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Suppl. plane HGK; therefore AB is parallel to CD. Wherefore,

two straight lines, &c. Q. E. D. c 6. 2. Sup.

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IF two straight lines which meet each other be parallel to two straight lines which meet each other,

though not in the same plane with the first two, the ': first two and the other two will contain equal angles.

Let the two straight lines AB, BC, which meet each other, be parallel to the two straight lines DE, EF, which meet each other, 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, AD is equal and parallel to

B a 33. 1.

BE; and because BC is equal and
parallel to EF, CF is equal and pa-
rallel to BE. Therefore AD and

A
CF are equal and parallel to BE;
b 8. 2. Sup. therefore AD is equal and parallel

to CF; therefore AC is equal and
parallel to DF, Now because AB,
BC are equal to DE, EF, and the
base AC to the base DF, the angle

E c 8. 1. ABC is equal to the angle DEF.

Therefore, if two straight lines, &c.
Q. E. D.

D

PROP. X. PROB.

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 re- Book II. quired 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 drawa AD perpendicular to BC; then if AD be also a 12. 1. perpendicular to the plane BH the thing required is already done. But if it be not, from the point D drawb, in the

A

b 11. 1.

ES plane BH, the straight line DE at right angles to BC; and from the point A draw

H AF perpendicular to DE;

F and through F drawc GH

c 31. 1. parallel to BC. Because BC is at right angles to ED

B and DA, BC is at right anglesd to the plane passing

d 4.2. Sup. through ED, DA. But GH is parallel to BC; wherefore GH is at right anglese to the plane passing through ED, DA, e 7.2. Sup. and is perpendicular to every straight line meeting it in that f Def. 1. plane. But AF, which is in the plane passing through ED, DA, 2. Sup. meets GH; therefore GH is perpendicular to AĚ. But AF is perpendicular to DE. Hence AF is perpendicular to each of the straight lines GH, DE; therefore AF is perpendicular to the plane BH, which passes through ED, GHd. 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.

Cor. If it be required to erect a perpendicular to a plane, from a point C in the plane, take a point A above the plane, and draw AF perpendicular to the plane; then from C draw a line parallel to AF, and it will the perpendicular requiredc.

PROP. XI. THEOR.

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 only one perpendicular to a plane from a point above the plane.

Suppl. 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; let a plane pass through

BA, AC; the common section of this with the given plane a 3. 2. Sup. is a straight line passing through Aa. Let DAE be their

common section; then the straight lines AB, AC, DAE are in one plane. Because CA is at right angles to the given plane, it will make right angles

c with every straight line meeting

B
b Def. 1. it in that planeb. But DAE,
2. Sup. which is in that plane, meets

CA; therefore CAE is a right
angle. For the same reason
BAE is a right angle. Where-
fore the angle CAE is equal to

A

E the angle BAE; and they are in one plane, which is impossible. Again, from a point above a plane there can be only one perpendicular to that plane;

for, if there could be two perpendiculars, they would be pac 6, 2. Sup. rallel to each other, which is absurd. Therefore, from the

same point, &c. Q. E. D.

PROP. XII. THEOR.

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

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Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to each other.

If not, they will meet each other, if produced. Let them meet; then their common section will be a straight line GH, in which take any point K, and join AK, BK. Be

cause AB is perpendicular to the a Def. 1. plane EF, it is perpendiculara to 2. Sup.

the straight line BK which is in that
plane; therefore ABK is a right
angle. For the same reason BAK

F B

A

E

D

is a right angle. Wherefore the two angles ABK, BAK of Book II. the triangle ABK are equal to two right angles, which is impossibleb. Therefore the planes CD, EF, though produced, b 17. 1. do not meet; therefore they are parallel. Therefore, planes, c Def. 7. &c. Q. E. D.

2. Sup.

PROP. XIII. THEOR.

IF two straight lines which meet each other be pa. rallel to two straight lines which meet each other, but are not in the same plane with the first two, the plane which passes through the former lines is pa. rallel to the plane which passes through the latter.

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Let AB, BC, two straight lines meeting each other, be parallel to DE, EF that meet each other, but are not in the same plane with AB, BC; the planes passing through AB, BC, and DE, EF, will not meet, though produced.

From the point B draw BG perpendiculara to the plane a 10.2. Sup. which passes through DE, EF, and let it meet that plane in G; and through G draw GH parallel to EDb, and GK pa- b 31. 1. rallel to EF. Because BG is perpendicular to the plane passing through DE, EF, it will make right an

E gles with every straight

с line meeting it in that B

F plane. But the straight

K 2 Sup lines GH, GK in that

с plane meet BG; there. fore each of the angles BGH, BGK is a right

А angle. Because BA is paralleld to GH (for

d 8. 2. Sup. each of them is parallel to DE), the angles GBA, BGH are together equale to two right angles. But BGH is a right e 29. 1. angle; therefore also GBA is a right angle, therefore GB is perpendicular to BA. For the same reason GB is perpendicular to BC. Therefore, since the straight line GB stands at right angles to the two straight lines BA, BC, which meet

c Def. 1.

D II

Suppl. in B, GB is perpendiculars to the plane through BA, BC.

But it is also perpendicular to the plane through DE, EF; f 4.2. Sup. therefore BG is perpendicular to each of the planes through

AB, BC and DE, EF; therefore the plane through AB, BC

is parallel to the plane through DE, EFs. Wherefore, if Sup two straight lines, &c.' Q. E. D.

g 12.2.

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. For BG meets the two parallel planes passing through DE, EF, and AB, BC, and is drawn perpendicular to the plane through DE, EF, and has been proved to be perpendicular to the plane through AB, BC.

PROP. XIV. THEOR.

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

Let the parallel planes AB, CD be cut by the plane EFHG, and let their common sections with it be EF, GH; EF įs parallel to GH.

For the straight lines
EF, GH are in the same

H
plane EFHG, which cuts

D the planes AB, CD;

B
and they will not meet
though produced, be-
cause the planes in which
they are do not meet;
therefore EF, GH are A

С a Def.30.1. parallela. Therefore, if

E
two parallel planes, &c.

C
Q. E. D.

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