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BD are equal to the two ED and DB, each to each, and they contain right angles ; therefore

4. The base AD is equal to the base BE (I. 4). Again, because AB is equal to DE, and BE to AD; AB and BE are equal to ED and DA, each to each; and, in the triangles ABE and EDA, the base AE is common ; therefore

5. The angle ABE is equal to the angle EDA (I. 8). But ABE is a right angle, therefore

6. EDA is also a right angle, and ED perpendicular to

DA; but it is also perpendicular to each of the two BD and DC. Therefore ED is at right angles to each of the three straight lines BD, DA, and DC at the point D where they meet; wherefore

7. These three straight lines are all in the same plane (XI. 5). But AB is in the plane of BD and DA, because any three straight lines which meet in three points are in one plane (XI. 2); therefore

8. AB, BD, and DC are in one plane, and each of the angles ABD and BDC is a right angle; therefore 9. AB is parallel to CD (I. 28).

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

PROPOSITION 7.-Theorem. If two straight lines be parallel, the straight line drawn from any point in the one to any point in the other, is in the same plane with the parallels.

Let AB and CD be parallel straight lines, and E any point in the one, and F any point in the other.

Then the straight line which joins E and Fis in the same plane with the parallels.

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Construction. If this straight line be not in thé same plane with them, let it, if possible, be out of the plane, as EGF. In the plane AD of the parallels, draw the straight line EHF from E to F.

Demonstration. Because the two straight lines EHF and EGF have the same extremities and do not coincide,

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1. EHF and EGF include a space between them, which is impossible (I. Ax. 10); therefore

2. The straight line joining the points E and F is not out of

the plane of the parallels AB and CD. Wherefore 3. It is in that plane.

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

PROPOSITION 8.-Theorem.

If two straight lines be parallel, and one of them be at right angles to a plane, the other is also at right angles to the same plane.

Let AB and CD be two parallel straight lines, and let one of them AB be at right angles to a plane.

Then the other CD is also at right angles to the plane.

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Construction. Let AB and CD meet the plane in the points B and D, and join BD; therefore

1. The straight lines AB, CD, and BD are in one plane

(XI. 7). In the plane BDE draw DE at right angles to BD (I. 11); make DE equal to AB (I. 3), and join BE, AE, and AD.

Demonstration. Because AB is perpendicular to the plane, it is perpendicular to BE and BD (XI. Def. 3); therefore

2. Each of the angles ABD and ABE is a right angle. Because the straight line BD meets the parallel straight lines AB and CD; therefore

3. The angles ABD and CDB are together equal to two right

angles (I. 29). But ABD is a right angle ; therefore

4. CDB is also a right angle, and CD is perpendicular to BD. But it may be proved, as in the sixth proposition, that ED is perpendicular to DA, and it is also perpendicular to BD (constr.) Therefore ED is perpendicular to the plane of BD and DA (XI. 4); and therefore

5. ED makes right angles with every straight line meeting it

in that plane (XI. Def. 3). But DC is in the plane of BD and DA, because all three are in the plane of the parallels AB and CD; therefore

6. ED is at right angles to DC, and CD to DE. But CD is also at right angles to DB; therefore

7. CD is at right angles to the two straight lines DE and DB

at the point of their intersection D; wherefore

8. CD is at right angles to the plane of DE and DB, that is the same plane to which AB is at right angles (XI. 4).

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

PROPOSITION 9.-Theorem,

Two straight lines which are each of them parallel to the same straight line, but not in the same plane with it, are parallel to one another.

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Let AB and CD be each of them parallel to EF, but not in the same plane with it.

Then AB is parallel to CD.

Construction. In EF, take any point G, and from it, draw, in the plane of EF and AB, the straight line GH at right angles to EF (L 11); and in the plane of EF and CD, GK at right angles to EF.

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Demonstration. Because EF is perpendicular both to GH and GK; therefore

1. EF is perpendicular to the plane HGK passing through

them XI. 4). But EF is parallel to AB (hyp.), therefore

2. AB is at right angles to the plane HGK (XI. 8). For the same reason,

3. CD is at right angles to the plane HGK. Therefore

4. AB and CD are each of them at right angles to the plane

HGK. But if two straight lines are at right angles to the same plane, they are parallel (XI. 6); therefore 5. AB is parallel to CD.

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

PROPOSITION 10.-Theorem.

If two straight lines meeting one another be parallel to two others, which meet one another, but are not in the same plane with the first two; the first two and the other two contain equal angles.

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

Then the angle ABC is equal to the angle DEF. Construction. Make BA, BC, ED, and EF all equal to one another. Join AD, CF BE AC and DF.

Demonstration. Because BA is equal and parallel to ED, therefore

1. AD is equal and parallel to BE (I. 33). For the same reason, CF is equal and parallel to BE; therefore

2. AD and CF are each equal and parallel to BE; but straight lines that are parallel to the same straight line, but not in the same plane with it, are parallel to one another (XI. 9), therefore

3. AD is parallel to CF. Because AD is equal and parallel to CF (I. Ax. 1), and AC and DF join them towards the same parts ; therefore

4. AC is equal and parallel to DF (I. 33); and because AB and BC are equal to DE and EF, each to each, and the base AC to the base DF; therefore 5. The angle ABC is equal to the angle DEF (I. 8).

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

PROPOSITION 11.- Problem.

To draw a straight line perpendicular to a plane, from a given point above it.

Let A be the given point above the plane BH. It is required to draw from the point A a straight line perpendicular to the plane BH.

Construction. In the plane BH draw any straight line BC, and in the plane passing through BC and A, from the point A draw AD perpendicular to BC (I. 12). If AD be also perpendicular to the plane BH, what was required is done. But if it be not, from the point D, draw in the plane BH, the straight line DE at right angles to BC (I. 11) and in the plane passing through DE and A, from the point À, draw AF perpendicular to DE.

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