same EF. And because EF is perpendicular both to GH and GK, it is perpendicular (4. 2. Sup.) to the plane HGK passing through them: and EF is parallel to AB; therefore AB is at right angles (7. 2. Sup.) 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 (6. 2. Sup.) to one another. Therefore AB is pa

rallel to CD. Wherefore two straight lines, &c. Q. E. D,

PROP. IX. THEOR.

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.

B

Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF: A Because BA is equal and parallel to ED, therefore AD is (33. 1.) 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 (8. 2. Sup.) to one an other. Therefore AD is parallel to CF; and it is equal to it, and AC, DF join them towards the same parts; and therefore (33. 1.) 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 (8. 1.) to the angle DEF. Therefore, if two straight lines, &c. Q. E. D.

PROP. X. PROB.

D

F

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

H

In the plane d'aw any straight line BC, and from the point A draw (12. 1.) AD perpendicular to BC. If then AD be also perpendicu lar to the plane BH, the thing required is already done; but if it be not, from the point D draw (11. 1.), 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 (31. 1.) GH parallel to BC: and because BC is at right angles to. ED, and DA, BC is at right angles (4. 2. Sup.) 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 shall be at right (7. 2. Sup.) angles to the same plane; wherefore GH is at right angles to the plane through ED,DA, and is perpendicular (def. 1. 2.Sup.) 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 is also at right angles to the plane passing through them (4. 2. Sup.). 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. Which was to be done.

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 (7. 2. Sup.).

PROP. XI. THEOR.

From the same point in a 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 à plane from a point above it.

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 plane with the given plane is a straight (3. 2. Sup.) 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 given plane, it makes right angles with

B

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 perpendicular to that plane; for if there could be two, they would be parallel (6. 2. Sup.) to one another, which is absurd. Therefore, from the same point, &c. Q. E. D.

D

PROP. XII. THEOR.

E

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 are parallel to one another.

G

If not, they must meet one another when produced, and their com mon section must 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 (def. 1. 2. Sup.) to the straight line BK which is in that plane, c and 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 (17. 1.): Therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallel (def. 7.2. Sup.). Therefore planes, &c. Q. Ë. D.

PROP. XIII. THEOR.

B

If two straight lines meeting one another, be paralled to two straight lines which also meet one another, but are in the same plane with the first two: the plane which passes through the first two is parallel to the plane passing through the others.

Let AB, BC, two straight lines meeting one another, be 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.

From the point B draw BG perpendicular (10. 2. Sup.) to the plane which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to ED (31. 1.), and GK parallel to EF:

B

C

E

F

G

K

And because BG is perpendicular to the plane through DE, EF, it must make right angles with every straight line meeting it in that plane (1. def. 2. Sup.). But the straight lines GH, GK in that plane meet it: Therefore each of the angles BGH, BGK is a right angle: And because BA is parallel (8. 2. Sup.) to GH for each of them is parallel to DE), the angles GBA, BGH are together equal

D

H

(29. 1.) to two right angles: And BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA: For the same reason, GB is perpendicular to BC: Since, therefore, the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B; GB is perpendicular (4. 2. Sup.) to the plane through BA, BC: And it is perpendicular to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF: But planes to which the same straight line is perpendicular, are parallel (12. 2. Sup.) to one another: Therefore the plane through AB, BC, is parallel to the plane through DE, EF. Wherefore, if two straight lines, &c. Q. E.D.

COR. It follows from this demonstration, that if a straight line meet two parallel planes, and be perpendicular to one of them, it must be perpendicular to the other also."

PROP. XIV. THEOR.

If two parallel planes be cut by another plane, their common sections with it are parallels.

Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with

F

PROP. XV. THEOR.

If two parallel planes be cut by a third plane, they have the same inclination to that plane.

Let AB and CD be two parallel planes, and EH a third plane cutting them: The planes AB and CD are equally inclined to EH.

Let the straight lines EF and GH be the common section of the plane EH with the two planes AB and CD; and from K, any point in EF, draw in the plane EH the straight line KM at right angles to EF, and let it meet GH in L; draw also KN at right angles to EF in the plane AB: and through the straight lines KM, KN. let a plane be made to pass, cutting the plane CD in the line LO. And because EF and GH are the common sections of the plane EH with the two parallel planes AB and CD, EF is parallel to GH (14. 2 Sup.). But EF is at right angles to the plane that passes through KN and KM (4. 2. Sup.), because it is at right angles to the lines KM and KN:

therefore GH is also at right angles to the same plane (7. 2. Sup.), and it is therefore at right angles to the lines LM, LO which it meets in that plane. Therefore, since LM and LO are at right angles to LG, the common section of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EH (4. def. 2. Sup.). For the same reason the angle MKN is the inclination of the plane AB to the plane EH. But because KN and LO are parallel, being the common sections of the parallel planes AB and CD with a third plane, the interior angle NKM is equal to the exterior angle OLM (29. 1.); that is, the inclination of the plane AB to the plane EH. is equal to the inclination of the plane CD to the same plane EH. Therefore, &c. Q. E. D.