« ΠροηγούμενηΣυνέχεια »
rallel to one another; and the others by the revolution of a right angled parallelograms.
parallelogram about one of its sides XIV.
which remains fixed. A sphere is a solid figure described
XXII. by the revolution of a semicircle The axis of a cylinder is the fixed about its diameter, which remains straight line about which the paunmoved.
rallelogram revolves. XV.
XXIII. The axis of a sphere is the fixed The bases of a cylinder are the circles
straight line about which the seini. described by the two revolving opcircle revolves.
posite sides of the parallelogram. XVI.
XIV. The centre of a sphere is the same Similar cones and cylinders are those with that of the semicircle.
which have their axes and the diaXVII.
meters of their bases proportionals. The diameter of a sphere is any
XXV. straight line which passes through A cube is a solid figure contained by the centre, and is terminated both
six equal squares. ways by the superficies of the sphere.
A tetrahedron is a solid figure conA cone is a solid figure described by tained by four equal and equilateral
the revolution of a right angled tri- triangles. angle about one of the sides con
XXVII. taining the right angle, which side An octahedron is a solid figura conremains fized.
tained by eight equal and equilaIf the fixed side be equal to the other teral triangles. side containing the right angle, the
XXVIII. cone is called a right angled cone; A dodecahedron is a solid figrire conif it be less than the other side, an tained by twelve equal pentagons obtuse angled, and if greater, an which are equilateral and equianacute angled cone.
XXIX. The axis of a cone is the fixed straight An icosahedron is a solid figure con
line about which the triangle re- tained by twenty equal and equivolves.
lateral triangles. XX.
DEF. A. The base of a cone is the circle de- A parallelopiped is a solid figure con
scribed by that side containing the tained by six quadrilateral figures, right angle, which revolves.
whereof every opposite , two are XXI.
parallel. A cylinder is a solid figure described
PROP. I. THEOREM. One part of a straight line cannot be in a plane, and another part
above it. If it be possible, let AB, part of the be produced to D: straight line ABC, be in the plane, And let any plane and the part BC above it: Ard since pass through the the straight line AB is in the plane, it straight line AD, LA B D can be produced in that plane: Let it and be turned about
it until it pass through the point C; the same plane that have a common and because the points B, C are in segment AB, which is impossible. this plane, the straight line BC is in (Cor. 11. 1.) Therefore, one part, &c. it (7. Def. 1.): Therefore there Q. E. D. are two straight lines ABC, ABD in
PROP. II. THEOREM.
Two straight lines which cut one another are in one plane, and three
straight lines which meet one another are in one plane. Let two straight lines AB, CD, cut the straight line BC is in A one another in E; AB, CD are in one the same; and, by the plane: And three straight lines EC, hypothesis, EB is in it: CB, BE, which meet one another, are Therefore the three in one plane.
straight lines EC, CB, Let any plane pass through the BE are in one plane : But straight line EB, and let the plane be in the plane in which EC, turned about EB, produced, if neces- EB are, in the same are
B sary, until it pass through the point (1.11.) CD, AB: Therefore AB, CD, C: Then because the points E, C are are in one plane. Wherefore two in this plane, the straight line EC is straight lines, &c. Q. E. D. in it (7. Def. 1.): For the same reason
PROP. III. THEOREM.
If two planes cut one another, their common seclion is a straight line.
Let two planes AB, BC, cut one in the plane BC, the straight line another, and let the
DFB: l'hen two straight lines DEB, line DB be their
DFB have the same extremities, and common section :
therefore include a space betwixt DB is a straight
them: which is impossible (10. Ax. 1.): line : If it be not,
Therefore BD the common section of from the point D
the planes AB, BC cannot but be a to B, draw, in the
straight line. Wherefore, if two plane AB, the straight line DEB, and planes, &c. Q. E. D.
PROP. IV. THEOREM.
If a straight line stand at right angles to each of two straight lines in the
point of their intersection, it shall also be al right angles to the plane which passes through them, that is, to the plane in which they are:
Let the straight line EF stand at ED all equal to one another; and right angles to each of the straight through E draw, in the plane in which lines AB, CD in E, the point of their are AB, CD, any straight line GEH; intersection: EF is also at right angles and join AD, CB; then from any to the plane passing through AB, CD. point F in EF, draw FA, FG, FD,
Take the straight lines AE, EB, CE, FC, FH, FB: And because the two
straight lines AE, ED are equal to angle FBC: Again it was proved the two BE, EC, and that they con- that AG is equal to BH, and also AF taiu equal angles (10. Ax. 1.) AED, to FB; FA, then, and AG, are equal BEC, the base AD is equal (4.1.) to to FB and BH, and the angle FAG the base BC, and the angle DAE to has been proved equal to the angle the angle EBC: And the angle AEG FBH; therefore the base GF is equal is equal to the angle BEH(10. Ax. 1.); (4. 1.) to the base FH: Again, betherefore the triangles AEG, BEH cause it was proved that GE is equal have two angles of one equal to two to EH, and EF is common; GE, EF angles of the other, each to each, and are equal to HE, EE; and the base the sides AE, EB, adjacent to the GF is equal to the base FH: thereequal angles, equal to one another; fore the angle GEF is equal (8. 1.) to wherefore they shall have their other the angle HEF ; and consequently sides equal (26. 1.): GE is therefore each of these angles is a right (10. equal to EH, and AG to BH: And def.) angle. Therefore FE makes because AE is equal to EB, and FE right angles with GH, that is, with common and at right angles to them, any straight line drawn through E in the base AF is equal (4. 1.) to the base the plane passing through AB; CD. FB; for the same reason, CF is equal In like manner, it may be proved, to FD: and because
that FE makes right angles with every AD is equal to BC,
straight line which meets it in that and AF to, FB, the
plane. But a straight line is at right two sides FA, AD, A
angles to a plane when it makes right
C are equal to the two
angles with every straight line which FB, BC, each to each;
meets it in that plane (3. def. 11.): and the base DF was
H Therefore EF is at right angles to the proved equal to the D
plane in which are AB, CD. Wherebase FC; therefore
fore, if a straight line, &c. Q. E. D. the angle FAD is equal (8. 1.) to the
PROP. V. THEOREM. If three straight lines meet all in one point, and a straight line stands 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 right angles to each of the straight lines BD, BE, it is also at right angles lines BC, BD, BE, in B the point (4. 11.) to the plane passing through where they meet ; BC, BD, BE are them; and therefore makes right anin one and the same plane.
gles (3. def. 11.) with every straight If not, let, if it be possible, BD and line meeting it in BE be in one plane, and BC be above that plane; but BF it; and let a plane pass through AB, which is in that BC, the common section of which plane meets it: with the plane, in which BD and Be Therefore the angle Are, shall be a straight (3. 11.) line; ABF is a right anlet this be BF: Therefore the three gle; but the angle straight lines AB, BC, BF, are all in ABC, by the hypoone plane, viz. that which passes thro' thesis, is also a
B AB, BC; and because AB stands at right angle; there
fore the angle ABF is equal to the BE: Wherefore the three straight angle ABC, and they are both in the lines BC, BD, BE are in one and the same plane, which is impossible : same plane. Therefore, if three Therefore the straight line BC is not straight lines, &c. Q. E. D. above the plane in which are BD and
PROP. VI. THEOREM.
If two straight lines be at right angles to the same plane, they shall be
parallel to one another.
Let the straight lines AB, CD be base AD is equal (4. 1.) to the base at right angles to the same plane; BE: Again, because AB is equal to AB is parallel to CD.
DE, and BE to AD; AB, BE are Let them meet the plane in the equal to ED, DA; and in the tripoints B, D, and draw the straight angles ABE, EDA, the base AE is line BD, to which draw DE at right common: therefore the angle ABE is angles, in the same plane; and make equal (8. 1.) to the angle EDA: But DE equal to AB, and join BE, AE, ABE is a right angle; therefore EDA AD). Then, because AB A
is also a right angle, and ED perpenis perpendicular to the
dicular to DA: But it is also perpenplane, it shall make
dicular to each of the two BD, DC: right (3. def. 11.) angles
Wherefore ED is at right angles to with every straight line.
each of the three straight lines BD, which meets it, and is B
DA, DC in the point in which they in that plane: But BD,
meet: Therefore these three straight BE, which are in that
lines are all in the same plane (5. 11.):
E plane, do each of them
But AB is in the plane in which are meet AB. Therefore each of the an. BD, DA, because any three straight gles ABD, ABE is a right angle: For lines which meet one another are in the same reason, each of the angles one plane (2. 11.): Therefore AB, BD, CDB, CDE is a right angle: And be- DC are in one plane: And each of the cause AB is equal to DE, and BD angles ABD, BDC is a right angle ; common, the two sides AB, BD are therefore AB is parallel (28. 1.) to equal to the two ED, DB; and they CD. Wherefore, if two straight lines, contain right angles; therefore the &c. Q. E. D.,
PROP VII. THEOREM.
If two straight lines be parallel, the straight line drawn from any point
in the one to any point in the other, is in the same plane with the parallels.
Let AB, CD be parallel straight draw the straight line EHF from lines, and take any point E in the one, E to F; and and the point F in the other : The since EGF also A straight line which joins E and F is is a straight in the same plane with the parallels. line, the two H
If not, let it be, if possible, above straight lincs the plane, as EGF; and in the plane EHF, EGF inABCD in which the parallels are, clude a space
between them, which is impossible: CD are, and is therefore in that plane. (10. Ax. 1.) Therefore the straight line Wherefore, if two straight lines, &c. joining the points E, F is not above Q. E. D. the plane in which the parallels AB,
PROP. VIII. 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 therefore the base AD is equal (4. lines, and let one of them AB be at 1.) to the base BE: Again, because right angles to a plane; the other CD AB is equal to De, and BE to AD; is at right angles to the same plane. the two AB, BE are equal to the two
Let AB, CD meet the plane in the ED, DA; and the base AE is compoints B, D, and join BD: Therefore mon to the triangles ABE, EDA; (7.11.) AB, CD, BD are in one plane. wherefore the angle ABE is equal In the plane to which AB is at right (8. 1.) to the angle EDA: And ABE angles, draw DE at right angles to is a right angle; and therefore EDA BD, and make DE equal to AB, and is a right angle, and ED perpendicular join BE, AE, AD. And because AB to DĂ: But it is also perpendicular is perpendicular to the plane, it is per- to BD; therefore ED is perpendicular pendicular to every straight line which (4. 11.) to the plane which passes meets it, and is in that plane:(3. Def, through BD, DA, and shall (3. def. 11.) Therefore each of the angles 11.) make right angles with every ABD, ABE, is a right angle: And straight line meeting it in that plane: because the straight line BD meets But DC is in the plane passing through the parallel straight lines AB, CD, the BD, DA, because all three are in the angles ABD, CDB are together equal plane in which are the parallels AB, (29. 1.) to two right angles: And ABD CD: Wherefore ED is at right angles is a right angle; therefore also CDB to DC; and therefore CD is at right is a right angle, and CD
angles to DE: But CD is also at perpendicular to BD:
right angles to DB; CD then is at And because AB is equal A c
right angles to the two straight lines to DE, and BDcommon,
DE, DB in the point of their interthe two AB, BD are
section D; and therefore is at right equal to the two ED,
angles (4. 11.) to the plane passing DB, and the angle ABDB
through DE, DB, which is the same is equal to the angle
plane to which AB is at right angles. EDB, because each of
Therefore, if two straight lines, &c. them is a right angle; I
Q. E. D.
PROP. IX. THEOREM.
Two straight lines which are each of them parallel to the same straight
line, and not in the same plane with it, are parallel to one another. Let AB, CD be each of them par- draw in the plane passing through EF, allel to EF, and not in the same plane AB, the straight line GH at right with it; AB shall be parallel to CD. angles to EF; and in the plane passe
In EF take any point G, from which ing through EF, CD, draw GK at