11.

each to each; and the base AE is common to the triangles ABE, EDA; wherefore the angle ABE is equal to the angle EDA: but ABE is *& a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: but it is also perpendicular tot BD; therefore ED is +Const. perpendicular to the plane which passes through *4. 11. BD, DA; and therefore makes. right angles * 3 Def. with every straight line meeting it in that plane: but DC is in the plane passing through BD, DA, because all three are in the plane in which are the parallels AB, CD; wherefore ED is at right angles to DC; and therefore CD is at right angles to DE: but CD is also at right angles to DB; therefore CD is at right angles to the two straight lines DE, DB in the point of their intersection D; and therefore is at right angles* *4. 11. to the plane passing through DE, DB, which is the same plane to which AB is at right angles. Therefore, if two straight lines, &c. Q. E. D.

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, 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, A H from which† draw, in the plane passing through EF, E AB, the straight line GH at right angles to EF; and in the plane passing through EF,

CK

B

+11. 1.

D

CD draw GK at right angles to the same EF. And because EF is perpendicular both to GH and GK, EF is perpendicular to the plane *4. 11. HGK passing through them : and EF is parallel to AB; therefore AB is at right angles to the *8. 11.

*6.11.

plane HGK. For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK. But if two straight lines are at right angles to the same plane, they are parallel to one another: therefore AB is parallel to CD. Wherefore, two straight lines, &c. Q. E. D.

pa

If two straight lines meeting one another be rallel 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, that 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 all equal to one another; and join AD, CF, BE, AC, DF: then, because BA is equal and parallel to ED, *33.1. therefore AD is* both equal and parallel to BE. For the same reason, CF is equal and parallel to BE. Therefore AD and CF are each of them equal and parallel to BE. But straight lines that are parallel to the same straight line, and not in the same plane with *9.11. it, are parallel * to one another :

therefore AD is parallel to CF; *1Ax.1. and it is equal to it; and AC,

DF join them towards the same 33. 1. parts; and therefore* AC is equal and parallel to DF. And because AB, BC are equal to DE, EF, each to each, and the base AC to the base DF; the angle ABC is

8.1.

B

equal to the angle DEF. Therefore, if two straight lines, &c. Q. E. D.

PROP. XI. PROB.

To draw a straight line perpendicular to a plane, from a given point above it.

Let A be the given point above the plane BH; it is required to draw from the point A a straight line perpendicular to the plane BH.

In the plane draw any straight line BC, and from the point A draw* AD perpendicular to* 12. 1. BC. If then AD be also perpendicular to the plane BH, the thing required is already done; but if it be not, from the point D draw,* in the *11. 1. plane BH, the straight line DE at right angles to BC; and from the point A draw AF perpendicular to DE: AF shall be perpendicular to the plane BH.

*

Through F draw* GH parallel to BC: and *31.1. because BC is at right angles to ED and DA, BC is at right angles to the plane passing *4. 11. through ED, DA: and GH is parallel to BC: but, if two straight lines be parallel, one of which is at right angles to a plane, the other is at right angles to the same plane; wherefore GH is at right angles to the plane through ED, DA; and is perpendicular * to every straight line meeting

*

E A
G F

8. 11.

B DC

* 3 Def.

11.

it in that plane: but AF, which is in the plane through ED, DA, meets it; therefore GH is perpendicular to AF; and consequently AF is perpendicular to GH; and AF is perpendicular to DE; therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stand at right angles to each of two straight lines in the point of their intersection, it is also at right angles to the plane passing through +4.11. them: but the plane passing through ED, GH is the plane BH; therefore AF is perpendicular to the plane BH: therefore, from the given

point A, above the plane BH, the straight line AF is drawn perpendicular to that plane. Which was to be done.

PROP. XII. PROB.

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

Let A be the point given in the plane; it is required to erect a straight line from the point A at right angles to the plane.

From any point B above the

*11.11. plane draw* BC perpendicular to

*31.1. it; and from A draw* AD_parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at

D.

B

right angles to the given plane, the other AD is *8.11. also at right angles to it: therefore a straight line has been erected at right angles to a given plane, from a point given in it. Which was to be done.

From the same point in a given plane there cannot be two straight lines at right angles to the plane, upon the same side of it: and there can be but one perpendicular to a plane from a point above 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 upon the same side of it. Let a plane pass through BA, AC; the common section of this with the *3 11. given plane is a straight * line passing through A let DAE be their common section: therefore the straight lines AB, AC, DAE are in one plane: and because CA is at right angles to the

11.

given plane, it makes right angles† with every + 3 Det. straight line meeting it in that plane: but DAE, which is in that plane, meets CA; therefore CAE is a right angle. For the same reason, BAE is a right angle. Where

fore the angle CAE is equal to the angle +11 Ax. BAE; and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane: for, if there could be two, they would be parallel to one another, which is absurd. There- 6. 11. fore, from the same point, &c. Q. E. D.

PROP. XIV. THEOR.

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

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

let

If not, they shall meet one another when produced: them meet; their common section is a straight line GH, in C which take any point K, and join AK, BK. Then, because AB is perpendicular to the plane EF, it is perpendicular to the straight

G KH

* 3 Def.

line BK which is in that plane: therefore ABK 11. is a right angle. For the same reason, BAK is a right angle: wherefore the two angles ABK, BAK of the triangle ABK are equal to two right angles, which is impossible therefore the *17.1. planes CD, EF, though produced, do not meet one another; that is, they are parallel. There- *8 Def. fore planes, &c. Q. E. D.

*

*

:

11.