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: But ABE is a right angle; therefore also EDA is a right angle, and ED perpendicular to DA: But ED is also perpendicular to each of the two BD, DC; therefore ED is at right angles to each of the three straight lines BD, DA, DC, in the point in which they meet; therefore (11. 5) these three straight lines are all in the same plane: But AB is in the plane in which are BD, DA, because any three straight lines which meet one another, are in one plane (11. 2); 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.

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

PROP. VII. THEOR.

If two straight lines be parallel, any straight line, drawn from a point in the one to a 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 the one and any point F in the other the straight line which joins E and F shall be in the same plane with AB, CD.

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 line EHF from E to F: Then, since EGF is also a straight line, the two straight lines EGF, EHF include a space between them-which is impossible : Therefore the

straight line joining the points E, F is not without the

plane in which are the parallels AB, CD, that is, it is

in that plane.

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

PROP. VIII. THEOR.

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

A

Let AB, CD meet the plane in the points B, D, and join BD; therefore (11. 7) AB, BD, CD are in one plane: in the plane to which AB is at right angles, draw DE at right angles to BD; make DE equal to AB, and join BE, AE, AD: Then, because AB is perpendicular to the plane, it is perpendicular to every straight line which meets it in that 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 equal

to two right angles: But of these ABD is a right angle, therefore also CDB is a right angle, and CD is perpendicular to DB: Now, because the two AB, BD are equal to the two ED, DB, each to each, and each of the angles ABD, EDB is a right angle therefore the base AD is equal to the base EB: Again, because the two AB, BE are equal to the two 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: But ABE is a right

angle; therefore also EDA is a right angle, and ED is perpendicular to DA: But ED is also perpendicular to BD; therefore (11. 4) ED is perpendicular to the plane which passes through BD, DA, and makes right angles with every straight line which meets it in that plane: 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 also 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 DB, DE, in the point of their intersection D, and is therefore at right angles to the plane passing through DB, DE, that is, to the plane to which AB is at right angles.

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

PROP. IX. THEOR.

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, from which draw, in the plane passing through EF and AB,

E

H

C K

the straight line GH at right angles to EF; and in the plane passing through EF and CD, the straight line GK at right angles to the same EF: Then, because EF is at right angles both to GH and GK, EF is at right angles to the plane GHK which passes through them (11. 4): And EF is parallel to AB; therefore also (11. 8) AB is at right angles to the plane GHK: In like manner, CD is at right angles to

the plane GHK: Therefore AB and CD are each of them at right angles to the plane GHK, and therefore (11. 6) they are parallel to one another, that is, AB is parallel to CD.

Wherefore, Two straight lines &c. Q.E. D.

PROP. X. THEOR.

If two straight lines meeting one another be 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, 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, BE, CF, AC, DF: Then, because AB is both equal and parallel to DE, therefore AD is both equal and parallel to BE: In like manner, CF is equal and parallel to BE: Therefore AD and CF are each of them equal and parallel to BE, and therefore (11. 9) AD is parallel to CF: And AD is also equal to CF, being each of them equal to BE; therefore also AC is both 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-therefore the angle ABC is equal to the angle DEF.

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

PROP. XI. PROB.

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

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

Draw any straight line BC in the plane BH, and from the point A draw AD perpendicular to BC: Then, if AD be also perpendicular to the plane BH, the thing required is done: But, if not, from D draw, in the plane BH, the straight line DE at right angles to BC; and from A draw AF perpendicular to DE, and through F draw GH parallel to BC: Then, because BC is at right angles to ED and DA, BC is at right angles to the plane passing through them (11. 4): And GH is parallel to BC; therefore also (11. 8) GH is at right angles to the plane through ED, DA, and makes right angles with every straight line which meets it in that plane: Therefore GH makes right angles with AF, and AF with GH: But AF also makes right angles with DE; therefore AF makes right angles B

with each of the straight lines GH, DE, in the point of their intersection, F: Therefore also (11. 4) it is at right angles to the plane through GH, DE, that is, to the plane BH: Wherefore, from the given point A, without the plane BH, the straight line AF has been drawn perpendicular to the plane. Q. E. F.