Imágenes de páginas
PDF
EPUB

THEOREM XVII.

602. The locus of all points from which the two perpendiculars onto the same sides of two fixed planes are equal, is a plane determined by one such point and the intersection line of the two given planes.

[merged small][merged small][merged small][merged small][merged small][ocr errors]

HYPOTHESIS. Let AB and CD be the two given planes, and K one such point.

CONCLUSION. Perpendiculars dropped on to AB and CD from any point in the plane determined by K and the intersection line BC are equal.

[ocr errors]

PROOF. Take P any point in the plane KCB. Draw PH 1 plane AB, and PF plane CD. Call G the point where the plane FPH cuts the line BC. Join FG, PG, GH.

Draw KC || PG. Because the perpendiculars from K are given equal, therefore KC is equally inclined to the two planes. But PG has the same inclination to each as KC;

(591. Parallels intersecting the same plane are equally inclined to it.)

[merged small][ocr errors][merged small][ocr errors][merged small]

THEOREM XVIII.

603. If three lines not in the same plane meet at one point, any two of the angles formed are together greater than the third.

[blocks in formation]

PROOF. The theorem requires proof only when the third angle considered is greater than each of the others. In the plane of the greatest BAC, make X BAF Through F draw a line BFC cutting AB in B, and AC in C. Join DB and DC.

BAD. Make AF
Make AF = AD.

Δ BAD = Δ ΒΑΡ,

(124. Triangles having two sides and the included angle respectively equal are

[blocks in formation]

But from ABCD we have BD + DC > BC.

(156. Any two sides of a triangle are together greater than the third.)

And, taking away the equals BD and BF,

DC > FC,

in As CAD and CAF, we have & CAD > ☀ CAF.

(161. If two triangles have two sides respectively equal, but the third side greater in the first, its opposite angle is greater in the first.)

Adding the equal s BAD and BAF gives

¥ BAD + X CAD > × BAC.

THEOREM XIX.

04. If the vertices of a convex polygon be joined to a point not in its plane, the sum of the vertical angles of the triangles so made is less than a perigon.

[merged small][ocr errors][merged small]

PROOF. The sum of the angles of the triangles which have the common vertex S is equal to the sum of the angles of the same number of triangles having their vertices at O in the plane of the polygon. But

× SAB + 4 SAE > & BAE,

¥ SBA + 4 SBC > & ABC, etc.

(603. If three lines not in a plane meet at a point, any two of the angles formed are together greater than the third.)

Hence, summing all these inequalities, the sum of the angles at the bases of the triangles whose vertex is S, is greater than the sum of the angles at the bases of the triangles whose vertex is 0; therefore the sum of the angles at S is less than the sum of the angles at O, that is, less than a perigon.

BOOK VIII.

TRI-DIMENSIONAL SPHERICS.

605. If one end point of a sect is fixed, the locus of the other end point is a Sphere.

606. The fixed end point is called the Center of the sphere.

[blocks in formation]

607. The moving sect in any position is called the Radius of the sphere.

608. As the motion of a sect does not change it, all radii are equal.

609. The sphere is a closed surface; for it has two points on every line passing through the center, and the center is midway between them.

610. Two such points are called Opposite Points of the sphere, and the sect between them is called a Diameter.

611. The sect from a point to the center is less than, equal to, or greater than, the radius, according as the point is within, on, or without, the sphere.

For, if a point is on the sphere, the sect drawn to it from the center is a radius; if the point is within the sphere, it lies on some radius; if without, it lies on the extension of some radius.

612. By 33, Rule of Inversion, a point is within, on, or without, the sphere, according as the sect to it from the center is less than, equal to, or greater than, the radius.

THEOREM I.

613. The common section of a sphere and a plane is a circle.

A

Take any sphere with center O.

Let A, B, C, etc., be points common to the sphere and a plane, and OD the perpendicular from O to the plane. Then

OA = OB = OC etc.,

being radii of the sphere,

ABC, etc., is a circle with center D.

(598. Equal obliques from a point to a plane meet the plane in a circle whose center is the foot of the perpendicular from the point to the plane.)

« AnteriorContinuar »