Greatest of Isoperimetrical Figures.

which it will do, for the angle TNN', being less than the right angle TNL is acute.

Join N'T, and, by art. 42,

NT NT.

But since the concave broken line TNM is included by TN'M', we have

TN+N'M > TN + NM',

whence, omitting TN equal to TN,

309. Corollary. As the circle is a polygon of an infinite number of sides, that is, of a greater number of sides than any other regular polygon, it is greater than any polygon of a finite number of sides which has a perimeter equal to the circumference of the circle.

SOLID GEOMETRY.

CHAPTER XV.

PLANES AND SOLID ANGLES.

310. Theorem. Three points not in the same straight line determine the position of the plane in which they are situated.

Demonstration. For if any plane, passing through two of the points, is swung around the line joining these two points, until it comes to a position in which it passes through the third point, it must remain in this position. For swinging any farther must remove it from this third point.

it

311. Corollary. Only one plane can be drawn through three points not in the same straight line.

312. Theorem. The common intersection of two planes, which cut each other, is a straight line.

Demonstration. For, if any two of the points common to the two planes be joined by a straight line, this straight line must, by art. 14, be in both of the planes; and no point out of this straight line can, by art. 311, be in the two different planes at the same time.

313. When two planes cut each other, they form an angle, the magnitude of which does not depend

Intersection and Angle of two Planes.

upon the extent, but merely upon the position of the planes.

314. Theorem. The angle of two planes, which cut each other, is measured by the angle of two lines drawn perpendicular to the common intersection of the two planes at the same point, one in one of the planes, and one in the other.

Demonstration. In order to show the legitimacy of this measure we have only to prove that the angle of the two lines is proportional to the angle of the two planes.

Let AB (fig. 147) be the common intersection of the two planes; and let AC and AD be the two lines drawn in these planes perpendicular to the common intersection AB.

Let a third plane be drawn having also the common intersection AB with the two given planes, and let AE be drawn in this plane perpendicular to AB. We are to prove that the angle of the planes DAB and CAB is to that of the two planes EAB and CAB as DAC is to EAC.

For this purpose, suppose the angles of the planes to be to each other as any two whole numbers, and let the angle of the two planes CAB and a AB be their common divisor, Aa being perpendicular to AB. The angle CAa must be a common divisor of the two angles CAE and CAD; and it is shown by precisely the same reasoning so often adopted, that the angles of the planes are to each other as CAD to CAE.

315. Corollary. When the angle CAD is a right angle, the planes are perpendicular to each other.

316. Definitions. A straight line is perpendicular

Line Perpendicular to a Plane.

to a plane, when it is perpendicular to every straight line drawn through its foot in the plane.

Reciprocally, the plane, in this case, is perpendicular to the line.

The foot of the perpendicular is the point in which it meets the plane.

317. Theorem. When a straight line is perpendicular to two straight lines drawn through its foot in a plane, it is perpendicular to every other straight line drawn through its foot in the plane, and, consequently, is perpendicular to the plane.

Demonstration. Let CAC', DAD' (fig. 148) be the two lines to which AB is perpendicular, and let EAE' be any other line drawn in the plane, we are to prove that BA is perpendicular to EAE'.

Take AC equal to AC', and AD equal to AD', join DC D'C'.

The two triangles DAC and D'AC' are equal, by art. 51, for AC AC', AD AD' and the included angle DAC = D'AC', by art. 23; and, therefore, the angle AC'D' = ACD and D'C DC.

The two triangles AC'E' and ACE are equal, by art. 53; for AC AC', the angle ACE ACE and the angle

CAE its vertical angle C'AE'; hence AE

ᏟᎬ .

=CE.

Join BC, BC, BD, BD, BE and BE!

The two triangles BCD and BC'D' are equal, by art. 61; for DC D'C', the sides BD and BD' are equal, by art. 38, being oblique lines drawn to DD' at equal distances from the perpendicular, and in like manner BC = BC; hence the angle BCD = BC'D'.

Oblique Lines drawn to a Plane.

The two triangles BCE and BCE are equal, by art. 51; for BC= B'C', CE CE' and the angle BCE BCE'; hence BE

BE'.

The points B and A of the line BA are consequently at equal distances E and E' from the extremities of EE'; and, therefore, BA is, by art. 42, perpendicular to EE'.

318. Corollary. The perpendicular BA is less than any oblique line BE, and measures 'the distance of the point B, from the plane.

319. Theorem. Oblique lines drawn from a point to a plane at equal distances from the perpendicular are equal; and of two oblique lines unequally distant the more remote is the greater.

Demonstration. a. The oblique lines BC, BD, BE &c. (fig. 149) at the equal distances AC, AD, AE &c. from the perpendicular BA are equal; for the triangles BAC, BAD, BAE &c. are equal, by art. 51, since the angles BAC, BAD, BAE &c. are equal, being right angles, the sides AC, AD, AE &c. are equal, and the side BA is

common.

b. Since the oblique line BC is drawn to the line AC at a distance AC greater than AC from the perpendicular BA, it is, by art. 41, greater than BC or its equal BD or BE.

320. Corollary. All the equal oblique lines BC, BD, BE &c. terminate in the circumference CDE, drawn with A as a centre, and a radius equal to AC.

321. Theorem. If a line is perpendicular to a plane, every line which is parallel to this perpendicular, is likewise perpendicular to the plane.