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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 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. 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 BA, the straight line AF is drawn perpendicular to that plane. Which was to be done.

PROP. XII. PROB.

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

Let A be the point given in the plane; it is required to erect a straight line from the point A at right angles

D B to the plane.

From any point B above the plane draw (11. 11.) BC perpendicular to it; and from A draw (31. 1.) AD parallel to BC. Because, therefore, AD, CB are two parallel straight lines, and one of them BC is at right angles A

С to the given plane, the other AD is also at right angles to it (8. 11.). Therefore a straight line has been erected at right angles to a given plane from a point given in it. Which was to be done.

PROP. XIII. 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 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 (3. 11.) 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 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

B

С
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 impossi-
ble. Also, from a point above a plane,
there can be but one perpendicular to

D
A

E that plane; for, if there could be two,

V.

they would be parallel (6. 11.) to one another, 'which is absurd. Therefore, from the same point, &c. Q. E. D.

PROP. XIV. THEOR. 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.

If not, they shall meet one another when produced; let them meet; their common section shall be a

G straight line GH, in which take any point K, and join AK, BK: then because AB is

K perpendicular to the plane EF, it is perpendicular (3. def. 11.) to the straight line BK С which is in that plane. Therefore ABK is a right angle. For the same reason, BAK

F 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,

E EF though produced, do not meet one another; that is, they are parallel (8. def.

D 11.). Therefore, planes, &c. Q. E. D.

H

PROP. XV. THEOR.

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 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 (11. 11.) to the plane which passes through "DE, EF, and let it meet that plane in G; and through G draw GH parallel (31. 1.) to ED, and GK parallel to EF; and because BG is perpendicular to the plane through DE, EF, it shall make right an

E gles with every straight line meeting it in that plane (3. def. 11.). B

G F But the straight lines GH, GK,

С in that plane meet it: therefore

K each of the angles BGH, BGK is a right angle: and because BA is parallel (9. 11.) to GH (for each A

D of them is parallel to DE, and they are not both in the same

H

* See Note.

plane with it) the angles GBA, BGH are together equal (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. 11.) to the plane through AB, 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 (14. 11.) 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.

PROP. XVI. 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 it be EF, GH; EF is parallel to GH.

For, if it be not, EF, GH shall meet, if produced, either on the side of FH, or EG; first, let them be produced on the side of FH, and meet in the point K; therefore, since EFK is in the plane AB, every point in EFK is in that plane;

K and K is a point in EFK; therefore K is in the plane AB: for the same reason K is also in the plane CD:

F wherefore the planes AB, CD pro

H duced meet one another; but they

B do not meet, since they are parallel

D by the hypothesis ; therefore the straight lines EF, GH do not meet

С when produced on the side of FH; A in the same manner it may be proved,

E that EF, GH do not meet when pro

G duced on the side of EG : but straight lines which are in the same plane and do not meet, though produced either way, are parallel : therefore EF is parallel to GH. Wherefore, if two parallel planes, &c. Q. E. D.

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PROP. XVII. THEOR.

If two straight lines be cut by parallel planes, they shall be cut in the same ratio.

Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D; as AE is to EB, so is CF to FD.

Join AC, BD, AD, and let AD meet the plane KL in the point X: and join EX, XF: because the two parallel planes

* See Note.

KL, MN are cut by the plane EBDX, the common sections EX, BD, are parallel (16. 11.). For the same reason, because the two parallel planes GH, KL are cut by

С

H the plane AXFC, the common G

A sections AC, XF, are parallel : and because EX is parallel to BD, a side of the triangle ABD, as AE to EB, so is (2. 6.) AX to XD. Again, because XF is pa

L rallel to AC, a side of the triangle

E

F
ABC, as AX to XD, so is CF to K
FD: and it was proved that AX
is to XD, as AE to EB; there-

N fore (11. 5.), as AE to EB, so is

B CF to FD. Wherefore, if two M straight lines, &c. Q. E. D.

PROP. XVIII. THEOR.

If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane.

Let the straight line AB be at right angles to a plane CK; every plane which passes through AB shall be at right angles to the plane CK.

Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG in the plane DE at right D

G A H angles to CE; and because AB is perpendicular to the plane CK, therefore it is also perpendicular

K to every straight line in that plane meeting it (3. def. 11.); and consequently it is perpendicular, to CE: wherefore ABF is a right angle; but GFB is likewise a

С

F B right angle: therefore AB is parallel (28. 1.) to FG. And AB is at right angles to the plane CK: therefore FG is also at right angles to the same plane (8. 11.). But one plane is at right angles to another plane when the straight lines drawn in one of the planes at right angles to their common section, are also at right angles to the other plane (4. def. 11.): and any straight line FG in the plane DE, which is at right angles to CE the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D.

PROP. XIX. THEOR.

IF two planes cutting one another be each of them perpendicular to a third plane; their common section shall be perpendicular to the same plane.

Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common section of the first two; BD is perpendicular to the third plane.

If it be not, from the point D draw in the plane AB, the straight line DE at right angles to AD, the common section of the plane AB with the third plane; and in the plane BC draw DF at right angles to CD the common section of the plane BC with the third plane. And because the plane AB is perpendicular to

B the third plane, and DE is drawn in the plane AB at right angles to AD their common section, DE is perpendicular to the third plane (4. def. 11.). In the same manner, it may be

E F proved that DF is perpendicular to the third plane. Wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side of it, which is impossible (13. 11.): therefore, from the point D there cannot be any straight line at right an

D gles to the third plane, except BD the common section of the planes AB, BC. BD therefore is A

с perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D.

PROP. XX. THEOR.

If a solid angle be contained by three plane angles, any two of them are greater than the third.*

Let the solid angle at A be contained by the three plane angles, BAC, CAD, DAB. Any two of them are greater than the third.

If the angles BAC, CAD, DAB be all equal, it is evident that' any two of them are greater than the third. But if they be not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, make, in the plane which passes through BA, AC, the angle BAE equal (23. 1.) to the angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And because DA is

* See Note.

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