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Then, because the two sides AE, ED are equal to the two sides BE, EC, each to each,

[Construction. and that they contain equal angles AED, BEC; [I. 15. therefore the base AD is equal, to the base BC, and the angle DAE is equal to the angle EBC.

[I. 4. And the angle AEG is equal to the angle BEH; [I. 15. therefore the triangles AEG, BEH have two angles of the one equal to two angles of the other, each to each; and the sides EA, EB adjacent to the equal angles are equal to one another;

[Construction. therefore EG is equal to EH, and AG is equal to BH.

[I. 26. And because EA is equal to EB. [Construction. and EF is common and at right angles to them, [Hypothesis. therefore the base AF is equal to the base BF. [I. 4. For the same reason CF is equal to DF.

And since it has been shewn that the two sides DA, AF are equal to the two sides CB, BF, each to each, and that the base DF is equal to the base CF; therefore the angle DAF is equal to the angle CBF. [I. 8.

Again, since it has been shewn that the two sides FA, AG are equal to the two sides FB, BH, each to each, and that the angle FAG is equal to the angle FBH; therefore the base FG is equal to the base FH. [I. 4.

Lastly, since it has been shewn that GE is equal to HE, and EF is common to the two triangles FEG, FEH; and the base FG has been shewn equal to the base FH; therefore the angle FEG is equal to the angle FEH. [I. 8. Therefore each of these angles is a right angle. [I. Defin. 10.

In like manner it may be shewn that EF makes right angles with every straight line which meets it in the plane passing through AB, CD. Therefore EF is at right angles to the plane in which are AB, CD.

[XI. Definition 3. Wherefore, if a straight line &c. Q.E.D.

PROPOSITION 5. THEOREM. If three straight lines meet all at one point, and a straight line stand at right angles to each of them at that point, the three straight lines shall be in one and the same plane.

Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, at Ď the point where they meet : BC, BD, BE shall be in one and the same plane.

For, if not, let, if possible, BD and BE be in one plane, and BC without it; let a plane pass through AB and BC; the common section of this plane with the plane in which are BD and BE is a straight line ;

[XI. 3. let this straight line be BF. Then the three straight lines AB, BC, BF are all in one plane, namely, the plane which passes through AB and BC.

And because AB stands at right angles to each of the straight lines BD, BE,

[Hypothesis. therefore it is at right angles to the plane passing through them;

[XI. 4. therefore it makes right angles with every straight line meeting it in that plane.

[XI. Definition 3. But BF meets it, and is in that plane ; therefore the angle ABF is a right angle. But the angle ABC is, by hypothesis, a right angle; therefore the angle ABC is equal to the angle ABF;[Ax. 11. and they are in one plane; which is impossible. [Axiom 9. Therefore the straight line BC is not without the plane in which are BD and BE, therefore the three straight lines BC, BD, BE are in one and the same plane.

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

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PROPOSITION 6. THEOREM. If two straight lines be at right angles to the same plane, they shall be parallel to one another. · Let the straight lines AB, CD be at right angles to the same plane: AB shall be parallel to CD.

Let them meet the plane at the points B, D; join BD; and in the plane draw at right angles to BD; [1.11. make DE equal to AB; [I. 3. and join BE, AE, AD.

Then, because AB is perpendicular to the plane,

[Hypothesis. it makes right angles with every straight line meeting it in that plane. [XI. Def. 3.

But BD and BE meet AB, and are in that plane, therefore each of the angles ABD, ABE is a right angle. For the same reason each of the angles CDB, CDE is a right angle.

And because AB is equal to ED, [Construction. and BD is common to the two triangles ABD, EDB, the two sides AB, BD are equal to the two sides ED, DB, each to each; and the angle ABD is equal to the angle EDB, each of them being a right angle;

[Axiom 11. therefore the base AD is equal to the base EB. [I. 4.

Again, because AB is equal to ED, [Construction. and it has been shewn that BE is equal to DA; therefore the two sides AB, BE are equal to the two sides ED, DA, each to each; and the base AE is common to the two triangles ABE, EDA; therefore the angle ABE is equal to the angle EDA. [I. 8. But the angle ABE is a right angle, therefore the angle EDA is a right angle, that is, ED is at right angles to AD.

But ED is also at right angles to each of the two BD, CD; therefore ED is at right angles to each of the three straight lines BD, AD, CD, at the point at which they meet; therefore these three straight lines are all in the same plane. [XI. 5.

But AB is in the plane in which are BD, DA;

[XI. 2. therefore AB, BD, CD are in one plane. And each of the angles ABD, CDB is a right angle; therefore 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 one to any point in the other, is in the same plane with the parallels.

Let AB, CD be parallel straight lines, and take any point E in one and any point F in the other: the straight fine which joins E and È shall be in the same plane with the parallels.

For, if not, let it be, if possible, without the plane, as

A

B EGÉ; and in the plane ABCD, in which the parallels are, draw the straight line EHF from E to F.

Then, since EGF is also a straight line, [Hypothesis. the two straight lines EGF, EHF include a space between them; which is impossible.

[Axiom 10. Therefore the straight line joining the points E and F is not without the plane in which the parallels AB, CD are; therefore it is in that plane.

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

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PROPOSITION 8. THEOREM. If two straight lines be parallel, and one of them be at right angles to a plane, the other also shall be at right angles to the same plane.

Let AB, CD be two parallel straight lines; and let one of them AB be at right angles to a plane: the other CD shall be at right angles to the same plane.

Let AB, CD meet the plane at the points B, D; join BD; therefore AB, CD, BD are in one plane.

[XI. 7. In the plane to which AB is at right angles, draw DE at right angles to BD;

[I. 11. make DE equal to AB; [I. 3. and join BE, AE, AD.

Then, because AB is at right angles to the plane, [Hypothesis, it makes right angles with every straight line meeting it in that plane;

[XI. Definition 3. therefore each of the angles ABD, ABE is a right angle. And because the straight line BD meets the parallel straight lines AB, CD, the angles ABD, CDB are together equal to two right angles.

[I. 29. But the angle ABD is a right angle,

[Hypothesis. therefore the angle CDB is a right angle ; that is, CD is at right angles to BD.

And because AB is equal to ED, [Construction. and BD is common to the two triangles ABD, EDB; the two sides AB, BD are equal to the two sides ED, DB, each to each; and the angle ABD is equal to the angle EDB, each of them being a right angle;

[Axiom 11. therefore the base AD is equal to the base EB. [I. 4.

Again, because AB is equal to ED, [Construction. and BE has been shown equal to DA,

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