Whence BD does not lie above the plane MN; and in the same way it does not lie below it. It is therefore in the plane MN.

The same demonstration applies to every other line perpendicular to AB at B; and hence the proposition is true.

PROPOSITION IV.

From the same point there can be drawn only one perpendicular to a plane, whether that point be in the plane or without it.

(1.) Let A be a point in the plane MN: there can only be drawn from A one perpendicular to the plane MN.

For, if possible, let there be two, viz., AB, AC, and through them draw the plane PQ, cutting MN in AQ.

Then since BA is perpendicular to the plane MN, BAQ is a right angle (Prop. 11.); and since CA is perpendicular to MN, CAQ is a right

M

B

N

angle. The angles BAQ, CAQ in the same plane PQ are therefore equal; the less to the greater, which is impossible. Whence from A situated in the plane MN only one perpendicular can be drawn. (2.) Let A be a point without the

plane MN, only one line can be drawn from A perpendicular to MN.

For, if possible, let there be two, viz., AB, AC, and through them draw a plane PQ, cutting MN in BCQ.

Then since AB, AC are perpendicular to the plane MN, the angles ABC, ACB of the triangle ABC are two right angles (Frcp. II.); which is

P

M

N

impossible (Euc. 1. 17). Whence both lines cannot be perpendicular to the plane MN.

Whence, whether the point be in the plane or without it, only one line through the point can be perpendicular to the plane.

PROPOSITION V.

Through the same point only one plane can be drawn perpendicular to a line, whether that point be in the line or without it.

(1.) Let A be a point in the line BC: only one plane prpendicular to BC can be drawn through A.

For, if possible, let there be two planes MN, MP drawn through A perpendicular to BC; and let AM be their intersection; and draw any other plane BQ through BC, and not passing through AM, cutting MN and MP in AD and AE.

Then since BC is perpendicular to the planes MN, MP, the angles BAD, BAE are right angles (Prop. 11.); and they

M

B

A

N

E

are in one plane BQ, which is impossible. Whence MN and MP cannot both be perpendicular to BC.

(2.) Let A be without the line BC; only one plane can be drawn through A perpendicular to BC.

For, if possible, let there be two, MN, MP, the common section of which is AM; and let a plane be drawn through BC and A, cutting MN, MP in AD, AE. Then since EA is in the plane MP perpendicular to CB, the angle AEB is a right angle; and since DA is in the plane MN perpendicular to CB, the angle BDA is a right angle. But AD, AE, and BC are in the same plane. Wherefore the interior angle ADE of the triangle ADE is equal to the exterior; which is impossible (Euc. 1. 16.). Wherefore only one plane through A can be perpendicular to BC.

PROPOSITION VI.

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M

(1.) If two planes be perpendicular to a line, they are parallel to one another; and

(2.) If one of two parallel planes be perpendicular to a line, the other is also perpendicular to it.

(1.) Let the planes MN, PQ be perpendicular to AB: they are parallel to one another.

For, through AB draw any two planes AD, AF, cutting MN in AC and AE, and PQ in BD and BF.

M

Then since AB is perpendicular to MN, the angles BAC, BAE are right angles (Prop. II.); and since AB is P perpendicular to PQ, the angles ABD, ABF are right angles.

Then, since in the same plane AD, the lines AC, BD make right angles with AB, they are parallel. Similarly

A

B

E

N

AE, BF are parallel. Wherefore the two lines CA, AE which meet in A being parallel to the two DB, BF which meet in B, the planes MN, PQ which contain them are parallel (Prop. VII. Chap. 1.).

(2.) Let MN, PQ be parallel planes, and one of them MN be perpendicular to the line AB: then the other, PQ, will also be perpendicular to AB.

For draw any two planes AD, AF through AB cutting the planes in AC, AE and BD, BF as before.

Then since the planes MN, PQ are parallel, the line AC is parallel to BD and AE to BF. Whence the two angles EAB, ABF are together equal to two right angles; and one of them EAB is a right angle (Prop. II.); and consequently the other, ABF is a right angle, or BF is perpendicular to AB. In the same manner BD is perpendicular to AB; and hence (Def. 5) the plane PQ through BD, BF is perpendicular to AB.

VOL. II.

M

PROPOSITION VII.

The traces of the profile angle are perpendicular to the edge of the dihedral angle.

Before proceding further, let the student refer to Defs. 3, 4, 7, and (Prop. II. Chap. 11.); he will then see that this proposition is only the proper technical form M of the following :—

A plane perpendicular to the edge of a dihedral angle cuts the faces in lines which are perpendicular to that edge.

Let MN, PQ be the faces of the dihedral angle MNQP, PN its edge, and RNQS a profile plane whose traces are R RN, NQ: then RN, NQ will be perpendicular to PN.

For, since PN is perpendicular to the R plane RQ, it is perpendicular to every line in it drawn through N: that is RN, NQ are both perpendicular to PN.

PROPOSITION VIII.

IN'

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(1.) The profile traces at all points in the edge of a dihedral angle are parallel, those in one face to one another, and those in the other to one another; and

(2.) All the profile angles of the same dihedral angle are equal to one another.

(Preceding figure.)

(1.) Let RQ, R'Q' be two profile planes to the dihedral angle MNQP; then the trace R'N' is parallel to the trace RN, and the trace N'Q' to the trace NQ.

For RN, R'N' being in the same plane, and both perpendicular to PN, they are parallel to one another.

In the same manner N'Q' is parallel to NQ; and that which is proved for any one point N' is true for all other points.

(2.) The profile angle R'N'Q' is equal to the profile angle RNQ. For, since the two lines R'N', N'Q' which meet in N' are parallel to the two RN, NQ which meet in N, the angle R'N'Q' contained between the first two is equal to the angle RNQ contained between the other two (Prop. VII. Chap. 1.).

COR. All the profile planes of the same dihedral angle are parallel, that is, the plane R'Q' is parallel to RQ (Prop. VII. Chap. I.).

SCHOLIUM. It might have been inferred from this demonstration that all planes perpendicular to the same straight line are parallel; but this is usually made the subject of a separate proposition, and it is so done in this course.

PROPOSITION IX.

(1.) If two dihedral angles be equal, their profile angles are equal;

and

(2.) If the profile angles of two dihedral angles be equal, the dihedral angles are themselves equal.

(1.) Let PABQ, P'A'B'Q' be two equal dihedral angles, and PQ, P'Q' the profile planes; PB, BQ the profile traces of the first dihedral angle, and P'B', B'Q' those of the second: then the angles PBQ, P'B'Q' will be equal.

For let the dihedral angle P'A'B'Q' be transposed to the position such that A'B' shall coincide with AB, the point B' coinciding with B, and the plane A'P' with the plane AP.

Then, since the face A'P' coincides with AP, and the dihedral angles are equal (Hypoth.), the face A'Q' will coincide with the face AQ.

Again, since A'B' coincides with AB, the point B' with the point B, and the face A'P' with the face AP, the perpendiculars B'P', BP in those coincident faces must coincide. In like manner, B'Q' must coincide with BQ; and hence the angles P'B'Q', PBQ will be equal. (2.) If the profile angles P'B'Q', PBQ be equal, the dihedral angles P'A'B'Q', PABQ will be equal.

For so transpose the dihedral angle P'A'B'Q' that the edge A'B' shall coincide with AB, the point B' with B, and the plane A'B'P' with APB: then if the plane A'Q' do not coincide with AQ, let it take some other position as AS, and B'Q' the position BS.

Then, as in the preceding case, B'P' coincides with BP; and since A'B'Q' is a right angle, ABS is a right angle, and BS is in the plane PQ (Prop. III.).

Whence since the dihedral angles PABS, P'A'B'Q' are equal, the profile angles P'B'Q', PBS are equal (preceding case); and P'B'Q' is equal to PBQ by hypothesis. Whence PBS is equal to PBQ, the less to the greater, which is impossible. Whence the dihedral angles P'A'B'Q', PABQ are not unequal; that is, they are equal.

SCHOLIUM. The properties in this and the preceding proposition are tacitly assumed by Euclid, and by most other writers; but

very little reflection is necessary to convince us that such a course is altogether unwarranted under the circumstances of the case. It should be laid down as a general rule in mathematics, that propositions which admit of proof from principles already laid down, ought never to be assumed as independently true. It is tantamount to multiplying our axioms without necessity; and moreover with great risk of involving contradictions amongst them.

PROPOSITION X.

Two dihedral angles have the same ratio to one another that their profile angles have to one another.

Let PABM, MBAQ be two dihedral angles, whose profile angles are PBM, MBQ: then

PABM: MABQ :: PBM : MBQ.

[In the figure, the dihedral angles are placed contiguously, so as to have one face common to both of them, viz., BM. When this is not so given, we must conceive them to be so placed by transposing one of them nearly as in the preceding proposition.]

For since PB, MB, QB are perpendicular to AB (Def. 4) they are in one plane (Prop. III.). In that plane take any number of angles, PBP, P,BP, etc., each equal to PBM; and any number QBQ,

QBQ2, etc., each equal to MBQ; and, finally, draw the planes ABP1, ABP, etc., and ABQI, ABQ2, etc.

Then PBP, is the profile angle of the dihedral angle PBAP1, PBP2, of PBAP2, etc.; and as the profile angles were made equal, the dihedral angles are also equal (Prop. Ix.). Wherefore, whatever number of profile angles each equal to PBM there be in MBP,,* the same number of equal dihedral angles there will be in MBAP; that is, equimultiples MBP, and MBAP, have been taken of MBP and MBAP. In like manner, whatever multiple MBQm is of MBQ, the same multiple will MBAQ be of MBAQ.

Again, if the profile angle MBP, be greater than MBQm the dihedral angle MBAP, will be greater than MBAQm; if equal, equal; and if less, less.

Wherefore (Euc. v. Def. 5.)

PBAM: QBAM:: PBM : QBM.

SCHOLIUM. It is essential to the acquisition of facility in the application of these principles to give close attention to some special cases that arise.

*This notation is used to signify that there are n dihedral angles with their corresponding profile angles taken, without specifying any particular number. The same with respect to m in the other set of multiples.