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Construction. Let any plane pass through the straight line AB, and let the plane be turned about AB until it pass through the point C.

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Demonstration. Because the points E and C are in this plane, therefore

1. The straight line CD is in it (I. Def. 7, and XI. 1), but the straight line AB is in the same plane (constr.); therefore

2. The straight lines AB and CD are in one plane. Again, because the points B and C are in the same plane with the point E in AB; therefore

1. The straight line BC is in this plane (I. Def. 7). But it has been proved that the straight lines AB and CD are in it; therefore 2. The three straight lines AB, BC, and CD, are in one plane.

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

PROPOSITION 3.-Theorem.

If two planes cut one another, their common section is a straight line..

Let two planes AB and BC cut one another, and let the line BD be their common section.

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Construction. If BD be not a straight line, from the point D to B, draw, in the plane AB, the straight line DEB (I. Post. 1), and in the plane BC, the straight line DFB.

Demonstration. Because the two straight lines DEB and DFB have the same extremities B and D, and do not coincide, therefore

1. DEB and DFB include a space between them, which is impossible (I. Ax. 10). Therefore

2. BD, the common section of the planes AB and BC, must be in a straight line.

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

PROPOSITION 4.-Theorem.

If a straight line stand at right angles to each of two straight lines at the point of their intersection, it is at right angles to the plane in which they are.

Let the straight line EF stand at right angles to each of the straight lines AB and CD, at E, the point of their intersection.

Then EF is at right angles to the plane of AB and CD, that is, the plane in which they are,

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Construction. Make EA, EB, EC, and ED all equal to one another (I. 3). Through E draw, in the plane of AB and CD, any straight line GH. Join AD and CB, and let GH meet them in Ğ and H. From any point F, in EF, draw FA, FG, FD, FC, FH, and FB.

Demonstration. Because the two sides AE and ED are equal to the two sides BE and EC, each to each, and they contain equal angles AED and.BEC (I. 15), therefore

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1. The base AD is equal to the base BC, and the angle DAE

to the angle EBC (I. 4).

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But the angle AEG is equal to the angle BEH (I. 15). Therefore the two triangles AEG and BE H have two angles of the one equal to two angles of the other, each to each, and the sides AE and EB, adjacent to the equal angles, equal to one another, therefore

2. Their other sides are equal, that is, GE is equal to EH, and

AG to BH (I. 26) ; and because AE is equal to EB, FE common, and the angle AEF equal to the angle BEF (hyp.), therefore

3. The base AF is equal to the base FB (I. 4). For the same reason, CF is equal to FD. And because AD is equal to BC, and AF to FB, the two sides FA and AD are equal to the two FB and BC, each to each. But the base DF has been proved equal to the base FC, therefore

4. The angle FĄD is equal to the angle FBC (I. 8). But it has been proved that GA is equal to BH, and also AF to FB; therefore

5. FA and AG are equal to FB and BH, each to each; and the angle FAG has been proved equal to the angle FBH, therefore

6. The base GF is equal to the base FH (I. 4). But it has been proved that GE is equal to EH, and EF is common. Therefore GE and EF are equal to HE and EF, each to each ; and the base GF is equal to the base FH, therefore

7. The angle GEF is equal to the angle HEF (I. 8), and each of these angles is a right angle (I. Def. 10); therefore

8. The straight line FE makes right angles with GH, that is,

with any straight line drawn through E in the plane of AB and CD.

In like manner,

may be proved that FE makes right angles with every other straight line meeting it in that plane. But a straight line is at right angles to a plane when it makes right angles with every straight line meeting it in that plane (XI. Def. 3) ; therefore 9. EF is at right angles to the plane of AB and CD.

Wherefore, if a straight line, &c. Q.E.D.

PROPOSITION 5.-Theorem.

If three straight lines meet all in one point, and a straight line stands at right angles to each of them at that point; these three straight lines are in the same plane.

Let the straight line AB stand at right angles to each of the straight lines BC, BD, and BE at the point B where they meet.

Then the straight lines BC, BD, and BE are in the same plane.

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Construction. If they be not in the same plane, let, if possible, BD and BE be in one plane, and BC out of it. Let a plane pass through AB and BC, and let the straight line BF be the common section of this plane, and the plane of BD and BE (XI. 3).

Demonstration. Because the three straight lines AB, BC, and BF are all in one plane (hyp.), viz., that which passes through AB and BC: and AB stands at right angles to each of the straight lines BD and BE; therefore AB is at right angles to the plane in which they are (XI. 4); wherefore

1. AB makes right angles with every straight line meeting it

in that plane (XI. Def. 3) ; but BF, which is in that plane, meets it; therefore

2. The angle ABF is a right angle. But the angle ABC is also a right angle (hyp.); therefore

3. The angle ABF is equal to the angle ABC, and they are

both in the same plane ;

which is impossible (1 Ax. 9); therefore

4. The straight line BC is not out of the plane of BD and

BE; Therefore

The three straight lines BC, BD, and BE are in the same plane.

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

PROPOSITION 6.-Theorem.

If two straight lines be at right angles to the same plane, they are parallel.

Let the straight lines AB and CD be at right angles to the same plane.

Then AB is parallel to CD.

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Construction. Let the straight lines meet the plane in the points B and D. Join BD, and draw DE at right angles to it in the same plane (I. 11). Make DE equal to AB (I. 3), and join BE, AE, and AD.

Demonstration. Because AB is perpendicular to the plane ; therefore

1. AB makes right angles with every straight line meeting it

in that plane (XI. Def. 3) ; but BD and BE in that plane each meet AB; therefore

2. Each of the angles ABD and ABE is a right angle ; for the same reason

3. Each of the angles CDB and CDE is a right angle. Because AB is equal to DE, and BD common, the two sides A B and

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