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PROPOSITION XXIV. THEOREM.

567. The acute angle which a straight line makes with its projection upon a plane is the least angle which it makes with any line of the plane.

M

A

B

N

Let BA meet the plane MN at A, and let AC be its projection upon the plane MN, and AD any other line drawn through A in the plane. To prove that BAC is less than ▲ BAD.

Proof.

Take AD equal to AC, and draw BD.

In the ABAC and BAD,

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Ex. 613. From a point A, 4 inches from a plane MN, an oblique line AC 5 inches long is drawn to the plane and made to turn around the perpendicular AB dropped from A to the plane. Find the area of the circle described by the point C.

Ex. 614. From a point A, 8 inches from a plane MN, a perpendicular AB is drawn to the plane; with B as centre, and a radius equal to 6 inches, a circle is described in the plane; at any point C of this circumference a tangent CD 24 inches long is drawn. Find the distance from A to D.

Ex. 615. Describe the relative position to a given plane of a line if its projection on the plane is equal to its own length.

PROPOSITION XXV. THEOREM.

568. Between two straight lines not in the same plane, there can be one common perpendicular, and only one.

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Let AB and DC be two lines not in the same plane.

To prove that there can be one common perpendicular between AB and DC, and only one.

Proof. Through any point A of AB draw AG || to DC, and let MN be the plane determined by AB and AG. Since AG is to DC, MN is | to DC.

§ 522

Through DC pass the plane PQL to MN, intersecting the plane MN in D'C'. Then D'C' is | to DC.

§ 525

D'C' must cut AB at some point C', otherwise AB would bell to D'C' (§ 103), and hence I to DC (§ 521). But this is impossible; for AB and DC are not in the same plane. Hyp. Draw C'C1 to MN. Then C'C is to AB. § 501 But C'C' is in the plane PQ (§ 552) and is .. C'C is 1 to DC.

.. C'C is 1 to AB and DC.

to D'C'.

§ 501

§ 107

Again, C'C is the only to both AB and DC.

For, if pos

sible, let EA be any other line

to AB and DC.

Then EA

is to AG (§ 107), and hence
Draw EH to D'C'. Then EH is 1 to MN (§ 551),
But this is impossible.

have two Is from E to MN.

Hence, C'C is the only common to DC and AB.

to MN.

§ 507

and we

§ 511

Q. E. D.

POLYHEDRAL ANGLES.

569. DEF. The opening of three or more planes which meet at a common point is called a polyhedral angle.

S

570. DEF. The common point S is the vertex of the angle, and the intersections of the planes SA, SB, etc., are its edges; the portions of the planes included between the edges are its faces, and the angles formed by the edges are its face angles.

B

571. The magnitude of a polyhedral angle depends upon the relative position of its faces, and not upon their

extent.

572. In a polyhedral angle, every two adjacent edges form a face angle, and every two adjacent faces form a dihedral angle. These face angles and dihedral angles are the parts of the polyhedral angle.

573. DEF. A polyhedral angle is convex, if every section made by a plane that cuts all its edges is a convex polygon.

574. DEF. A polyhedral angle is called trihedral, tetrahedral, etc., according as it has three faces, four faces, etc.

575. DEF. A trihedral angle is called rectangular, bi-rectangular, tri-rectangular, according as it has one, two, or three right dihedral angles.

576. DEF. A trihedral angle is called isosceles if it has two of its face angles equal.

577. DEF. Two polyhedral angles can be made to coincide and are equal if their corresponding parts are equal and arranged in the same order.

S

S'

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578. A polyhedral angle is designated by its vertex, or by its vertex and all the faces taken in order. Thus the poly

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hedral angle in the margin may be designated by S, or by S-ABCD.

579. If the faces of a polyhedral angle S-ABCD are produced through the vertex S, another polyhedral angle S-A'B'C'D' is formed, sym

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Also the dihedral angles SA, SB, etc., are equal, respectively, to the dihedral ▲

angles SA', SB', etc. § 547

B

(The second figure shows a pair of vertical dihedral angles.)

The edges of S-ABCD are arranged from left to right (counter clockwise) in the order SA, SB, SC, SD, but the edges of S-A'B'C'D' are arranged from right to left (clockwise) in the order SA', SB', SC', SD'; that is, in an order the reverse of the order of the edges in S-ABCD.

Two symmetrical polyhedral angles, therefore, have all their parts equal, each to each, but arranged in reverse order.

B'

In general, two symmetrical polyhedral angles are not superposable. Thus, if the trihedral angle S-A'B'C' is made to turn 180° about XY, the bisector of the angle A'SC, then SA' will coincide with SC, SC' with SA, and the face A'SC' with ASC; but the dihedral angle SA, and hence the dihedral angle SA', not being equal to SC, the plane A'SB' will not coincide with BSC; and, for a similar reason, the plane C'SB' will not coincide with ASB. Hence, the edge SB' takes some position SB" not coincident with SB; that is, the trihedral angles are not superposable.

B

C

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580. The sum of any two face angles of a trihedral angle is greater than the third face angle.

S

B

In the trihedral angle S-ABC, let the angle ASC be greater than ASB or BSC.

To prove
Proof. In ASC draw SD, making ▲ ASD equal to Z ASB.
Through any point D of SD draw ADC in the plane ASC.
Take SB equal to SD.

▲ ASB + ≤ BSC greater than ▲ ASC.

Pass a plane through the line AC and the point B.

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:: ≤ ASB + ≤ BSC is greater than ▲ ASD + ≤ DSC. That is, ASB + ≤ BSC is greater than ▲ ASC.

Q. E. D.

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