# The Elements of Euclid for the Use of Schools and Colleges/Book XI

*BOOK XI*.

DEFINITIONS.

1. A solid is that which has length, breadth, and thickness.

2. That which bounds a solid is a superficies.

3. A straight line is perpendicular, or at right angles, to a plane, when it makes right angles with every straight line meeting it in that plane.

4. A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane.

5. The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point at which the first line meets the plane to the point at which a perpendicular to the plane drawn from any point of the first line above the plane, meets the same plane.

6. The inclination of a plane to a plane is the acute angle contained by two straight lines drawn from any the same point of their common section at right angles to it, one in one plane, and the other in the other plane.

7. Two planes are said to have the same or a like inclination to one another, which two other planes have, when the said angles of inclination are equal to one another.

8. Parallel planes are such as do not meet one another though produced. 9. A solid angle is that which is made by more than two plane angles, which are not in the same plane, meeting at one point.

10. Equal and similar solid figures are such as are contained by similar planes equal in number and magnitude. [*See the Notes*.]

11. Similar solid figures are such as have all their solid angles equal, each to each, and are contained by the same number of similar planes.

12. A pyramid is a solid figure contained by planes which are constructed between one plane and one point above it at which they meet.

13. A prism is a solid figure contained by plane figures, of which two that are opposite are equal, similar, and parallel to one another; and the others are parallelograms.

14. A sphere is a solid figure described by the revolution of a semicircle about its diameter, which remains fixed.

15. The axis of a sphere is the fixed straight line about which the semicircle revolves.

16. The centre of a sphere is the same with that of the semicircle.

17. The diameter of a sphere is any straight line which passes through the centre, and is terminated both ways by the superficies of the sphere.

18. A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed.

If the fixed side be equal to the other side containing the right angle, the cone is called a right-angled cone; if it be less than the other side, an obtuse-angled cone; and if greater, an acute-angled cone.

19. The axis of a cone is the fixed straight line about which the triangle revolves. 20. The base of a cone is the circle described by that side containing the right angle which revolves.

21. A cylinder is a solid figure described by the revolution of a right-angled parallelogram about one of its sides which remains fixed.

22. The axis of a cylinder is the fixed straight line about which the parallelogram revolves.

23. The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram.

24. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals.

25. A cube is a solid figure contained by six equal squares.

26. A tetrahedron is a solid figure contained by four equal and equilateral triangles.

27. An octahedron is a solid figure contained by eight equal and equilateral triangles.

28. A dodecahedron is a solid figure contained by twelve equal pentagons which are equilateral and equiangular.

29. An icosahedron is a solid figure contained by twenty equal and equilateral triangles.

A. A parallelepiped is a solid figure contained by six quadrilateral figures, of which every opposite two arc parallel.

*THEOREM*.

*One part of a straight line cannot he in a plane, and another other part without it*.

If it be possible, let *AB*, part of the straight line *ABC*, be in a plane, and the part BC without it.

Then since the straight line *AB* is in the plane, it can be produced in that plane; let it be produced to *D*; and let any plane pass through the straight line,and be turned about until it pass through the point *C*.

Then, because the points *B* and *C* are in this plane, the straight line *BC* is in it. [I. *Definition* 7.

Therefore there are two straight lines *ABC*, *ABD* in the same plane, that have a common segment *AB*;

but this is impossible. [I. 11, *Corollary*.

Wherefore, *one part of a straight line* &c. q.e.d.

*THEOREM*.

*Two straight lines which cut one another are in one plane; and three straight lines which meet one another are in one plane*.

Let the two straight lines *AB*, *CD* cut one another at *E*: *AB* and *CD* shall be in one plane; and the three straight lines *EC*, *CB*, *BE* which meet one another, shall be in one plane.

Let any plane pass through the straight line *EB*, and let the plane be turned about *EB*, produced if necessary, until it pass through the point *C*.

Then, because the points *E* and *C* are in this plane, the straight line *EC* is in it; [I. *Definition* 7.

for the same reason, the straight line *BC* is in the same plane;

and, by hypothesis, *EB* is in it.

Therefore the three straight lines *EC*, *CB*, *BE* are in one plane.

But *AB* and *CD* are in the plane in which *EB* and *EC* are; [XI. 1.

therefore *AB* and *CD* are in one plane.

*two straight lines*&c. q.e.d.

*THEOREM*.

*If two planes cut one another their common section is a straight line*.

Let two planes *AB*, *BC* cut one another, and let *BD* be their common section: *BD* shall be a straight line.

If it be not, from *B* to *D*, draw in the plane *AB* the straight line *BED*, and in the plane *BC* the straight line *BFD*. [Postulate 1.

Then the two straight lines *BED*, *BFD* have the same extremities, and therefore include a space between them;

but this is impossible. [*Axiom* 10.

Therefore *BD*, the common section of the planes *AB* and *BC* cannot but be a straight line.

Wherefore, *if two planes* &c. q.e.d.

*THEOREM*.

*If a straight line stand at right angles to each of two straight lines at the point of their intersection, it shall also he at right angles to the plane which passes through them, that is, to the plane in which they are.*

Let the straight line *EF* stand at right angles to each of the straight lines *AB*, *CD*, at *E*, the point of their intersection: *EF* shall also be at right angles to the plane passing through *AB*, *CD*.

Take the straight lines *AE*, *EB*, *CE*, *ED*, all equal to one another; join *AD*, *CB*; through *E* draw in the plane in which are *AB*, *CD*, any straight line cutting *AD* at *G*, and *CB* at *H*: and from any point *F* in *EF* draw *FA*, *FG*, *FD*, *FC*, *FH*, *FB*. Then, because the two sides *AE*, *ED* are equal to the two sides *BE*,*EC*, each to each, [*Construction*.

and that they contain equal angles *AED*, *BEC*; [I. 15.

therefore the base *AD* is equal to the base *BC*, and the angle *DAE* is equal to the angle *EBC*. [I. 4.

And the angle *AEG* is equal to the angle *BEH*; [I. 15.

therefore the triangles *AEG*, *BEH* have two angles of the one equal to two angles of the other, each to each;

and the sides *EA*, *EB* adjacent to the equal angles are equal to one another; [*Construction*.

therefore *EG* is equal to *EH*, and *AG* is equal to *BH*. [I. 26.

And because *EA* is equal to *EB*, [Construction.

and *EF* is common and at right angles to them, [*Hypothesis*.

therefore the base *AF* is equal to the base *BF*. [I. 4.

For the same reason *CF* is equal to *DF*.

And since it has been shewn that the two sides *DA*, *AF* are equal to the two sides *CB*, *BF*, each to each,

and that the base *DF* is equal to the base *CF*;

therefore the angle *DAF* is equal to the angle *CBF*. [I. 8.

Again, since it has been shewn that the two sides *FA*, *AG* are equal to the two sides *FB*, *BH*, each to each,

and that the angle *FAG* is equal to the angle FBH*;*
therefore the base

*FG*is equal to the base

*FH*. [I. 4.

Lastly, since it has been shewn that *GE* is equal to *HE*, and *EF* is common to the two triangles *FEG*, *FEH*;

and the base *FG* has been shewn equal to the base FH;

therefore the angle *FEG* is equal to the angle *FEH*. [I. 8.

Therefore each of these angles is a right angle. [I, Defn. 10.

In like manner it may be shewn that *EF* makes right angles with every straight line which meets it in the plane passing through *AB*, *CD*.

Therefore *EF* is at right angles to the plane in which are *AB*, *CD*. [XI. Definition 3.

*if a straight line*&c. q.e.d.

*THEOREM*.

*If three straight lines meet all at one point, and a straight line stand at right angles to each of them at that point, the three straight lines shall he in one and the same plane.*

Let the straight line *AB* stand at right angles to each of the straight lines *BC*, *BD*, *BE*, at *B* the point where they meet: *BC*, *BD*, *BE* shall be in one and the same plane.

For, if not, let, if possible, *BD* and *BE* be in one plane, and *BC* without it; let a plane pass through *AB* and *BC*; the common section of this plane with the plane in which are *BD* and *BE* is a straight line; [XI. 3.

let this straight line be *BF*.

Then the three straight lines *AB*, *BC*, *BF* are all in one plane, namely, the plane which passes through *AB* and *BC*.

And because *AB* stands at right angles to each of the straight lines *BD*, *BE*, [*Hypothesis*.

therefore it is at right angles to the plane passing through them; [XI. 4.

therefore it makes right angles with every straight line meeting it in that plane. [XI. *Definition* 3.

But *BF* meets it, and is in that plane;

therefore the angle *ABF* is a right angle.

But the angle *ABC* is, by hypothesis, a right angle;

therefore the angle *ABC* is equal to the angle *ABF*; [*Ax*. 11.

and they are in one plane; which is impossible. [*Axiom* 9.

Therefore the straight line *BC* is not without the plane in which are *BD* and *BE*,

therefore the three straight lines *BC*, *BD*, *BE* are in one and the same plane.

*if three straight lines*&c. q.e.d.

*THEOREM*.

*If two straight lines be at right angles to the same plane, they shall be parallel to one another*.

Let the straight lines *AB*, *CD* be at right angles to the same plane: *AB* shall be parallel to *CD*.

Let them meet the plane at the points *B*,*D*; join *BD*; and in the plane draw *DE* at right angles to *BD*; [1. 11.

make *DE* equal to *AB*; [I. 3.

and join *BE*,*AE*,*AD*.

Then, because *AB* is perpendicular to the plane, [*Hypothesis*.

it makes right angles with every straight line meeting it in that plane. [XI. *Def*. 3.

But *BD* and *BE* meet *AB*, and are in that plane,

therefore each of the angles *ABD*, *ABE* is a right angle. For the same reason each of the angles *CDB*, *CDE* is a right angle.

And because *AB* equal to *ED*, [*Construction*.

and *BD* is common to the two triangles *ABD*, *EDB*,

the two sides *AB*, *BD* are equal to the two sides *ED*, *DB*, each to each;

and the angle *ABD* is equal to the angle *EDB*, each of them being a right angle; [*Axiom* 11.

therefore the base *AD* is equal to the base *EB*. [I. 4.

Again, because *AB* is equal to *ED*, [*Construction*.

and it has been shewn that *BE* is equal to *DA*;

therefore the two sides *AB*, *BE* are equal to the two sides *ED*, *DA*, each to each;

and the base *AE* is common to the two triangles *ABE*, *EDA*;

therefore the angle *ABE* is equal to the angle *EDA*. [I 8.

But the angle *ABE* is a right angle,

therefore the angle *EDA* is a right angle,

that is, *ED* is at right angles to *AD*.

But *ED* is also at right angles to each of the two *BD*, *CD*;

therefore *ED* is at right angles to each of the three straight lines *BD*, *AD*, *CD*, at the point at which they meet;

therefore these three straight lines are all in the same plane. [XI. 5.

But *AB* is in the plane in which are *BD*, *DA*; [XI. 2.

therefore *AB*, *BD*, *CD* are in one plane.

And each of the angles *ABD*, *CDB* is a right angle;

therefore *AB* is parallel to *CD*. [I. 28.

Wherefore, *if two straight lines* &c. q.e.d.

*THEOREM*.

*If two straight lines he parallel, the straight line drawn from any point in one to any point in the other, is in the same plane with the parallels.*

Let *AB*, *CD* be parallel straight lines, and take any point *E* in one and any point *F* in the other: the straight line which joins *E* and *F* shall be in the same plane with the parallels.

For, if not, let it be, if possible, without the plane, as *EGF*; and in the plane *ABCD*, in which the parallels are, draw the straight EGF,EHF from *E* to *F*.

Then, since EGF is also a straight line, [Hypothesis.

the two straight lines *EGF*, *EHF* include a space between them; which is impossible. [*Axiom* 10.

Therefore the straight line joining the points *E* and *F* is not without the plane in which the parallels *AB*, *CD* are;

therefore it is in that plane.

*if two straight lines*&c. q.e.d.

*THEOREM*.

*If two straight lines he parallel, and one of them be at right angles to a plane, the other also shall he at right angles to the same plane.*

Let *AB*, *CD* be two parallel straight lines; and let one of them *AB* be at right angles to a plane: the other *CD* shall be at right angles to the same plane.

Let *AB*, *CD* meet the plane at the points *B*, *D*; join *BD*; therefore *AB*, *CD*, *BD* are in one plane. [XI. 7.

In the plane to which *AB* is at right angles, draw DE at right angles to *BD*; [I. 11.

make *DE* equal to *AB*; [I 3.

and join *BE*,*AE*,*AD*.

Then, because *AB* is at right angles to the plane, [*Hypothesis*.

it makes right angles with every straight line meeting it in that plane; [XI. *Definition* 3.

therefore each of the angles *ABD*, *ABE* is a right angle.

And because the straight line *BD* meets the parallel straight lines *AB*, *CD*,

the angles *ABD*, *CDB* are together equal to two right angles, [I. 29.

But the angle *ABD* is a right angle, [*Hypothesis*. therefore the angle *CDB* is a right angle;

that is, CD is at right angles to BD.

And because is equal to *ED*, [*Construction*.

and *BD* is common to the two triangles *ABD*, *EDB*;

the two sides *AB*, *BD* are equal to the two sides *ED*, *DB*, each to each;

and the angle *ABD* is equal to the angle *EDB*, each of them being a right angle; [*Axiom* 11.

therefore the base *AD* is equal to the base EB. [I. 4.

Again, because *AB* is equal to *ED*, [*Construction*.

and *BE* has been shewn equal to *DA*,

the two sides *AB*, *BE* are equal to the two sides *ED* *DA*, each to each;

and the base *AE* is common to the two triangles *ABE*, *EDA*;

therefore the angle ABE is equal to the angle *ADE*. [I. 8.

But the angle *ABE* is a right angle;

therefore the angle *ADE* is a right angle;

that is, *ED* is at right angles to *AD*.

But *ED* is at right angles to *BD*, [*Const*.

therefore *ED* is at right angles to the plane which passes through *BD*, *DA*, [XI. 4.

and therefore makes right angles with every straight line meeting it in that plane. [XI. *Definition* 3.

But *CD* is in the plane passing through *BD*, *DA*, because all three are in the plane in which are the parallels *AB*, *CD*;

therefore *ED* is at right angles to *CD*,

and therefore *CD* is at right angles to *ED*.

But *CD* was shewn to be at right angles to *BD*;

therefore *CD* is at right angles to the two straight lines *BD*, *ED*, at the point of their intersection *D*,

and is therefore at right angles to the plane passing through *BD*, *ED*, [XI. 4,

that is, to the plane to which *AB* is at right angles.

Wherefore, *if two straight lines* &c. q.e.d.

*THEOREM*.

*Two straight lines which are each of them parallel to the same straight line, and not in the same plane with it, are parallel to one another*.

Let *AB* and *CD* be each of them parallel to *EF*, and not in the same plane with it: *AB* shall be parallel to *CD*.

In *EF* take any point *G*; in the plane passing through *EF* and *AB*, draw from *G* the straight line *GH* at right angles to *EF*;

and in the plane passing through *EF* and *CD*, draw from *G* the straight line *GK* at right angles to *EF*. [I. 11.

Then, because *EF* is at right angles to *GH* and *GK*, [*Construction*. *EF* is at right angles to the plane *HGK* passing through them. [XI. 4.

And *EF* is parallel to *AB*; [*Hypothesis*.

therefore *AB* is at right angles to the plane *HGK*. [XI. 8.

For the same reason *CD* is at right angles to the plane *HGK*.

Therefore *AB* and *CD* are both at right angles to the plane *HGK*.

Therefore *AB* is parallel to *CD*. [XI. 6.

Wherefore, *if two straight lines* &c. q.e.d.

*THEOREM*.

*If two straight lines meeting one another he parallel to two others that meet one another, and are not in the same plane with the first two, the first two and the other two shall contain equal angles*.

Let the two straight lines *AB*, *BC*, which meet one another, be parallel to the two straight lines *DE*, *EF*, which meet one another, and are not in the same plane with *AB*, *BC*: the angle *ABC* shall be equal to the angle *DEF*.
Take *BA*, *BC*, *ED*, *EF* equal to one another, and join *AD*, *BE*, *CF*, *AC*, *DF*.

Then, because *AB* is equal and parallel to *DE*,

therefore *AD* is equal and parallel to *BE*. [1. 33.

For the same reason, *CF* is equal and parallel to *BE*.

Therefore *AD* and *CF* are each of them equal and parallel to *BE*.

Therefore AD is parallel to CF, [XI. 9

and AD is also equal to CF, [Axiom 1.

Therefore *AC* is equal and parallel to *DF*. [I. 33.

And because *AB*, *BC* are equal to *DE*, *EF*, each to each,

and the base *AC* is equal to the base *DF*,

therefore the angle *ABC* is equal to the angle *DEF*. [I. 8.

Wherefore, *if two straight lines* &c. q.e.d.

*PROBLEM*.

*To draw a straight line perpendicular to a given plane from a given point without it*.

Let *A* be the given point without the plane *BH*: it is required to draw from the point *A* a straight line perpen- dicular to the plane *BH*.

Draw any straight line *BC* in the plane *BH*, and from the point *A* draw *AD* perpendicular to *BC*. [1. 12.

Then if *AD* be also perpendicular to the plane *BH*, the thing required is done.

But, if not, from the point *D* draw, in the plane *BH*, the straight line *DE* at right angles to *BC',' [I. 11.*
and from the point

*A*draw

*AF*perpendicular to

*DE*. [1. 12.

*AF*shall be perpendicular to the plane

*BH*.

Through *F* draw *GH* parallel to *BC*. [I. 31. Then, because *BC* is at right angles to *ED* and *DA*, [*Constr*.*BC* is at right angles to the plane passing through *ED* and *DA*. [XI. 4.

And *GH* is parallel to *BC*; [*Construction*.

therefore *GH* is at right angles to the plane passing through *ED* and *DA*; [XI. 8.

therefore *GH* makes right angles with every straight line meeting it in that plane. [XI. *Definition* 3.

But *AF* meets it, and is in the plane passing through *ED* and *DA*;

therefore *GH* is at right angles to *AF*,

and therefore *AF *is at right angles to *GH*.

But is also at right angles to DE; [Construction.

therefore *AF* is at right angles to each of the straight lines *GH* and *DE* at the point of their intersection;

therefore *AF* is perpendicular to the plane passing through *GH* and *DE*, that is, to the plane *BH*. [XI. 4.

Wherefore, *from the given point A, without the plane BH, the straight line AF has been drawn perpendicular to the plane, q.e.f.*

*PROBLEM*.

*To erect a straight line at right angles to a given plane, from a given point in the plane*.

Let *A* be the given point in the given plane: it is required to erect a straight line from the point *A*, at right angles to the plane.
From any point *B* without the plane, draw *BC* perpendicular to the plane; [XI. 11.

and from the point *A* draw *AD* parallel to *BC*, [I. 31.

AD shall be the straight line required.

For, because *AD* and *BC* are two parallel straight lines, [*Constr*.

and that one of them BC is at right angles to the given plane, [*Construction*.

the other *AD* is also at right angles to the given plane. [XI. 8,

*a straight line has been erected at right an-gles to a given plane, from a given point in it*. q.e.f.

*THEOREM*.

*From the same point in a given plane, there cannot be two straight lines at right angles to the plane, on the same side of it; and there can he hut one perpendicular to a plane from, a point without the plane.*

For, if it be possible, let the two straight lines *AB*,*AC* be at right angles to a given plane, from the same point *A* in the plane, and on the same side of it.
Let a plane pass through *BA*, *AC*;

the common section of this with the given plane is a straight line; [XI. 3.

let this straight line be *DAE*.

Then the three straight lines *AB*, *AC*, *DAE* are all in one plane.

And because *CA* is at right angles to the plane, [*Hypothesis*.

it makes right angles with every straight line meeting it in the plane. [XI. *Definition* 3.

But *DAE* meets *CA*, and is in that plane; therefore the angle *CAE* is a right angle. For the same reason the angle *BAE* is a right angle. Therefore the angle *CAE* is equal to the angle *BAE*; [*Ax*.ll.

and they are in one plane; which is impossible. [*Axiom* 9.

Also, from a point without the plane, there can be but one perpendicular to the plane.

For if there could be two, they would be parallel to one another, [XI. 6.

which is absurd.

*from the same point*&c. q.e.d.

*THEOREM*.

*Planes to which the same straight line is perpendicular are parallel to one another*.

Let the straight line *AB* he perpendicular to each of the planes *CD* and *EF*: these planes shall be parallel to one another.

For, if not, they will meet one Smother when produced; let them meet, then their common section will be a straight line;

let *GH* be this straight line; in it take any point *K*, and join *AK*, *BK*.

Then, because *AB* is perpendicular to the plane *EF*, [*Hyp*.

it is perpendicular to the straight line *BK* which is in that plane; [XI. Definition 8.

therefore the angle *ABK* is a right angle.

For the same reason the angle *BAK* is a right angle.

Therefore the two angles *ABK*, *BAK* of the triangle *ABK are equal to two right angles;*
which is impossible. [I. 17

Therefore the planes

*CD*and

*EF*, though produced do not meet one another;

that is, they are parallel. [XI.

*Definition*8.

Wherefore, *planes* &c. q.e.d.

*THEOREM*.

*If two straight lines which meet one another, be parallel to two other straight lines which meet one another, but are not in the same plane with the first two, the plane pass-ing through these is parallel to the plane passing through the others*. Let *AB*,*BC*, two straight lines which meet one another, be parallel to two other straight lines *DE*, *EF*, which meet one another, but are not in the same plane with *AB*,*BC*: the plane passing through *AB*, *BC*, shall be parallel to the plane passing through *DE*, *EF*.

From the point *B* draw *BG* perpendicular to the plane passing through *DE*, *EF*, [XI. 11.

and let it meet that plane at G;

through *G* draw *GH* parallel to *ED*, and *G* parallel to *EF*. [1.31.

Then, because *BG* is perpendicular to the plane passing through *DE*, *EF*, [Construction.

it makes right angles with every straight line meeting it in that plane; [XI. *Definition* 3.

but the straight lines *GH* and *GK* meet it, and are in that plane;

therefore each of the angles *BGH* and *BGK* is a right angle.

Now because *BA* is parallel to *ED*, [Hypothesis.

and *GH* is parallel to *ED*, [*Construction*.

therefore *BA* is parallel to *GH*; [XI. 9.

therefore the angles *ABG* and *BGH* are together equal to two right angles. [I. 29.

And the angle *BGH* has been shewn to be a right angle;

therefore the angle *ABG* is a right angle.

For the same reason the angle *CBG* is a right angle.

Then, because the straight line *GB* stands at right angles to the two straight lines *BA*, *BC*, at their point of intersection *B*,

therefore *GB* is perpendicular to the plane passing through *BA*,*BC*. [XI. 4.

And *GB* is also perpendicular to the plane passing through *DE*, *EF*. [*Construction*.

But planes to which the same straight line is perpendicular are parallel to one another; [XI. 14.

therefore the plane passing through *AB*, *BC* is parallel to the plane passing through *DE*, *EF*.

Wherefore, *if two straight lines* &c. q e.f.

*THEOREM*.

*If two parallel planes be cut by another plane, their common sections with it are parallel*.

Let the parallel planes *AB*, *CD* be cut by the plane *EFHG*, and let their common sections with it be *EF*, *GH*: *EF* shall be parallel to *GH*.

For if not, *EF* and *GH*, being produced, will meet either towards *F*, *H*, or towards *E*, *G*. Let them be produced and meet towards *F*, *H* at the point *K*.

Then, since *EFK* is in the plane *AB*, every point in *EFK* is in that plane; [XI. 1.

therefore *K* is in the plane *AB*.

For the same reason *K* is in the plane *CD*.

Therefore the planes *AB*, *CD*, being produced, meet one another.

But they do not meet, since they are parallel by hypothesis.

Therefore *EF* and *GH*, being produced, do not meet towards *F*, *H*.

In the same manner it may be shewn that they do not meet towards *E*, *G*.

But straight lines which are in the same plane, and which being produced ever so far both ways do not meet are parallel;

therefore *EF* is parallel to *GH*.

*if two parallel planes*&c. q.e.d.

*THEOREM*.

*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* is 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.

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.

*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*, at right angles to *CE*. [I. 11.

Then, because *AB* is, at right angles to the plane *CK*, [*Hypothesis*. therefore it makes right angles with every straight line meeting it in that plane; [XI. *Definition* 3. 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; [*Construction*, therefore *FG* is parallel to *AB*. [I. 28. And *AB* is at right angles to the plane *CK*; [*Hypothesis*. 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.

*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, [*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 *DE* 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.

*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;

and the angle *BAD* is equal to the angle *BAE*; [*Constr*.

therefore the base *BD* is equal to the base *BE*. [I. 4.

And because *BD*, *DC* are together greater than *BC*, [1. 20.

and one of them *BD* has been shewn equal to *BE* a part of *EC*.

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.

*if a solid angle*&c. q.e.d.

*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*.

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 *BDC*.

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, [1. 32.

therefore the nine angles of these triangles, namely, the angles *CBA*, *BAC*, *ACB*, *ACD*,*CDA*, *CAD*, *ADD*, *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.

Next, let the solid angle at *A* be contained by any number of plane angles *BAC*, *CAD*, *DAE*, *EAF*, *FAB*: these shall be together less than four right angles.

Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be *BC*, *CD*, *DE*, *EF*, *FB*.

Then, because the solid angle at *B* is contained by the three plane angles *CBA*, *ABF*, *FBC*, any two of them are together greater than the third, [XL 20.

therefore the angles *CBA*, *ABF* are together greater than the angle *FBC*.

For the same reason, at each of the points *C*, *D*, *E*, *F*, the two plane angles which are at the bases of the triangles having the common vertex *A*, are together greater than the third angle at the same point, which is one of the angles of the polygon *BCDEF*.

Therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon.

Now all the angles of the triangles are together equal to twice as many right angles as there are triangles, that is, as there are sides in the polygon BCDEF; [I. 32.

and all the angles of the polygon, together with four right angles, are also equal to twice as many right angles as there are sides in the polygon; [I. 32, *Corollary* 1.

therefore all the angles of the triangles are equal to all the angles of the polygon, together with four right angles. [*Ax*. 1.

But it has been shewn that all the angles at the bases of the triangles are together greater than all the angles of the polygon;

therefore the remaining angles of the triangles, namely, those at the vertex, which contain the solid angle at *A*, are together less than four right angles.

Wherefore, *every solid angle* &c. q.e.d.