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Book XI. Let AB, CD be two parallel straight lines, and let one of
them AB be at right angles to a plane; the other CD is at right angles to the same place.
Let AB, CD meet the plane in the points B, D, and join 8 7. II. BD: Therefore & AB, CD, BD are in one plane. In the plane
to which AB is at right angles, draw DE at right angles to BD, and make DE equal to AB, and join BE, AE, AD. And because AB is perpendicular to the plane, it is perpen
dicular to every straight line which meets it, and is in that a 3. def. 11. plane: 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 b 29. I. equal o to two right angles : And ABD is a right angle;
therefore also CDB is a right angle, and CD perpendicular to
C the angle EDB, because each of them
is a right angle; therefore the base AD C 4. I.
is equal to the base BE: Again, be-
D mon to the triangles ABE, EDA; d 8. 1.
wherefore the angle ABE is equal to
But it is also perpendicular to BD; therefore ED is perpendia e 4. 11. cular to the plane which passes through BD, DA, and shall ! f 3. def. 11. make right angles 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, OD: 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 ; CD then is at right angles to the two straight lines DE, DB in the point of their intersection D; and therefore is ac right anglese 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.
THE O R.
WO 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.
a 4. II.
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, from which draw, in the plane palling through EF, AB, the straight line GH at right angles to EF; and in the plane passing through EF, CD, draw GK at right angles to the same EF. And A H because EF is perpendicular both
-B to GH and GK, EF is perpendi.. cular to the plane HGK pafling through them: And EF is parallel
F to AB; therefore AB is at right
b 8. 11. angles to the plane HGK. For the fame reason, CD is likewise at right C K
D angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK. But if two straight lines be at right angles to the same plane, they shall be parallel to one another. Therefore AB is c 6, 11. parallel to CD. Wherefore two straight lines, &c. Q. E. D.
two others that meet one another, and are not in the same plane with the first two; the first two and the other fwo 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 is equal to the angle DEF. Take Ba, BC, ED, EF all equal to one another; and join
Book XI. AD, CF, BE, AC, DF: Because BA is equal and parallel to
line, and not in the same plane with it, b . are parallel to one another. Therefore c1. Ax. 1. AD is parallel to CF; and it is equal to it, and AC, DF join them towards
cause AB, BC are equal to DE, EF, and the base AC to the 28. 1. base DF; the angle ABC is equal d to the angle DEF. There
fore, if two straight lines, &c. Q. E. D.
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 12. 1.
A draw AD perpendicular to BC. If then AD be also per. pendicular to the plane BH, the thing required is already
done ; but if it be not, from the b II. I. point D drawb, in the plane BH,
DE; and through F draw GH
is at right angles to ED and DA, d 4. 11.
BC is at right anglesd to the plane
D с two straight lines be parallel, one of which is at right angles to a plane, the other shall be at right e angles to the same plane;
wherefore GH is at right angles to the piane through ED, DA, 1 3. def. II. and is perpendicular to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets
e 8. 11.
it : Therefore GH is perpendicular to AF; and consequently Book XI. AF is perpendicular to GH; and AF is perpendicular to DE; m therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stands at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane pafling through 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,
O erec a straight line at right angles to a given
plane, from a point given in the plane.
a II. II.
Let A be the point given in the plane ; it is required to erect a straight line from the point A at right
B angles to the plane.
From any point B above the plane draw.. BC perpendicular to it ; and from. A drawb AD parallel to BC. Be
b 31. 1. cause, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles to the given plane,
А 'C the other AD is alio at right angles. to its. Therefore a straight line has been erected at right angles to c 8. 11. a given plane from a point given in it. Which was to be done.
ROM 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 per. pendicular to a plane froni a point above the plane.
For, if it be possible, let the two itraight lines AC, AB, be at right angles to a given plane from the fame point A in the plane, and upon the fame side of it; and let a plane pass through BA, AC, the common section of this with the given plane is a straight
Book XI. . line pafling through A: Let DAE be their common section ;
And because CA is at right angles to the given plane, it shall
D A Ε impossible. Also, from a point a
bove a plane, there can be but one perpendicular to that plane ; 16. 11. for, if there could be two, they would be parallel to one
another, which is abfurd. Therefore, from the same point, &c.
Q. E, D.
cular, are parallel to one another.
Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another.
If not, they shall meet one another when produced ; let them meet; their common section shall be a Itraight line GH, in which
G take any point K, and join AK, BK :
Then, because AB is perpendicular to
to two right angles, which is impor, b 17. I. fible: Therefore the planes CD, EF,