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Demonstration. Because CA is at right angles to the given plane,

therefore

V

A

E

1. CA makes right angles with every straight line meeting it in that plane (XI. Def. 3) ;

but DE, which is in that plane, meets CA; therefore

2. CAE is a right angle.

For the same reason, BAE is a right angle; wherefore the angle CAE is equal to the angle BAE (I. Ax. 11), therefore

3. CAE and BAE cre in one plane,

which is impossible (I. Ax. 9). Therefore two perpendiculars cannot be drawn from the same point in a plane, on the same side of it. Also, from a point above a plane, there can be but one perpendicular to it. For, if there could be two, they would be parallel to one another (XI. 6), which is absurd.

Therefore, from the same point, &c. Q.E.D.

PROPOSITION 14.—Theorem.

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 and EF.

Then the planes CD and EF are parallel to one another.

Construction. If not, they shall meet one another when produced. Let them be produced and meet; and let the straight line GH be their common section. In GH, take any point K, and join AK and BK.

Demonstration. Because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK meeting it in that plane (XI. Def. 3); therefore

1. ABK is a right angle.

For the same reason, BAK is a right angle; therefore

that is,

which is impossible (I. 17); therefore

2. The two angles ABK and BAK of the triangle ABK, are equal to two right angles,

3. The planes CD and EF, though produced, do not meet one another;

4. The planes CD and EF are parallel to one another (XI. Def. 8).

Therefore, planes to which, &c. Q.E.D.

PROPOSITION 15.-Theorem.

Two planes are parallel, if two straight lines which meet each other in the one plane, be parallel to two straight lines which meet each other in the other plane.

Let AB and BC, two straight lines which meet each other, be parallel to two straight lines DE and EF, which meet each other, but are not in the same plane with AB and BC.

Then the plane of AB and BC is parallel to the plane of DE and EF.

[graphic][subsumed][subsumed][subsumed]

Construction. From the point B draw BG perpendicular to the

plane of DE and EF (XI. 11); and let it meet that plane in G. Through G draw GH parallel to ED, and GK parallel to EF (I. 31).

[graphic][subsumed][subsumed][subsumed]

Demonstration. Because BG is perpendicular to the plane of DE and EF; therefore

1. BG makes right angles with every straight line meeting it in that plane (XI. Def. 3);

but the straight lines GH, GK meet it in that plane; therefore 2. Each of the angles BGH, BGK is a right angle.

And because BA is parallel to GH, for each of them is parallel to DE, and they are not both in the same plane with it (XI. 9), therefore

3. The angles GBA, BGH are together equal to two right angles (I. 29);

and BGH is a right angle; therefore also

4. GBA is a right angle, and GB perpendicular to BA;

for the same reason, GB is perpendicular to BC. Because the straight line GB stands at right angles to the two straight lines BA and BC meeting one another in B; therefore

5. GB is perpendicular to the plane of BA and BC (XI. 4.); but it is perpendicular to the plane of DE and EF (constr.), therefore

6. BG is perpendicular to each of the planes of AB and BC, and of DE and EF;

but planes to which the same straight line is perpendicular are parallel to one another (XI. 14); therefore

7. The plane of AB and BC is parallel to the plane of DE

and EF.

Wherefore, two planes are, &c. Q.E.D.

PROPOSITION 16.-Theorem.

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

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

Then EF is parallel to GH.

[graphic][subsumed]

Construction. For, if they be not parallel, EF and GH shall meet if produced either on the side of FH or EG. Let them be produced on the side of FH, and meet in the point K.

Demonstration. Because the straight line EFK is in the plane AB (XI. 1), therefore

1. The point K is in the plane AB;

for the same reason, the point K is also in the plane CD; wherefore 2. The planes AB and CD being produced, meet one another. But they do not meet, because they are parallel (hyp.); therefore 3. The straight lines EF and GH do not meet when produced on the side of FH.

1

In the same manner, it may be proved that EF and GH do not meet when produced on the side of EG. But straight lines which are in the same plane, and do not meet, though produced either way, are parallel (I. Def. 35); therefore

4. EF is parallel to GH.

Wherefore, if two parallel planes, &c. Q.E.D.

PROPOSITION 17.-Theorem.

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

Let the straight lines AB and CD be cut by the parallel planes GH, KL, and MN in the points A, E, and B; Č, F, and D.

Then as AE is to EB, so is CF to FD.

M

C

Construction. Join AC, BD, and AD, and let AD meet the plane KL in the point X. Join EX and XF.

Demonstration. Because the two parallel planes KL and MN are cut by the plane BX; therefore

1. The common sections EX and BD are parallel (XI. 16) ; for the same reason, because the two parallel planes GH and KL are cut by the plane XC; therefore

2. The common sections AC and XF are parallel (XI. 16). And because EX is parallel to BD, a side of the triangle ABD; therefore

3. As AE to EB, so is AX to XD (VI. 2).

Again, because XF is parallel to AC, a side of the triangle ADC; therefore

4. As AX is to XD, so is CF to FD.

But it was proved that AX is to XD, as AE is to EB; therefore 5. As AE is to EB, so is CF to FD (V. 11).

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

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