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.

M

B'

A

E

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 CPD 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 Ρ 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;

(8.) Two equal, and each greater than the third;

(y) 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 (8.) 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: B

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

(1.) The sum of BAD and BAC is greater than BAD; and there fore greater also than the third angle

DAC.

Again, to prove that BAD, DAC are together greater than BAC, make BAE in the plane BAC equal to the angle BAD, and the line AE to AD; draw any line BC through E in the plane ABC, to meet AB, AC in B, C ; and join BD, CD.

Then BA, AD are equal to BA, AE and the angle BAD to the angle BAE; hence BD is equal to BE.

But BD and DC are together greater than BC; and hence CD is greater than CE. Whence DA, AC are equal to EA, AC, but the base DC greater than the base EC; and hence the angle DAC is greater than EAC; and consequently the two BAD and DAC together greater than BAE and EAC together, that is than BAC.

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

Again, it has been shown that EAC, the difference between BAC, and one of the other equal angles BAD, is less than DAC the third angle.

CASE (8.) Let all three angles BAD, DAC, CAB be unequal, and let BAC be the greatest of them.

(1.) Take the most unfavourable case to the truth of the proposition, viz., BAD and DAC compared with BAC. Make the angle BAE in the plane BAC equal to BAD, and AE equal to AD; and complete the construction as in the preceding

case.

B

E

Then by reasoning exactly similar to that in the preceding case, it is shown that BAD and DAC together are greater than BAE and EAC together; that is than BAC.

If either of the angles BAD, DAC be interchanged with BAC, the sum will be greater than before, and the third angle less; and hence, the sum of those two will be greater than the third angle.

(2.) By similar construction to the preceding, and corresponding reasoning, the difference EAC of the angles BAC and BAD is less than the third angle DAC. And this is equally the case whether BAD be the greater or the less of the two angles BAD, DAC.

PROPOSITION II.

If three points be taken in the edges of a trihedral angle equidistant from the vertex, and a plane be described through these points: and if also a perpendicular to this plane be drawn from the summit, it will be the centre of the circle which passes through the three points in the edges.

Let SA, SB, SC be the three edges of a trihedral angle, and these distances be all equal; let a plane MN be drawn through ABC,

and a perpendicular SD from S to MN: then D is the centre of the circle through A, B, C, in the

plane MN.

For join AD, BD, CD.

Then in the triangles SDA, SDB the two sides SD, SA are equal to the two SD, SB; and the opposite angles to SA, SB right angles; therefore the remaining sides and angles are equal, each to each. That is, DB is equal to DA. In like manner DC is equal to DA; and hence A, B, C are equidistant from D, or D is the centre of the circle ABC.

S

M

A

น

S

COR. 1. Each of the edges makes equal angles with the plane ABC: for these angles SAD, SBD, SCD are the corresponding angles of equal triangles.

COR. 2. Each of the edges makes equal angles with the perpendicular SD.

COR. 3. If AB, BC, CA be joined, and perpendiculars DG, DF, DE be drawn from D to each of them, these perpendiculars bisect those sides.

COR. 4. If SE, SF, SG be drawn, the angles SED, SFD, SGD will be the profile angles of the dihedral angles formed by the faces of the trihedral angle with the plane MN.

PROPOSITION III.

The plane angles of every salient polyhedral angle are together less than four right angles.

(1.) Let the angle be trihedral, and represented by the figures of the preceding proposition. Make the same construction with respect to A, B, C, the plane MN, perpendicular SD, etc.

Then E, F, G will fall between the extremities of the lines AC, CB, BA respectively. Whence (Prop. v., Chap. III.) ADC is greater than ASC, CDB than CSB, and BDA than BSA. Wherefore the three angles at S are together less than the three angles at D together.

A

But the three angles at D are together equal to four right angles ;* and hence the three angles at S are less than four right angles.

(2.) Let the angle at S be tetrahedral, having ASB, BSC, CSD, DSA for its four faces or plane angles: these

will together be less than four right angles.

For, produce two of the alternate faces ASB, CSD to meet in SE.

Then (Prop. 1.) the two angles ASE, DSE are together greater than the angle ASD of the trihedral angle S. EDA. Wherefore also, the two angles BSE, CSE are together greater than BSA, ASD, DSC; and hence again the three angles of the trihedral angle S. BEC at S are greater

than the four at S of the tetrahedral angle S. ABCD.

B

But the angles of S. BEC are less than four right angles by the preceding case; and hence the four angles at S of the tetrahedral angle S. ABCD are together less than four right angles.

GENERALLY. The pentrahedral angle is reduced in the same manner to the tetrahedral, and its plane angles shown to be less; and hence less than those of the tetrahedral angle, and thence again less than four right angles.

PROPOSITION IV.

If from any point within a trihedral angle, perpendiculars be drawn to the faces, these perpendiculars will be the edges of a new trihedral angle, which has the following relations to the original one :— (1.) Its edges will be perpendicular to the faces of the original; (2.) Its faces will be perpendicular to the edges of the original; (3.) Its plane angles will be the supplements of the opposite profile angles of the original; and

(4.) Its profile angles will be the supplements of the plane angles of the original.

(1.) This is but a repetition of the hypothesis, for the purpose of enumerating the connected properties seriatim.

(2.) Let S. ABC be the original trihedral angle; and from any points within it let perpendiculars

sa, sb, sc to the faces BSC, CSA, ASB be drawn: .the planes bsc, csa, asb will be perpendicular to the edges SA, SB, SC, respectively.

For let the plane bsc cut the planes BSA, ASC in c A, b A.

Then since sc is perpendicular to the plane BSA, the plane bsc through it is perpendicular to BSA (Prop. XVI., Chap. 11.). In like manner the plane bsc through bs is perpendicular to the plane ASC.

Wherefore the plane bsc, being

B

e..

*In Fig. 2, the angle CDA is equal to two right angles; and in Fig. 3, the

angle CDA is the reverse angle.