angle ABC, by the hypothesis, is also a right angle; therefore the angle ABF is equal to the angle ABC, and they are both in the same plane; which is impossible: therefore the straight line BC is not above the plane in which are BD and BE. Wherefore the three straight lines BC, BD, BE, are in one and the same plane.

PROPOSITION VI. 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 is parallel to CD.

E

Let them meet the plane in the points B, D, and draw the straight line BD, to which draw DE at right angles, in the same plane; and make DE equal to AB, and join BE, AE, AD. Then, because AB A is perpendicular to the plane, it shall make right (I. Def. 1) angles with every straight line which meets it, and is in that plane; but B 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; and they contain right angles; therefore the base AD is equal (Pl. Ge. I. 4) to the base BE. Again, because AB is equal to DE, and BE to AD, and the base AE common to the triangles ABE, EDA, the angle ABE is equal (Pl. Ge. I. 8) to the angle EDA; 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, DC; 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 (I. 5); but AB is in the plane in which are BD, DA, because any three straight lines which meet one another are in one plane (1.2). 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."

PROPOSITION VII. THEOREM.

If two straight lines be parallel, and one of them is 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 is at right angles to the same plane.

For, if CD be not perpendicular to the plane to which AB is perpendicular, let DG be perpendicular to it. Then (1.6)

브

B

D

DG is parallel to AB; DG and DC therefore are both parallel to AB, and are drawn through the same point D; which is impossible.

PROPOSITION VIII. THEOREM.

Two straight lines which are each of them parallel to the same straight line, though not both 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.

H

B

E

G

K

D

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; and in the plane passing A through EF, 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 HGK passing through them; and EF is parallel to AB; therefore AB is at right angles (I. 7) to the 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 (I. 6) to one another. Therefore AB is parallel to CD.

PROPOSITION IX. THEOREM.

If two straight lines meeting one another be parallel to two others that meet one another, though 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 is equal to the angle DEF.

Take BA, BC, ED, EF, all equal to one another; and join AD, CF, BE,

AC, DF; because BA is equal and parallel to ED, therefore AD is (Pl. Ge. I. 33) 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, though not in the same plane with it, are parallel (I. 8) to one another; therefore AD is parallel to CF; and it is equal to it, and AC, DF, join them towards the same parts; and therefore AC is equal and parallel to DF; and because AB, BC, are equal to DE, EF, and the base AC to the base DF, the angle ABC is equal (Pl. Ge. I. 8) to the angle DEF.

PROPOSITION X. THEOREM.

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.

G

E

F

D

C

In the plane draw any straight line BC, and from the point A draw AD perpendicular to 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 plane BH, the straight line DE at right angles to BC; and from the point A draw AF perpendicular to DE; and through F draw GH parallel to BC; and because BC is at right angles to ED and DA, BC is at right angles (I. 4) to the plane passing 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

B

shall be at right (I. 7) angles to the same plane; wherefore GH is at right angles to the plane through ED, DA, and is perpendicular (I. Def. 1) to every straight line meeting 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 also perpendicular to DE; 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 passing through them; and the plane passing through ED, GH, is the plane BH; therefore AF is perpendicular to the plane BH; so that, from the given point A, above the plane BH, the straight line AF is drawn perpendicular to that plane.

COR.-If it be required from a point C in a plane to erect a perpendicular to that plane, take a point A above the plane, and draw AF perpendicular to the plane; then, if from C a line be drawn parallel to AF it will be the perpendicular required; for, being parallel to AF, it will be perpendicular to the same plane to which AF is perpendicular.

PROPOSITION XI. THEOREM.

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 AC, AB, be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let

a plane pass through BA, AC; the common section of this with the given plane is a straight line passing through A (1.3). Let

DAE be their common section; therefore D A ET the straight lines AB, AC, DAE, are in one plane; and because CA is at right angles to the given plane, it shall make right angles with every 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; wherefore the angle CAE is equal to the angle BAE; and they are in one plane; which is impossible. Also, from a point above a plane, there can be but one per pendicular to that plane; for if there could be two, they would be parallel (I. 6) to one another; which is absurd.

PROPOSITION XII. THEOREM.

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

G

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 straight line GH, in which take any point K, and join AK, BK; then, because AB is perpendicular to the plane EF, it is perpendicular (I. Def. 1) c to the straight line BK which is in that plane; therefore ABK 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 (Pl. Ge. I. 17); therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallel (I. Def. 6.)

PROPOSITION XIII. THEOREM.

A

E

If two straight lines meeting one another be parallel to two straight lines which meet one another, but are not in the same plane with the first two, the plane which passes through these is parallel to the plane passing through the others.

Let AB, BC, two straight lines meeting one another, bè parallel to DE, EF, that meet one another, but are not in the same plane with AB, BC; the planes through AB, BC, and DE, EF, shall not meet, though produced.

E

} G

2

From the point B draw BG perpendicular (I. 10) to the plane which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED, and GK parallel to EF. And because BG is perpendicular to the plane through DE, EF, it shall make right angles with every straight line meeting it in that

B

A