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If two plains ( AB and CB), intersecting each other, be perpendi

cular to a third (ADC); their common section (DB) is perpendicular to the same plain.

If DB be not perpendicular to the plain?
ADC, draw from the point D in the plain
AB, the right line DE perpendicular to AD

F (11. 1 Eu.); and in the plain CB, from the point D, draw DF perpendicular to CD (by the sume); and because the plain AB is perpendicular to the plain ADC, and DE is drawn in the plain AB, perpendicular to AD A their common section, DE is perpendicular - to the plain ADC (Def. 4. 2 Sup). In like manner DF may be proved to be perpendicular to the plain ADC. Therefore two right lines DE and DF are at right angles to the same plain ADC, on the same side of it, at the same point D, which is absurd (8. 2. Sup.). Therefore there cannot be any right line perpendicular to the plain ADC at the point D, except DB; therefore DB is perpendicular to the plain ADC.


Two right lines (AB and CD), parallel to the same right line

(EF), which is not in the same plain with them, are parallel to each other.

From any point G in EF, let two perpendiculars GH and GK to EF


B be drawn (11. 1 Eu.), one in the plain passing by EF and AB, and

G the other in that passing by EF and

E CD, meeting AB and CD in H and K; and EF being perpendicular to Gü and GK (Constr.), is perpendicular to the plain passing by them (2. 2 Sup.) ; whence AB and CD being parallel to EF, are perpendicular to the same plain (5.2 Sur.), and therefore parallel to each other (4. 2. Sup).


If two right lines (AB and AC), meeting each other (as in 1, be

parallel to two others (DE and DF), likewise meeting each other (as in D), though not in the same plain with them; the first two and the last two contain equal angles.


Let AB and DE be taken equal to each other, and also AC and DF, and let BC, EF, BE, AD and CF be drawn. And since AB and DE are equal and parallel, BE and AD are equal and parallel (33. 1 Eu.); in like manner CF may be shewn to be equal and parallel to AD; therefore BE and CF are equal and parallel to each other (Ax. 1. 1 and 30. 1 Eu.), and therefore BC and EF are equal (33. 1 Eu.) ; whence the triangles ABC and DFF being mutuaiij quilateral, the angles BAC and Eur are equal (8. 1 Eu).




Plains (AB and CD), to which the same right line (EF) is per

pendicular, are parallel.


For if not, let them, being produced, meet, and let their common section be the right line GH, to any point wherein as K, having drawn EK and FK, the angles KEF and KFE are each of them right angles Hype and Def. 3. 2 Sup.), and so two angles of the triangle KEF are equal to two right angles, which is absurd (17. 1 Eu.); therefore the plains AB and CD being produced, do not meet each other, and are therefore parallel (Def. 7. 2 Sup).

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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.


From A, let fall the perpendicular AG on the plain EF (6. 2 Súp.), and from the point G, wherein it meets the same, A draw GH parallel to "DE ·and GK to


K к 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

B right angle (Constr. and Def. 3. 2 Sun.), GAB is a right angle (29. i Eu.); in like manner miglit 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 bow a plain may be found, passins 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 Dr parallel to AB and AC (31. I Eu). The plain passing by the right lives DE and DF (Past. to B. 2 Sup.), is parallel to the girea plain BC (14. 2 Sup.), as was required to be found,


If two parallel plains ( AB and CD) be cut by a third (EF); their

common sections (EG and HF) are parallel.

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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 FA (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).

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Parallel plains (as AB, CD and EF), cut right lines fas GH

and KL) proportionally.


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Let the right line GH meet the par

T allel plains in M, N and 0, and KL the

NI same plains in P, Q and R; MN is to NO, as PQ is to QR.

Join MP and OR, also MR meeting the plain CD in S, and join SN and SQ; and because the parallel plains CD and EF are cut by the plain MRO, the common sections NS and OR are parallel (15,2 Sup.); in like manner, since the parallel plains AB and Cl) are cut by the plain MPR, the common sections MP and SQ are parallel (by the scime): 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).


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. If the angles BAC, CAD and DAB be

A equal, it is evident that any two of them are greater than the third ; if not, let BAC be

B 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. i 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, heing not less than either of the other two, is with either of them, greater than the other.

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