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then will af be perpendicular to the plane BG, as was required.

For, through the point F, draw the line ng parallel to the line BC (I. 27.)

Then since the right line bc is perpendicular to each of the right lines DA, DE, it will also be perpendicular to the plane which passes through those lines (VII. 3:)

And because the lines bc, HG are parallel to each other, and one of them, bc, is perpendicular to the plane ADF, the other, HG, will also be perpendicular to that plane (VII. 5.)

But if a line be perpendicular to a plane it will be perpendicular to all the lines which meet it in that plane (VII. Def. 2.); whence the line ng is perpendicular

to AF.

And since the line AF is perpendicular to each of the lines HG, ED, at their point of intersection F, it will also be perpendicular to the plane BG (VII.3), as was to be shewn.

PRO P. X. THEOREM.

Planes to which the same right line is perpendicular, are parallel to each other.

H

A

Let the right line AB be perpendicular to each of the planes CD, EF; then will those planes be parallel to each other.

For if they be not, let them be produced till they meet each other; and in the line GH, which is their common fection, take any point K; and join KA, KB:

Then, because the line AB is perpendicular to the plane EF (Uy Hyp.), it will also be perpendicular to the line BK, which lies in that plane (VII. Def. 2.); and the angle ABK will be a right angle.

And, for the same reason, the line AB, which is perpendicular to the plane Dc (by Hyp.), will be perpendicular to the line AK; and the angle BAK will also be a right angle.

The angles ABK, BAK are, therefore, equal to two right angles, which is absurd (I. 28.); and consequently the planes can never meet, but must be parallel to each other (VII. Def. 6.), as was to be shewn.

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If two right lines which meet each other, be parallel to two other right lines which meet each other, though not in the same plane with them, the planes which pass through those lines will be parallel.

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Let the right lines AB, BC, which meet each other in B, be parallel to the right lines DE, EF, which meet each other in E, though not in the same plane with them ; then will the plane ABC be parallel to the plane DEF.

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For through the point B draw BG perpendicular to the plane DFE (VII. 9.); and make GH parallel to De, and GK TO EF (I. 27.)

Then because BG is at right angles with the plane dhe, it will also be at right angles with each of the lines GH, GK which meet it in that plane (Def. 2.)

And since gh is parallel to de or AB (by Conft. and VII. 6.), and BG intersects them, the angles BGH, GBA are, together, equal to two right angles (I. 25.)

But the angle byh has been shewn to be a right angle; whence the angle gba is also a right angle; and consequently go is perpendicular to BA.

And, in the same manner, it may be fhewn, that GB is perpendicular to BC.

The right line GB, therefore, being perpendicular to each of the right lines BA, BC, will also be perpendicu. lar to the plane ACB through which they pass (VII. 3.)

But planes to which the same right line is perpendicu. lar are parallel to each other (VII. 10.); whence the plane ACB is parallel to the plane DFE.

Q. E. D.

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PRO P. XII. THEOREM.

If

any two parallel planes be cut by another plane, their common sections will be parallel.

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Let the two parallel planes AB, cd be cut by the plane EGHF; then will their common sections EF, GH be parallel to each other.

For if EF, GH be not parallel, they may be produced till they meet, either on the side FH, or the

fide EG.

Let them be produced on the side FH, and meet each other in the point K.

Then, since the whole line EFK is in the plane AB, or the plane produced, the point k must be in that plane.

And because the whole line ghk is in the plane cd, or the plane produced, the point k must also be in that plane.

Since, therefore, the point k is in each of the planes AB, CD, those planes, if produced, will meet in that point.

But the two planes are parallel to each other, by hypothesis; whence they meet, and are parallel, at the same time, which is absurd.

The lines EF, GH, therefore, do not meet on the side FH; and, in the fame manner, it may be proved, that they do not meet on the side EG; confequently they are parallel to each other.

Q. E. D.

PRO P. XIII. THEORE M.

If a right line be perpendicular to a plane, every plane which passes through it will also be perpendicular to that plane.

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Let the right line AB be perpendicular to the plane CK; then will every plane which passes through that line be also perpendicular to ck.

For let Éd be any plane which passes by the line AB; and in this plane draw any right line GF perpendicular to the common section ce (1. 11.)

Then, because the line AB is perpendicular to the plane CK (by Hyp.), it will also be perpendicular to the line ce; and the angle ABF will be a right angle (VII. Def. 2.)

And since the angles ABF, GFB are each of them a right angle, and the lines AB, GF are in the same pláne, they will be parallel to each other (VII. 4.)

Since, therefore, these lines are parallel to each other, and one of them, AB, is perpendicular to the plane ck,

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