The Elements of Euclid for the Use of Schools and Colleges

Then since the straight line AB is in the plane, it can be produced in that plane; let it be produced to D; and let any plane pass through the straight line AD,and be turned

about until it pass through the point C.

Then, because the points B and C are in this plane, the straight line BC is in it.

[I. Definition 7. Therefore there are two straight lines ABC, ABD in the same plane, that have a common segment AB; but this is impossible.

[I. 11, Corollary.

Q.E.D.

Wherefore, one part of a straight line &c.

PROPOSITION 2. THEOREM.

Two straight lines which cut one another are in one plane; and three straight lines which meet one another are in one plane.

Let the two straight lines AB, CD cut one another at E: AB and CD shall be in one plane; and the three straight lines EC, CB, BE which meet one another, shall be in one plane.

Let any plane pass through the straight line EB, and let the plane be turned about EB, produced if necessary, until it pass through the point C.

Then, because the points E and Care in this plane, the straight line EC is in it; [I. Definition 7. for the same reason, the straight line BC is in the same plane; and, by hypothesis, EB is in it.

Therefore the three straight lines EC, CB, BE are in one plane.

But AB and CD are in the plane in which EB and EC

are;

therefore AB and CD are in one plane.

[XI. 1.

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, BC cut one another, and let BD be their common section: BD shall be a straight line.

If it be not, from B to D, draw

in the plane AB the straight line BED, and in the plane BC the straight line BFD.

[Postulate 1.

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Therefore BD, the common section of the planes AB and BC cannot but be 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 shall also be at right angles to the plane which passes through them, that is, to the plane in which they are.

Let the straight line EF stand at right angles to each of the straight lines AB, CD, at E, the point of their intersection: EF shall also be at right angles to the plane passing through AB, CD.

Take the straight lines AE, EB, CE, ED, all equal to one another; join AD, CB; through E draw in the plane in which are AB, CD, any straight line cutting AD at G, and CB at H; and from any point F in EF draw FA, FG, FD, FC, FH, FB.

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.

And because EA is equal to EB,

[I. 26. [Construction.

and EF is common and at right angles to them, [Hypothesis. therefore the base AF is equal to the base BF. For the same reason CF is equal to DF.

[I. 4.

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 В 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 [XI. 3.

line;

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.

But BF meets it, and is in that plane;

therefore the angle ABF is a right angle.

[XI. Definition 3.

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.

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 DE at right angles to BD; [I.11. make DE equal to AB;

and join BE, AE, AD.

[I. 3.

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;

therefore the base AD is equal to the base EB.

[Axiom 11.

[I. 4.

[Construction.

Again, because AB is equal to ED, 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.

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