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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 given point in the given plane: it is required to erect a straight line from the point A, at right angles to the plane.

ID

B

From any point B without the plane, draw BC perpendicular to it (11. 11), and from A draw AD parallel to BC: Then, because AD, BC are two parallel straight lines, and one of them BC is at right angles to the given plane, the other AD is also at right angles to it (11. 8): Wherefore a straight line

has been erected at right angles to a given plane, from a given point in the same. Q. E.F.

PROP. XIII. THEOR.

From the same point in a given plane, there cannot be drawn 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 possible, let the two straight lines AB, AC be drawn 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, and let the common section of this

D

with the given plane be the straight line DAE: Then 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 the line DAE, which meets it in that plane, and therefore CAE is a right angle:

In like manner, BAE is a right angle; therefore the angle CAE is equal to the angle BAE, and they are in the same plane-which is impossible.

Also, from a point above a plane, there can be drawn but one perpendicular to the plane; for, if there could be two, they would be parallel to one another (11. 6)— which is absurd.

Wherefore, From the same point &c.

Q. E. D.

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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 shall be parallel to one another.

For, if not, they shall meet one another when produced: Let their common section

be the straight line HG, in which

take any point K, and join AK, BK: Then, because AB is perpendicular to the plane EF, it is perpendicular to the straight line BK which meets it in that plane, and therefore ABK is a right angle:

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In like manner, BAK is a right angle: Therefore the two angles ABK, BAK, of the triangle ABK, are equal to two right angles-which is impossible: Therefore the planes CD, EF, though produced, do not meet one another, that is, they are parallel.

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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 through the others.

Let AB, BC, two straight lines meeting one another, be parallel to DE, EF, which meet one another, but are not in the same plane with AB, BC: the plane through AB, BC, shall be parallel to the plane through DE, EF.

From the point B, draw BG at right angles to the plane through DE, EF, and let it meet it in G; and through G, draw GH parallel to ED, and GK paral- B lel to EF: Then, because BG is perpendicular to the plane through DE, EF, it makes right angles with the lines GH, GK which meet it

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in that plane, and therefore each of the angles BGH, BGK is a right angle: And because (11. 9) AB is parallel to GH (being each of them parallel to DE), the angles ABG, BGH are together equal to two right angles; and BGH is a right angle; therefore also ABG is a right angle, and GB perpendicular to BA: In like manner, GB is perpendicular to BC: And, because GB stands at right angles to each of the two BA, BC in their point of intersection, B, therefore GB is perpendicular to the plane through BA, BC: And it is also 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 to one another

(11. 14); therefore the plane through AB, BC is parallel to the plane through DE, EF.

Wherefore, If two straight lines &c.

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Q. E. D.

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 shall be parallel to GH.

For, if not, EF, GH shall meet, if produced, either on the side of F, H, or E, G: First, let them be produced on the side of F, H, and meet in the point K: Then, since EFK is in the plane AB, every point in EFK is in that plane: and K is a point in EFK; therefore K is in the plane AB: In like manner, K

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is in the plane CD: Therefore the planes AB, CD, produced, meet one another: But they do not meet, since (Hyp.) they are parallel; therefore the straight lines EF, GH do not meet when produced on the side of F, H: Similarly, it may be proved, that EF, GH do not meet when produced on the side of E, G: 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.

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Q. E. D.

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: AE shall be to EB as CF to FD.

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

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H

MN are cut by the plane EBDX, the common sections EX, BD are parallel (11. 16): In like manner, because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel: Now because EX is parallel to BD, a side of the triangle ABD, therefore AE is to EB as AX to XD: And, because XF is parallel to AC, a side of the triangle ADC, therefore AX is to XD as CF to FD: But it was proved that AX is to XD as AE to EB; therefore (5. 11) AE is to EB as CF to FD.

Wherefore, If two straight lines &c.

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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 the plane CK: then 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 angles to CE: Then, because AB

D

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