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angles of the rectilineal figure BCDEF be re-entrant; or, which is the same, if any of the planes forming the solid angle at A, being produced, pass through that angle. *
PROP. XXII. THEOR.
If every two of three plane angles be greater than the third, and if the straight lines which contain them be all equal ; a triangle may be made, having its sides equal, each to each, to the straight lines that join the extremities of those equal straight lines.
Let ABC, DEF, GHK be three plane angles, every two of which are greater than the third : then, if AB, BC, DE, EF, GH, HK be all equal, and if AC, DF, GK be joined ; a triangle may be made of AC, DF, GK; that is, every two of these lines are together greater than the third.
If the angles at B, E, H be equal, AC, DF, GK are also (1. 4.) equal; and any two of them greater than the third.
But if these angles be not all equal, let ABC be not less than either of the two at E, H; therefore (I. 4. and 24.) the straight line AC is not less than either of the other two DF, GK; and it is plain that AC, together with either of the other two, must be greater than the third. Also DF and GK are greater than AC. For, make the angle ABL equal to the angle H, and make BL equal to AB, or GH; and join AL, LC. Then, because AB, BL are equal to GH, HK, and the angle ABL to H, the base AL is equal to GK. And, because (hyp.) the angles E, H are greater than the angle ABC, of which H is equal to ABL, therefore the remaining angle E is greater than LBC. Then, because the two sides LB, BC are equal to the two DE, EF, but the angle DEF greater than LBC, DF is greater (I. 24.) than LC: and it has been proved that GK is equal to AL: therefore DF and GK are greater than AL and LC. But AL and LC are greater (I. 20.) than AC: much more then are DF and GK greater than AC. Wherefore any two of these straight lines AC, DF, GK are greater than the third ; and, therefore, a triangle may be made, (1. 22.) the sides of which will be equal to AC, DF, GK. If every two, therefore, &c.
The reason of this is plain from the note to the second corollary to the 32d proposition of the first book.
LEMMA. If a solid angle be contained by three planes, and if on the intersections of those planes, three equal lines be taken, terminated in the vertex of the solid angle; each of these lines is greater than the radius of the circle described about the triangle formed by joining their other extremities.
Let the solid angle at A be contained by the three plane angles, BAC, CAD, DAB, and let AB, AC, AD be taken all equal ; join BC, CD, DB; about the triangle BCD describe (IV.5.) a circle, and draw the radius EC: EC is less than AC.
Let BC, CD, two sides of the triangle BCD, which do not pass through the centre E, be bisected in G and F, and join EA, ED, EF, EG, AF, AG. Then, since the triangles CAD, CED (hyp. and I. def. 30.) are isosceles, and since AF, EF are drawn from their vertices bisecting their common base CD, AFC, EFC are (III. 3. cor. 1.) right angles ; and therefore (XI. 4.) CF is perpendicular to the plane AFE. But the plane BCD passes through CF, therefore (XI. 18.) the plane BCD is perpendicular to AEF; and consequently the plane AEF is perpendicular to BCD. It would be proved in the same manner, that the plane AEG is also perpendicular to the plane BCD. Now AE is the common section of the planes AEF, AEG; and since each of these planes is perpendicular to the plane BCD, AE is also (XI. 19.) perpendicular to BCD; and therefore (XI. def. 1.) AEC is a right angle, and (I. 19.) AC is greater than CE. Therefore, if a solid
angle, &c. Cor. Hence, if there be three plane angles B, E, H, (see the figures for the next proposition,) which are such that a solid angle may be contained by them, and if BA, BC, ED, EF, HG, HK be taken all equal, and AC, DF, GK be drawn; AB is greater than the radius of the circle described about a triangle having its three sides equal to AC, DF, GK, each to each.
PROP. XXIII. PROB. To make a solid angle having the angles containing it, equal to three given plane angles, any two of which are greater than the third, and all three together less than four right angles.
Let B, E, H be given plane angles, any two of which are greater than the third, and all of them together less than four
right angles. It is required to make a solid angle contained by plane angles equal to B, E, H, each to each.
From the lines containing the angles, cut off BA, BC, ED, EF, HG, HK, all equal to one another; and join AC, DF, GK: then (XI. 22.) a triangle may be made of three straight lines
equal to AC, DF, GK. Let this (I. 22.) be the triangle LMN, AC being equal to LM, DF to MN, and GK to LN.
About LMN describe (IV. 5.) a circle, and draw the radii, LO, MO, NO; draw also OP (XI. 12.) perpendicular to the plane LMN. Then, by the corollary to the lemma, any of the radii LO, MO, NO is less than AB. Find (I. 47. cor. 3.) the side of a square equal to the difference of the squares of AB and LO; make OP equal to that side ; and join PL, PM, PN: the plane angles LPM, MPN, and NPL form the solid angle required.
For, since OP is (const.) perpendicular to the plane LMN, the angles, LOP, MOP, NOP, are (XI. def. 1.) right angles : and therefore, since, in the triangles OLP, OMP, ONP, the sides OL, OM, ON are equal, OP common, and the contained angles equal, the bases LP, MP, NP are (I. 4.) all equal. Also (const.) the square of AB is equal to the squares of LO, OP; and (I. 47.)
square of LP is also equal to the squares of LO, OP, because LOP is a right angle. Therefore (I. ax. 1.) the square of AB is equal to the square of PL, and (I. 46. cor. 3.) AB to PL: and hence all the straight lines LP, MP, NP, BA, BC, ED, &c., are equal. Then, in the triangles LPM, ABC, the sides AB, BC are equal to LP, PM, each to each; and (const.) AC is equal to LM: therefore (1. 8.) the angles ABC, LPM are equal: and it would be shown in a similar manner, that the angle E is equal to MPN, and H to NPL. The solid angle at P, therefore, being contained by three plane angles, which are equal to the three given angles B, E, H, each to each, is such as was required.
PROP. A. THEOR. If two solid angles be each contained by three plane angles, equal to one another, each to each : the planes in which the equal angles are, have the same inclination.
Let there be two solid angles at A and B; and let the angle at A be contained by the three plane angles CAD, CAE, EAD; and
the angle at B by the three FBG, FBH, HBG, which are equal to the other three, each to each : the planes in which the equal angles are, have the same inclination to one another.
In AC take any point K, and in the plane CAD, from K draw KD at right angles to AC; and in the plane CAE draw KL at right angles to the same AC: therefore (XI. def. 2.) the angle DKL is the inclination of the planes CAD, CAE. In BF take BM equal to AK, and from M draw in the planes FBG, FBH, MG, MN perpendicular to BF; therefore the angle GMN is the inclination of the planes FBG, FBH. Join LD, NG; and, because (hyp.) in the triangles KAD, MBG, the angles KAD, MBG are equal, as also the right angles AKD, BMG, and that the sides AK, BM, adjacent to the equal angles, are equal to one another; therefore (I. 26.) KD is equal to MG, and AD to BG. For the same reason, in the triangles KAL, MBN, KL is equal to MN, and AL to BN: and in the triangles LAD, NBG, LA, AD are equal to NB, BG, and (hyp.) they contain equal angles; therefore (I. 4.) LD is equal to NG. Lastly, in the triangles KLD, MNĞ, the sides DK, KL are equal to GM, MN, and LD to NG; therefore (I. 8.) the angle DKL is equal to GMN. But DKL is the inclination of the planes CAD, CAE, and GMN is the inclination of the planes FBG, FBH, which planes (XI. def. 2.) have therefore the same inclination. In the same manner also, it may be demonstrated, that the other planes in which the equal angles are, bave the same inclination to one another. Therefore, if two solid angles, &c.
PROP. B. THEOR. Two solid angles contained, each by three plane angles which are equal to one another, each to each, and alike situated, are equal to one another.
Let there be two solid angles at A and B, of which the solid angle at A is contained by the three plane angles CAD, CAE, EAD; and that at B, by the three plane angles FBG, FBH, HBG, which are equal to the other three, each to each : the solid angles at A and B are equal.
Let the solid angle at A be applied to the solid angle at B; and, first, the plane angle CAD being applied to FBG, 80 that the point A may coincide with B, and AC with BF; then AD coincides with BG, because the angle CAD is equal to FBG. And because (XI. A.) the inclination of the planes CAE, CAD is equal to the inclination of the planes FBH, FBG, the plane CAE coincides with FBH, because the planes
CAD, FBG coincide with one another. Also, because AC, BF coincide, and the angle CAE is equal to FBH; therefore AE coincides with BH, and AD with BG; wherefore the plane EAD coincides with HBG. Therefore the solid angle A coincides with the solid angle B; and consequently (I. ax. 8.) they are equal to one another. Two solid angles, therefore, &c.
Cor. If AE, BH be taken equal to each other; and EK be drawn perpendicular to the plane CAD, and HL to FBG; and if AK, BL be joined, EK is equal to HL, and the angle EAK to HBL. For, since the solid angles coincide, the points E, H coincide, because AE, BH are equal : also, since the planes CAD, FBG coincide, the perpendiculars EK, HL (XI. 13.) coincide, and are equal : and since AE, AK coincide with BH, BL, the angles EAK, HBL are equal. *
PROP. C. THEOR.
Solid figures contained by the same number of equal and similar planes, alike situated, and having none of their solid angles contained by more than three plane angles, are equal and similar to one another.
Let AG, KA be two solids contained by the same number of similar and equal planes, alike situated; viz., let the plane AC be similar and equal to KM; AF to KP; BG to LQ; GD to QN; DE to NO; and FH to PR: AG is equal and similar to KQ.
Because the plane angles BAD, BAE, EAD are (hyp.) equal to the plane angles LKN, LKO, OKN, each to each ; therefore (XI. B.) the solid angles at A and K are equal. In the same manner, the other solid angles of the figures are proved to be equal. If then AG be applied to KQ, first, the plane figure AC being applied to KM; AB coincid. ing with KL, AC must coincide with KM, because they are equal and similar: therefore ihe straight lines AD, DC, CB coincide with KN, NM, ML, each with each ; and the points A, D, C, B, with K, N, M, L. Also the solid angles at A and K coincide: wherefore the plane AF coincides with KP, and the figure AF with KP, because they are equal and similar. Therefore the straight lines AE, EF, FB, coincide with KO, OP, PL ; and the points E, F, with O, P. In the same manner, the figure AH coincides with KR, the straight line
This corollary is transferred to this proposition from the 35th of this book, for the purpose of facilitating the demonstration of the latter.