cular to the plane, it shall make right angles with every straight line which meets it, and is in that plane (c); but BD, BE, which are in that plane, do each of them meet AB; 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 is equal to DE, and BD common, the two sides AB, BD are equal to the two ED, DB, each to each; and they contain right angles; therefore the base AD is equal to the base BE (d): again, because AB is equal to DE, and BE to AD; AB, BE are equal to ED, DA, each to each; and, in the triangles ABE, EDA, the base AE is common; therefore the angle ABE is equal to the angle EDA (e); but ABE is a right angle; therefore EDA is also a right angle, and ED perpendicular to DA: but it is also perpendicular to each of the two BD, DU; wherefore ED is at right angles to each of the three straight lines BD, DA, DC, in the point in which they meet; therefore these three straight lines are all in the same plane (ƒ): but AB is in the plane in which are BD, DA, because any three straight lines which meet one another are in one plane (g); therefore AB, BD, DC are in one plane: and each of the angles ABD, BDC is a right angle; therefore AB is parallel to CD (h).

PROPOSITION VII.

THEOREM.-If two straight lines (AB, CD) be parallel, the straight line drawn from any point (E) in the one to any point (F) in the other, is in the same plane with the parallels.

DEMONSTRATION. If not, let

it be, if possible, above the plane, as EGF; and in the plane ABCD, in which the parallels are, draw the straight line EHF from E to F: and since EGF also is a straight line, the two straight lines EHF, EGF include a space between them; which is impossible (a): therefore the straight line joining the points E, F is not above the plane in which the parallels AB, CD are, and is therefore in that plane.

H

(a) I. Ax. 10.

PROPOSITION VIII.

THEOREM.-If two straight lines (AB, CD) be parallel, and one of them (AB) is at right angles to a plane, the other (CD) shall also be at right angles to the same plane.

one

DEMONSTRATION. Let AB, CD meet the plane in the points B, D, and join BD: therefore AB, CD, BD are in plane (a). In the plane to which AB is at right angles, draw DE at right angles to BD (b), and make DE equal to AB (c), and join BE, AE, AD. And because AB is perpendicular to the plane, it is perpendicular to every straight line which meets it and is in that plane (d); 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 (e): and ABD is a right angle; therefore also CDB is a right angle, and CD perpendicular to BD: and because AB is equal to DE, and BD common, the two AB, BD are equal to the two ED, DB, each to each; and the angle ABD is equal to the angle EDB, because each of them is a right angle; therefore the base AD is equal to the base BE (f): again, because AB is equal to DE, and BE to AD; the two AB, BE are equal to the two ED, DA, each to each; and the base AE is common to the triangles ABE, EDA; wherefore the angle ABE is equal to the angle EDA (g): but ABE is a right angle; and therefore EDA is a right angle, and ED perpendicular to DA: but it is also perpendicular to BD (h); therefore ED is perpendicular to the plane which passes through BD, DA (2); and therefore makes right angles with every straight line meeting it in that plane (d); 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 to the plane passing through DE, DB (i), which is the same plane to which AB is at right angles.

PROPOSITION IX.

THEOREM.-If two straight lines (AB, CD) are each of them parallel to the same straight line (EF), and not in the same plane with it, they are parallel to one another.

DEMONSTRATION. In EF,

take any point G, from which
draw, in the plane passing
through EF, AB, the straight
line GH at right angles to
EF (a); and in the plane pass-
ing through EF, CD, draw GK
at right angles to the same EF.
And because EF is perpendi-
cular both to GH and GK, EF C
is perpendicular to the plane
HGK passing through them (6):
and EF is parallel to AB; there-
fore AB is at right angles to
the plane HGK (c): for the

D

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 (d); therefore AB is parallel to CD.

PROPOSITION X.

THEOREM.-If two straight lines (AB, BC) meeting one another be parallel to two others (DE, EF) 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.

DEMONSTRATION. 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, therefore AD is both equal and parallel to BE (a): 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 it, are parallel to one another (b); therefore AD is parallel to CF; and it is equal to it (c); and AC, DF join them towards the same parts; and therefore AC is equal and parallel to DF (a). And because AB, BC are equal to DE, EF, each to each,

and the base AC to the base DF, the angle ABC is equal to the angle DEF (d).

PROPOSITION XI.

PROBLEM. To draw a straight line perpendicular to a

plane (BH), from a given point (A) above it.

SOLUTION. In the

plane, draw any
straight line BC,
and from the point
A, draw AD perpen-
dicular to BC (a): if G
then AD be also per-
pendicular to the
plane BH, the thing
required is already
done: but if it be not,
from the point D
draw, in the plane
BH, the straight line
DE at right angles
to BC (b); and from
the point A, draw
AF perpendicular to
DE: AF shall be

perpendicular to the

plane BH.

D

DEMONSTRATION. Through F, draw GH parallel to BC (c): and because BC is at right angles to ED and DA, BC is at right angles to the plane passing through ED, DA (d); 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 (e): wherefore GH is at right angles to the plane through ED, DA; and is perpendicular to every straight line meeting it in that plane (f): 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 them (d): 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.

PROPOSITION XII.

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

SOLUTION. From any point B above the plane, draw BC perpendicular to it (a); and from A, draw AD parallel to BC (b).

DEMONSTRATION. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles to the given plane, the other AD is also at right angles to it (c): therefore, a straight line has been erected at right angles to a given plane, from a point given in it.