, they are in the circumference of a circle, of which e is the centre.
SCHOLIUM. The same remark applies to the different positions of the cutting plane in respect of the cone that was made respecting the pyramid.
PROPOSITION XXV. If from two fixed points E, P any parallel straight lines be drawn,
riz., EA and PB, EC and PD, etc., to meet the fixed plane MN in A and B, C and D, etc. : then AB, CD, etc., will always pass through the same point in the plane, viz., the point F in which EP meets it.
For since EA, PB are parallel, they are in one plane. Also the points E, P being in that plane, the line EP is in that plane. The point F being in the line EP is also in that plane; and hence the intersection AB of the plane EABP with MN contains the point F. That is, the line AB passes through F.