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PROP. XIV. THEOR.

If two right lines (AB and AC), meeting each other, be parallel to two others (DE and DF), which meet each other, but are not in the same plain with the former two; the plain (BC), passing by the former two, is parallel to that (EF), passing by the others.

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G

K

B

H

'From A, let fall the perpendicular AG on the plain EF (6. 2 Sup.), and from the point G, wherein it meets the same, draw GH parallel to DE and GK to DF (31. 1 Eu.); and since AB and GH are each of them parallel to DE, they are parallel to each other (11. 2 Sup.), and therefore the angle AGH being a right angle (Constr. and Def. 3. 2 Sup.), GAB is a right angle (29. 1 Eu.); in like manner might GAC be shewn to be a right angle; therefore GA being perpendicular to AB and AC, is perpendicular to the plain BC passing by them (2. 2 Sup.); whence, the right line AG being likewise perpendicular to the plain EF (Constr.), the plains BC and EF are parallel (13.2 Sup).

Cor. Hence it appears how a plain may be found, passing through any given point, as D, parallel to a given plain BC; namely, by drawing in the given plain two right lines AB and AC, from any point therein A, and drawing from the given point D, two right lines DE and DF parallel to AB and AC (31. 1 Eu). The plain passing by the right lines DE and DF (Post. to B. 2 Sup.), is parallel to the given plain BC (14. 2 Sup.), as was required to be found.

PROP. XV. THEOR.

If two parallel plains (AB and CD) be cut by a third (EF); their common sections (EG and HF) are parallel.

For if not, let them, being produced, meet, as in K, and since the plains AB and CD, being produced, coincide with these right lines GE and FH (Def. 4 and 7. 1 Eu.), these plains being produced meet also in K, which is absurd (Hyp. and Def. 7. 2 Sup.); therefore the common sections GE and FH being produced towards E and H do not meet; in like manner it may be shewn, that they do not meet towards G and F, therefore they do not meet being produced either way, and are therefore parallel (Def. 34. 1 Eu).

PROP. XVI. THEOR.

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Parallel plains (as AB, CD and EF), cut right lines (as GH and KL) proportionally.

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common sections NS and OR are parallel (15,2 Sup.); in like manner, since the parallel plains AB and CI) are cut by the plain MPR, the common sections MP and SQ are parallel (by the same): whence, in the triangle MOR, the right line NS being parallel to OR, MN is to NO, as MS to SR (2. 6 Eu.) ; in like manner, in the triangle MPR, SQ being parallel to MP, MS is to SR, as PQ to QR (by the same); whence, the ratios of MN to NO, and of PQ to QR, being each equal to that of MS to SR, are equal to each other (11. 5 Eu).

PROP. XVII. THEOR.

Of three plain angles forming a solid angle, any two whatever are greater than the third.

Let the solid angle A be formed by three plain angles BAC, CAD and DAB. Any two of them are greater than the third.

B

D

A

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If the angles BAC, CAD and DAB be equal, it is evident that any two of them are greater than the third; if not, let BAC be that angle which is not less than either of the other two, and is greater than one of them BAD, and from the angle BAC take BAE equal to BAD (23. 1 Eu.), take AD and AE equal to each other, through E draw the right line BEC meeting AB and AC in B and C, and draw DB and DC; and because in the triangles BAD and BAE, the side AD is equal to AE (Constr.), AB common, and the included angles BAD and BAE equal (Constr.), BD is equal to BE (4. 1 Eu.); but BD and DC together are greater than BC (20. 1 Eu.), taking from each the equals BD and BE, there remains DC greater than EC (Ax. 5. 1 Eu.); whence, AD and AC being severally equal to AE and AC, the angle DAC is greater than the angle EAC (25. 1 Eu.), to which adding the equal angles BAD and BEA, the angles BAD and DAC together, are greater than BAE and EAC together, or than the whole angle BAC (Ax. 4. 1 Eu.) ; and the angle BAC, being not less than either of the other two, is with either of them, greater than the other.

PROP. XVIII. THEOR.

The plain angles, which constitute any solid angle, are together less than four right angles.

Let A be a solid angle, contained by any number of plain angles BAC, CAD, DAE, EAF and FAB, these are together less than four right angles.

B

E

Let the plains which contain the solid angle A be cut by another plain BCDEF; and of the three plain angles which contain the solid angle at B, the angles ABF and ABC are together greater than the third CBF (17.2 Sup.); for the same reason, of the three plain angles which contain each of the solid angles at C, D, E and F, the two which are at the bases of the triangles having their common vertex at A, are together greater than the third, which is one of the angles of the figure BCDEF; therefore all the angles at the bases of the triangles having their common vertex at A are together greater than all the angles of the figure BCDEF; but all the angles of the figure BCDEF are equal to twice as many right angles, except four, as the figure has sides (Cor. 1. 32. 1 Eu.), therefore all the angles at the bases of the triangles having their common vertex at A are greater than twice as many right angles, except four, as the figure has sides; and all the angles of these triangles are equal to twice as many right angles as the figure has sides (32. 1 Eu.), therefore the angles of these triangles which are at their common vertex A, being those which contain the solid angle A, are less than four right angles:

Cor. From this proposition it follows, that there can be no more than five solids contained by equilateral and equiangular plain figures, or as they are usually called, regular solids, namely, three contained by equilateral triangles, one by squares, and one by regular pentagons.

For a solid angle cannot be contained by two plain angles, three at least are required.

And since the three angles of an equilateral triangle are equal to two right angles (32. 1 Eu.), six such angles are equal to four right angles, and therefore cannot constitute a solid angle (by this prop.); and since six angles of an equilateral triangle are equal to four right angles, three, four, or five such angles are

less than four right angles, and can therefore constitute a solid angle, as is manifest from this proposition; but three such angles form the angle of a tetrahedron or equilateral pyramid, see Def. 31. 2 Sup. ; four such angles form the angle of an octohedron, see Def. 33. 2 Sup. ; and five such angles form the angle of an icosihedron, see Def. 35. 2 Sup.

Three angles of a square form the angle of a cube or hexahedron, see Def. 32. 2 Sup. ; four such angles are equal to four right angles, and therefore cannot constitute a solid angle.

And since the five angles of a regular pentagon are equal to six right angles (Cor. 1. 32. 1 Eu.), any one of its angles is equal to a right angle and a fifth of a right angle, and therefore three such angles are equal to three right angles and three fifths, and three such angles form the angle of a dodecahedron, see Def. 34. 2. Sup. ; but four such angles are equal to four right angles and four fifths, and therefore cannot form a solid angle (by this prop).

And since six angles of an equilateral triangle, or four angles of a square, are equal to four right angles, and four angles of a regular pentagon are greater than four right angles, therefore more than six angles of an equilateral triangle, or than four of a square, or than four of a regular pentagon, are greater than four right angles, and therefore cannot constitute a solid angle (by this prop).

And since the six angles of a regular hexagon are equal to eight right angles (Cor. 1. 32. 1 Eu.), three such angles are equal to four right angles, and therefore cannot constitute a solid angle (by this prop); neither therefore can any greater number.

And since three angles of a regular hexagon are equal to four right angles, three angles of a regular heptagon, or of any regular polygon of more than six sides, are greater than four angles, as also easily follows from Cor. 1. 32. 1 Eu., therefore all regular polygons of more than five sides are incapable of forming a solid angle, and therefore there can be no more regular solids than the five mentioned in this corollary.

Schol. It is manifestly supposed in this proposition, that when the solid angle is contained by more than three plain angles, any of the legs of the plain angles which form the solid angle, as AC, falls without the plain passing by the two adjacent legs AB and AD.

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