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For draw the perpendicular BH in the profile plane to the trace GC; and let D be any other point in the circle; and join BD, DH.

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Then (Euc. 111. 7, 8) HC is less than HD, and HD less than HG; and hence (Prop. xi., Chap. 11.) BD si greater than BC, and BG than BD. In like manner, BG is greater than BE: that is, BC is the least and BG the greatest line that can be drawn from B to the circle.

(2.) Let F be more remote from C than E is: then BF will be greater than BE.

For join HF.

Then (Euc. 111.7, 8), FH is greater than EH; and hence (Prop. xi., Chap. II.) BF is greater than BE.

(3.) Make CD equal to CE; then BD will be equal to BE.

For since HD is equal to HE (Euc. 111. 7, 8), BD is equal to BE (Prop. xi., Chap. 11.). .

(4.) The less line BE makes a greater angle with the plane MN than the greater BF does.

For since BF is greater than BE, HF is greater than HE (Prop. xi., Chap. 11.). Make HK equal to HE and join BK.

Then (Prop. XI., Chap. 11.) BK is equal to BE, and the angle BKH equal to BEH. But BŘH is exterior to BFK, and therefore greater than BFK or BFH. Wherefore, also, BEH is greater than BFH. And these are the profile angles of the lines BE, BF since the planes BFH, BEH are drawn through the common perpendicular BH from the common pointB, and are therefore themselves perpendicular to the plane MN.

SCHOLIUM. This proposition might have been enunciated with respect to the edges of a cone whose vertex is B and base the circle GDC.

PROPOSITION IV. If a line be inclined to a plane, and a plane revolve about the line to

meet the plane, and make dihedral angles with it: then, (1.) That formed when the trace of the revolving plane is perpendicu

lar to the line itself, is the least ; (2.) As the plane revolves from this position towards coincidence

(either way) with the profile plane, the dihedral angles increase in

magnitude ; (3.) Only two positions of the revolving plane, one on each side of the

profile plane, can make equal dihedral angles with the given plane.

Let AB be inclined to MN, and let AC be the profile trace of AB (Prop. 1., Schol. 2); and HG perpendicular to it, that is (Prop. xii.,

VOL. II.

Chap. 11.) perpendicular to the line AB itself, and let BAE be any other plane through AB: then

(1.) The dihedral angle formed by the plane BAGH with MN is less than any other as BAE.

For, from B draw in the profile plane of AB the line BC perpendicular to AC, and CE perpendicular to AE; and join BE.

Then since CEA is a right angle, CAE is less than a right angle; and hence CA is greater than CE. Wherefore also the angle BEC is greater than BAC (Prop. III.): that is, the angle BAC is the least that can be made under the conditions specified.

But since BC is perpendicular to MN and CA, CE perpendicular to lines GH, AE in MN, the lines BA, BE are also perpendicular to these lines (Prop. XIll., Chap. 11). Whence BAC, BEC are the profile angles of the dihedral angles made by the planes BAGH and BAE with the plane MN; and hence the former dihedral angle is less than the latter.

(2.) Let the plane BAF be nearer to the profile plane BAC than the plane BAE is : then the dihedral angle made by BAF with MN will be greater than that made by BAE.

For draw CE, CF perpendicular to AE, AF; and join BE, BF.

Then E, F are in the circumference of a circle whose diameter is AC (Euc. III, 31); and by hypothesis AF is nearer to AC than AE is. Whence AF is greater than AE; and consequently CF is less than CE (Euc. 11. 15). Wherefore, again, the angle BEC is less than BFC. Also, as in the preceding case, these are the profile angles of the dihedral angles made by BAE, BAF with MN. The conclusion therefore follows.

(3.) Let the planes BAD, BAE be such that their traces AD, AE with MN make equal angles with AC: then the planes BAD, BAE inake equal dihedral angles with MN.

For draw CD, CE perpendicular to AD, AE; and join BD, BE.

Then since the angles CDA and CEA are right angles, they are in the circle on AC; and since AD, AE make equal angles with the diameter AC (Constr.) they are equal. Whence CD is equal to CE; and hence again the angle BDC is equal to BEC.

Also, as before, these are the angles of the dihedral angles enunciated, and the conclusion follows, that two planes can be drawn through AB, which shall make equal dihedral angles with MN.

Nor can there be a third plane drawn through AB which shall make a dihedral angle with MN equal to either of these.

For any third plane must be either nearer to the profile plane, and therefore make a greater dihedral angle with MN, or more remote, and therefore make a less, by the preceding case.

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PROPOSITION V. Two points are given in a plane, and likewise a line perpendicular to the plane : assign the point in the perpendicular to which lines drawn from the given points shall contain the greatest angle.

[This is proposed in the form of a problem merely for the sake of brevity in the general enunciation. The conditions will be seen in the several cases of the specific enunciations to be more easily expressible by reference to the figures.]

(1.) Let AB be perpendicular to the plane MN, and C, D two given points in MN; and let the perpendicular AE from A upon the line joining CD meet CD in a point E not beyond the extremities C and D of the line CD: then the angle DAC will be greater than any other angle DBC having its vertex in AB.

For join BE, and then BE is greater than A E, since BAE is a right angle. Make EB' in EA produced equal to EB, and join B'D, B'C.

Then since B'E, ED are equal to BE, ED each to each, the angle B'ED equal to BED (both being right angles, Constr. and Prop. xIII., Chap. 11.); the base B'D is equal to the base BD. And in the same manner it is proved that B'C is equal BC.

Whence since CB', B'D are equal to CB, BD and the base CD common, the angle CB'D is equal to CBD (Euc. 1. 8).

But A is within the triangle CB'D; and hence the angle CAD is greater than CB'D; and therefore, also, greater than CBD. That is CAD is the greatest angle that can be formed under the conditions.

In nearly the same way it may be shown that the angles continually diminish as the point B recedes from A in either direction of the line AB from A.

If, also, E had fallen at C or D, the same conclusion is deducible by a slight modification of the reasoning.

(2.) The same general conditions as before, except that E falls in the prolongation of CD.

Let AE be the perpendicular on CD produced; produce AE indefinitely, and describe a circle through C, D to touch AE in P'; in AB take a point P, so that EP is equal to EP': then the angle CPD is greater than any other angle CBD described under the conditions.

For, make EB' in EA (produced if necessary) equal to EB; and join B'C, B'D.

Then, as before, the angle PED is a right angle (Prop. xii., Chap. II.), and therefore equal to P'ED. Whence the two sides and included

angle of PED are equal to the two sides and included angle of P'ED; and therefore PD is equal to P'D. In like manner PC is equal to P'C; and therefore (Euc. 1. 8) the angle CPD is equal to CP'D.

By the same mode of reasoning it is proved that the angle CBD is equal to CB'D.

Now EB' is a tangent to the circle CDP' and hence every other point B' than P' lies without the circle : whence B'D cuts the circle in some point H. Join HC.

Then the angles CP'D, CHD are equal, being in the same segment; and CHD is greater than CB'D, being the exterior angle of the triangle CHB'. Whence CP'D is greater than CB'D: that is, CPD is greater than CBD.

The same reasoning applies if B be taken between A and P. Whence the angle CPĎ thus formed is the greatest possible under the conditions on the side of the plane towards B.

There is a point p, however, on the other side of the plane MN in AB, such that CpD is equal to CPD; viz., at the same distance from A, or such that Ap is equal to AP.

From this we easily learn that the angles formed at points above P or below p continually diminish as their vertices recede from those points respectively upwards and downwards.

Also, that as the vertices approach to A from P and p, they also continually diminish to the limit CAD, which is the least angle that can be formed between P and p.

And, lastly, if the point of contact P should fall at A, then CAD, instead of being a minimum limit, will become the greatest angle.

SCHOLIUM. This proposition is of great importance in orthographic projection, which will be seen under that head. The first part also is the foundation of some of the propositions immediately following this. The proposition itself was first distinctly discussed in the “ Cambridge Mathematical Journal,” November, 1847.

CHAPTER IV.
TRIHEDRAL AND POLYHEDRAL ANGLES.

PROPOSITION I. Every two plane angles of a trihedral angle are together greater than

the third, and their difference is less than the third angle. The conditions that may subsist amongst the three angles are the following four:

(a.) All three angles equal;
(B.) Two equal, and each greater than the third ;
(v.) Two equal, and each less than the third; and

(8.) All unequal. CASE (a.) Let the three plane angles BAC, CAD, DAB of the trihedral angle at A be all equal: then

(1.) Any two of them BAD, DAC together are double of any one of them BAC, and therefore greater than BAC.

(2.) The difference between any two, as BAD, DAC is nothing; and therefore less than the third angle BAC.

CASE (B.) Let DAB, DAC be equal to B/ one another, and each greater than the third angle BAC: then

(1.) The sum of the two BAD, DAC is double of one of them; and each of these is greater than BAC: hence they are both together greater than BAC.

Again, the angles DAC and CAB together are greater than DAC, and hence greater than BAD.

(2.) The angles DAB, DAC being equal, their difference is nothing; and hence less than the third angle BAC.

Again, since DAC is greater than CAB, make CAH in the plane CAD equal to CAB: BK then the angle DAH is the difference between DAC and CAB. This is less than DAC; and therefore less than the third angle BAD.

CASE (.) Let two angles BAD, DAC be equal, and each less than the third BAC: then

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