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point which is supposed to be common to the two surfaces, will be circles having the same radii and centres with the spheres respectively.

2. If two spheres cut one another, they shall cut one another in a circle, the plane of which is perpendicular to the line joining their centres, and its centre in that line.

Let C, c be the centres of two spheres, and P, Q any two points common to the surfaces of both; join C P, Pc, C Q, Qc, and Ce; draw PO perpendicular to Cc; and join OQ. Then, because the triangles CPc, C Qc have the three sides of the one equal to the three sides

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of the other, each to each, the angle PCO is equal (I. 7.) to the angle QCO. And because the triangles PC O, QCO have two sides PC, CO of the one equal to two sides QC, CO of the other, each to each, and the included angles equal to one another; the remaining sides PO, QO are equal to one another, and the angle QOC is equal to PO C, that is, to a right angle. Therefore, the point Q is in the circumference of a circle which is described from the centre O, with the radius OP, in a plane (3. Cor. 1.) perpendicular to CO or Cc. And the same may be shewn of every other point Q in the common section of the two spheres. Therefore that common section is a circle, the plane of which is perpendicular to the line Cc, and its centre in that line, as above stated. In fact, if the plane P Cc cut the spheres in the circles A Pp and ap P, and if it be made to revolve about the line Cc, the circles A Pp, ap P will generate the two spheres (def. 21.), and the line OP (3. Cor. 2.) the circle of intersection.

3. If three spheres be described, whose centres A, B, C do not lie in the same straight line, the surfaces of the three cannot have more than two points in common, which points (if there be any such) lie in a straight line perpendicular to the plane of the centres, and at equal distances on either side of that plane.

From the points A, B, with the radii of the spheres described from A, B, respectively, in the plane ABC, de

line Q N, so that N may be the centre, and NQ the radius, of the circle (dotted in the figure) in which the spheres, whose centres are A and C, cut one another. Then, it is evident that the only points, common to the surfaces of the three spheres, are those which are common to the circumferences of these two circles. Let PM, QN be produced to meet one another in the point O; and let the planes of the circles cut one another in the line Vv, which passes through O, and is perpendicular to the plane ABC (18. Cor. 2.), because the planes of the circles, being perpendicular to the lines AB, AC respectively, are perpendicular, each, to the plane ABC, which passes through each of those lines (18.). Then, if M P be greater than M O, the circle, which has the centre M and the radius MP, will cut the line V v in two points V, v, which are equally distant from the point O (I. 12.). Join V N, v N. Then, because A N is at right angles to the plane Q N V, the angles AÑ V, AN v are right angles. And, because V, v are points in the surface of the sphere which has the centre A and radius AP or AQ, AV and 'A v are, each of them, equal to AQ. Therefore, because the right-angled (3.) triangles A N V, AN v have their hypotenuses AV, Av, and sides A N equal respectively to the hypotenuse AQ and side A Ñ of the right-angled triangle AN Q, other sides NV, N v are equal each to NQ (I. 13.); and, consequently, the circle which has the centre N and the radius NQ also passes through the same two points V, v. But, the circumferences of the two circles do not meet one another in any other point; for they can meet only in the line V v,

their

in which their planes cut one another. Therefore the points V, v, which are in a perpendicular to the plane A B C, and at equal distances on either side of that plane, and none else, are common to the surfaces of the three spheres; as above stated.

If the point V coincide with O, the point will likewise coincide with it; and in this case the three spherical surfaces will have only one point common to them, viz. the point O, which is in the plane of the three centres A, B, C.

If the centres of three spheres lie in the same straight line, their circles of intersection (if there are any) cannot meet one another, because their planes are perpendicular to this straight line, and therefore (11.) parallel; and accordingly the surfaces can have no point in common; unless each of them passes through the same point of the straight line in which the centres lie, for then each of them will touch the other two in that point.

PROP. 38. Prob. 3. Through a given point C, to draw a plane, which shall be perpendicular to a given straight line AB.

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G

E

A

B H

F

Through A draw a plane GH (38.) perpendicular to A D, and let it cut the given plane B C in the line AE: AE will be the straight line required. For, AD, being perpendicular to the plane GH, is perpendicular to the line A E, which meets it in that plane (def. 1.). Otherwise:

From any point D in AD draw DF perpendicular to the plane BC (36.): join A F; and, in the plane B C, draw AE perpendicular to A F; then A E (4.) will be the line required. Therefore, &c.

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From A draw the straight line AD perpendicular to the plane B C (36.); and through A draw the plane E F (38.) perpendicular to the straight line AD:

E

D

B

PROP. 45. Prob. 10.

To find the shortest distance between two given straight lines, A B and CD, which do not lie in the same plane.

Through A B draw the plane E F, parallel to the line CD (44.); and through the same A B draw the plane G H, perpendicular to the plane E F (41.). Then, because the straight line CD is parallel to the plane E F, which cuts the plane GH in the line AB, it cuts, or may be produced to cut, the plane G H in some point L. From L, draw LK perpendicular to AB: K L shall be the shortest distance required.

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For, the plane GH, being perpendicular to the plane E F, the straight line K L, which is drawn in the former plane, at right angles to the common section AB, must be perpendicular to the plane EF (18.), and therefore (10. and I. 14.) to the line CD, which is parallel to the plane EF. Now, because the line AB lies in the plane EF, CD cannot be nearer to AB than it is to the plane EF, to which it is parallel. But be

the plane E F is the plane required (11.). cause KL is perpendicular to the plane

Otherwise:

In the given plane B C, draw any two straight lines cutting one another; and, through the given point A, draw parallels to them respectively: the plane of these parallels will be the plane required (15.). Therefore, &c.

PROP. 44. Prob. 9. Through a given straight line AB, to draw a plane, which shall be parallel to a second given straight line CD.

If A B is parallel to CD, then any plane (10.) which passes through A B, and does not pass through CD, is parallel to C D.

But, if AB is not parallel to CD, through any point A of AB draw the straight line A E parallel to CD

(I. 48.): the plane A BAE is the plane

required (10.).

Therefore, &c.

E

D

B

EF, KL is the shortest distance of C D the shortest distance between the straight from the plane EF. Therefore KL is lines A B and C D.

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Therefore, &c.

Cor. Any point in the bisecting plane is at equal distances from the planes of the dihedral angle ABCD (I.13); for it may easily be shown that these distances are the sides of right-angled triangles, which have a common hypotenuse and the angles opposite to them equal to one another.

PROP. 47. Prob. 12.

and G H its perpendicular distance from
CD: through G draw G M, parallel to
BE or DC (I. 48), to meet AE in M,
and join C M. Then, because G M and
HC are in the same plane (I. def. 12.),
GH and M C are likewise in the same
plane; and, because CD is perpendicu-
lar both to CA and CE, it is perpen-
dicular also to CM and CF (3.): but, CD
is likewise perpendicular to G H; there-
fore (I. 14.) C M is parallel to HG, and
the quadrilateral CG is a rectangle.
Again, because FK, CL are parallel
and equal, and that the angle FCL is
a right angle, CK is a parallelogram
(I. 21.) and a rectangle. Therefore
(I. 22.) G H is equal to C M, and K L
to CF. But CF (I. 12. Cor. 3.) is
ABC, ABD, draw
less than C M, because C F is perpendicular to AB (I. 44.):
EF, EG perpendi-
cular to A E. Therefore, also, K L is less
than GH. And in the same manner it
may be shown that K L is less than any

other distance between the lines AB,
CD. If AB do not meet the plane DCE,
that is, if it be parallel to that plane,
AC is the shortest distance required.
Therefore, &c.

Cor. 1. The shortest distance' be

tween two straight lines is a straight line which is at right angles to each of

them. This is evident from each of the constructions above given. Indeed, if such were not the case, it could not be the shortest distance; for a perpendicular to the line which it does not meet at right angles, drawn from the point in which it meets the other, would be shorter (I. 12. Cor. 3.).

The problem shows, therefore, how to draw a straight line, which shall be at right angles to each of two given straight lines not lying in the same plane.

Cor. 2. If, in the second construction, the perpendiculars A C and BD be given, the length of the shortest distance KL may be found; for it is equal to the perpendicular C F, which is drawn from the right angle to the hypotenuse of a right-angled triangle having those perpendiculars for the two sides.

PROP. 46. Prob. 11.

To bisect a given dihedral angle ABCD.

Draw from any the same point B in the edge B C, BA and BD perpendicular to it (I. 44.) in the planes CBA and CB D respectively, and bisect the angle ABD by the straight line BE (I. 46.): the plane EBC will bisect the B given dihedral angle A B C D (17.).

E

given tetrahedron or triangular pyraTo circumscribe a sphere about a mid ABCD.

Bisect any one of the edges, A B, in E(I.43.): from E, in the adjoining faces

take F, G, the cen-
tres of the circles

circumscribing the
triangles А ВС,
ABD (III. 59.);
and therefore in the
straight lines E F,

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EG respectively from the points F, G, in the plane FEG, draw FO, GO, perpendicular to FE, GE (I. 44.) respectively, and meeting one another in O, and join OA: the sphere, which is described from the centre O, with the radius O A, shall pass through the points B, C, D, and circumscribe the

tetrahedron.

EF and EG are each of them at Join O B, O C, OD. Then, because right angles to E A, the plane FEG is perpendicular to E A (3.), and thereABD, which pass through EA. But fore (18.) to each of the planes ABC, OF and OG are perpendicular to the common sections FE and G E respectively. Therefore (18.) O F and O G are perpendicular to the planes A B C and ABD respectively. And because the points A, B, C, are equidistant from the foot of the perpendicular O F, they are likewise equidistant from the point O in the perpendicular (8.); therefore O B and O C are each of them equal to O A. And for the like reason OB (as before) and O D are each of them equal to OA. Therefore, the four distances, O A, O B, O C, O D, are equal to one another; and the sphere, which is described from the centre O, with the radius O A, is circumscribed about the tetrahedron ABCD.

Otherwise:

Bisect three of the edges at right angles by planes meeting one another in

the point O (38.): O is evidently the centre of a sphere, passing through the points A, B, C, D, and circumscribing the pyramid. Therefore, &c.

PROP. 48. Prob. 13.

To inscribe a sphere in a given tetrahedron or triangular pyramid A B C D. Bisect the dihedral angles at the edges BC, CD, DB of any the same face,(46.) and let the bisecting planes meet one another in the point 0: from O draw OP, OQ, OR, OS, perpendicular to the faces B CD, ABC, ABD, ACD, respectively (36): the sphere, which is described from the centre O, with the radius OP through the points Q, R, inscribed in the tetrahedron.

D

R

B

S

shall pass and be

For, the point O being in the plane which bisects the dihedral angle at B C, is equidistant (46. Cor.) from the planes ABC, BCD, that is, O Q is equal to OP; and, for the like reason, OR and OS are each of them equal to OP; therefore, the four distances, O P, OQ, OR, OS, are equal to one another: and the sphere, which is described from the centre O, with the radius OP, is inscribed in the tetrahedron A B C D.

Therefore, &c.

Cor. If the faces and altitude AL of the tetrahedron be given, the length of the radius OP may be found. For, since the pyramid, which has the alti tude A L and the base B C D, is equal to the sum of four pyramids having the common altitude OP, and for their bases the four faces of the first pyramid; or, which is the same thing, to a pyramid, which has the altitude OP, and a base equal to the sum of those faces (30.); it may easily be shown (32. Cor. 2. and II. 10. Cor.) that OP is to A L as the base B CD to the sum of the four faces.

Scholium.

In these problems, 12. and 13., it is assumed that, three planes being given, a point may be found which is common to the three, or that, any three planes being drawn, there is some point through which each of them passes, and which may be called their point of intersection. It is evident, however, that if the common section of two of the planes be parallel to the third plane (in which case,

also, by Prop. 10, the common sections are all parallel to one another), there can be no such point; for, since all the points which are common to the two first planes are found in their common section, if this common section does not cut the third plane (that is, in other words, if there is no point common to this common section and the third plane), there can be no point which is comthe contrary, the common section of mon to the three planes. When, on two of the planes is not parallel to the third, it may be produced to cut that plane, and the point in which it cuts it is the point which is common to the three planes.

pass

When a point is to be determined, therefore, by the intersection of three planes, in order to be satisfied with regard to the meeting of the planes, it is only necessary to consider the common section of any two of them, and examine whether it is or is not parallel to the third plane. Thus, in Problem 13, the question occurs, whether OC is, or not, parallel to the plane which is made to through DB; and it is easily perceived that it is not parallel; for, if the plane DBA be supposed to be turned round DB, in order to arrive at the position in which it will be parallel to O ̊C, before it arrives at that position it must evidently become parallel to A C, and DBO is necessarily more distant from the position of parallelism than DAB when parallel to AC, for it passes between the face ADB of the pyramid, and the base CDB. In the second solution of Prob. 12, the three planes, which are supposed to meet in O, are at right angles to the edges, and therefore the common section of any two of them, as, for instance, the two which are at right angles to AB, AC, is perpendicular to the face ABC of the pyramid (18. and 18. Cor. 2.), for which reason any plane which is parallel to it must be perpendicular to the same face, which is evidently not the case with the plane which is at right angles to the edge AD.

PROP. 49. Prob. 14.

Three plane angles AOB, BOC, COA, which form a solid angle O, being given; to find by a plane construction the angle contained by two of the planes, viz. AOB and BOC.

At the point B, in the plane BOC, and upon either side of the angle B OC, make the angles BOD, CO E equal to the angles BOA, COA, each to each

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