If two straight lines be cut by parallel planes, they shall be cut in the same ratio.

Let the straight lines AB and CD be cut by the parallel planes GH, KL, MN, at the points A, E, B, and C, F, D: AE shall be to EB as CF iš to FD.

Join AC, BD, AD; let

AD meet the plane KL at the point X; and join EX, XF.

Then, because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel; [XI. 16. and because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel. [XI. 16.

K

E

B

M

And, because EX is parallel to BD, a side of the triangle ABD,

therefore AE is to EB as AX is to XD.

[VI. 2.

Again, because XF is parallel to AC, a side of the triangle

ADC,

therefore AX is to XD as CF is to FD.

[VI. 2.

And it was shewn that AX is to XD as AE is to EB;

therefore AE is to EB as CF is to FD.

[V. 11.

Wherefore, if two straight lines &c. Q.E.D.

PROPOSITION 18. THEOREM.

If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane.

Let the straight line AB be at right angles to the plane CK: every plane which passes through AB shall be at right angles to the plane CK.

Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; [XI. 3.

take any point F in CE, from which draw FG, in the plane DE,

K

at right angles to

[I. 11.

CE.

CK,

Then, because AB is at right angles to the plane

[Hypothesis.

therefore it makes right angles with every straight line meeting it in that plane;

but CB meets it, and is in that plane; therefore the angle ABF is a right angle. But the angle GFB is also a right angle ; therefore FG is parallel to AB.

[XI. Definition 3.

[Construction. [I. 28. [Hypothesis.

And AB is at right angles to the plane CK; therefore FG is also at right angles to the same plane. [XI.8.

But one plane is at right angles to another plane, when the straight lines drawn in one of the planes at right angles to their common section, are also at right angles to the other plane; [XI. Definition 4.

and it has been shewn that any straight line FG drawn in the plane DE, at right angles to CE, the common section of the planes, is at right angles to the other plane CK; therefore the plane DE is at right angles to the plane CK.

In the same manner it may be shewn that any other plane which passes through AB is at right angles to the plane CK.

Wherefore, if a straight line &c. Q.E.D.

PROPOSITION 19. THEOREM.

If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane.

Let the two planes BA, BC be each of them perpendicular to a third plane, and let BD be the common section of the planes BA, BC: BD shall be perpendicular to the third plane.

For, if not, from the point D, draw in the plane BA, the straight line DE at right angles to AD, the common section of the plane BA with the third plane; [I. 11. and from the point D, draw in the plane BC, the straight line DF at right angles to CD, the common section of the plane BC with the third plane. [I. 11.

Then, because the plane BA is perpendicular to the third plane,

A

[Hypothesis.

and DE is drawn in the plane BA at right angles to AD their common section; [Construction. therefore DE is perpendicular to the third plane. [XI. Def. 4.

In the same manner it may be shewn that DF is perpendicular to the third plane.

Therefore from the point D two straight lines are at right angles to the third plane, on the same side of it; which is impossible.

[XI. 13. Therefore from the point D, there cannot be any straight line at right angles to the third plane, except BD the common section of the planes BA, BC;

therefore BD is perpendicular to the third plane. Wherefore, if two planes &c. Q.E.D.

PROPOSITION 20. THEOREM.

If a solid angle be contained by three plane angles, any two of them are together greater than the third.

Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB: any two of them shall be together greater than the third.

If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third.

If they are not all equal, let BAC be that angle which is not less than either of the other two, and is greater than one of them, BAD. At the point A, in the straight line BA, make, in the plane which passes through BA, AC, the angle BAE equal to the angle BAD; [I. 23.

make AE equal to AD;

[I. 3.

through E draw BEC cutting AB, AC at the points B, C; and join DB, DC.

Then, because AD is equal to AE,

[Construction

and AB is common to the two triangles BAD, BAE,

the two sides BA, AD are equal to the two sides BA, AE, each to each;

[Constr.

[I. 4.

and the angle BAD is equal to the angle BAE; therefore the base BD is equal to the base BE. And because BD, DC are together greater than

BC,

[I. 20.

and one of them BD has been shewn equal to BE a part of BC, therefore the other DC is greater than the remaining part EC.

And because AD is equal to AE,

[Construction.

and AC is common to the two triangles DAC, EAC, but the base DC is greater than the base EC; therefore the angle DAC is greater than the angle EAC.

[I. 25. And, by construction, the angle BAD is equal to the angle BAE;

therefore the angles BAD, DAC are together greater than the angles BAE, EAC, that is, than the angle BAC.

But the angle BAC is not less than either of the angles BAD, DAC;

therefore the angle BAC together with either of the other angles is greater than the third.

Wherefore, if a solid angle &c.

Q.E.D.

PROPOSITION 21. THEOREM.

Every solid angle is contained by plane angles, which are together less than four right angles.

First let the solid angle at A be contained by three plane angles BAC, CAD, DAB: these three shall be together less than four right angles.

In the straight lines AB, AC, AD take any points B, C, D, and join BC, CD, DB.

Then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of them are together greater than the third, [XI. 20. therefore the angles CBA, ABD are together greater than the angle DBC.

B

For the same reason, the angles BCA, ACD are together greater than the angle DCB,

and the angles CDA, ADB are together greater than the angle BDČ.

Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are together greater than the three angles DBC, DCB, BDC;

but the three angles DBC, DCB, BDC are together equal to two right angles. [I. 32. Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are together greater than two right angles.

And, because the three angles of each of the triangles ABC, ACD, ADB are together equal to two right angles,

[I. 32. therefore the nine angles of these triangles, namely, the angles CBA, BAC, ACB, ACD, CDA, CAD, ÄDB, DBA, DAB are equal to six right angles;

and of these, the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles,

therefore the remaining three angles BAC, CAD, DAB, which contain the solid angle at A, are together less than four right angles.