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The axis of the cylinder is the fixed straight line about which the rectangle revolves.

The bases, or ends, of the cylinder are the circular faces described by the two revolving opposite sides of the rectangle.

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24. A right cone is a solid figure described by the revolution of a rightangled triangle about one of the sides containing the right angle which remains fixed.

The axis of the cone is the fixed straight line about which the triangle revolves.

The base of the cone is the circular face described by that side which revolves.

The hypotenuse of the right-angled triangle in any one of its positions is called a generating line of the cone.

25. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals.

PROPOSITION 1. THEOREM.

One part of a straight line cannot be in a plane and another part outside it.

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If possible, let AB, part of the st. line ABC, be in the plane PQ, and the part BC outside it.

Then since the st. line AB is in the plane PQ,
.. it can be produced in that plane.

that plane. I. Post. 2.

Produce AB to D; and let any other plane which passes through AD be turned about AD until it passes also through C. Then because the points B and C are in this plane,

... the st. line BC is in it: 1. Def. 7. .. ABC and ABD are in the same plane and are both st. lines; which is impossible.

1. Def. 4. .. the st. line ABC has not one part AB in the plane PQ, and another part BC outside it.

Q.E.D.

Note. This proposition scarcely needs proof, for the truth of it follows immediately from the definitions of a straight line and a plane.

It should be observed that the method of proof used in this and the next proposition rests upon the following axiom :

If a plane of unlimited extent turns about a fixed straight line as an axis, it can be made io pass through any point in space.

PROPOSITION 2. THEOREM. Any two intersecting straight lines are in one plane : and any three straight lines, of which each pair intersect one another, are in one plane.

A

B

Let the two st. lines AB and CD intersect at E; and let the st. line BC be drawn cutting AB and CD at B and C.

Then (i) AB and CD shall lie in one plane.

(ii) AB, BC, CD shall lie in one plane. (i) Let any plane pass through AB; and let this plane be turned about AB until it passes through C.

Then, since C and E are points in this plane, .:. the whole st. line CED is in it. I. Def. 7 and xi. 1.

That is, AB and CD lie in one plane. (ii) And since B and C are points in the plane which contains AB and CD,

... also the st. line BC lies in this plane. Q.E.D.

COROLLARY. One, and only one, plane can be made to pass through two given intersecting straight lines.

Hence the position of a plane is fixed, (i) if it passes through a given straight line and a given point outside it;

Ax. p. 419. (ii) if it passes through two intersecting straight lines; XI. 2. (iii) if it passes through three points not collinear ; XI. 2. (iv) if it passes through two parallel straight lines. I. Def. 35.

PROPOSITION 3. THEOREM. If two planes cut one another, their common section is a straight line.

С

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Let the two planes XA, CY cut one another, and let BD be their common section.

Then shall BD be a straight line. For if not, from B to D in the plane XA draw the st. line BED;

and in the plane CY draw the st. line BFD. Then the st. lines BED, BFD have the same extremities;

.:. they include a space ;
but this is impossible.

Ax. 10. is the common section BD cannot be otherwise than a st.

line.

ALTERNATIVE PROOF.

Let the planes XA, CY cut one another, and let B and D be

two points in their common section. Then because B and D are two points in the plane XA, ... the st. line joining B, D lies in that plane. I. Def. 7. And because B and D are two points in the plane CY, .. the st. line joining B, D lies in that plane.

Hence the st. line BD lies in both planes,

and is therefore their common section. That is, the common section of the two planes is a straight line.

Q.E.D.

PROPOSITION 4. THEOREM. Alternative Proof.]

If a straight line is perpendicular to each of two straight lines at their point of intersection, it shall also be perpendicular to the plane in which they lie.

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Let the straight line AD be perp. to each of the st. lines AB, AC at A their point of intersection. Then shall AD be perp. to the plane in which AB and AC lie. Produce DA to F, making AF equal to DA.

Draw any st. line Bc in the plane of AB, AC, to cut
AB, AC at B and C;
and in the same plane draw through A any st. line AE to cut
BC at E.
It is required to prove that AD is perp. to AE. XI. Def. 1.

Join DB, DE, DC; and FB, FE, FC.
Then in the A' BAD, BAF,
because DA= FA,

Constr.
and the common side AB is perp. to DA, FA;
BD=BF.

I. 4. Similarly CD=CF. Now if the A BFC be turned about its base BC until the vertex F comes into the plane of the A BDC,

then F will coincide with D, since the conterminous sides of the triangles are equal. 1. 7

.. EF will coincide with ED,

that is, EF=ED.

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