be the half-side of this polygon; C its centre. In the isoperimetrical circle, let the angle DOE be equal to ACI, and consequently, the arc DE be equal to the half-side AI. The polygon P is to the circle C as the triangle ACI is to the sector ODE; hence PC:: AI.CI: DE.OE: CI: OE. From the point E draw a tangent EG meeting OD produced in G the similar triangles ACI, GOE will give the proportion CI: OE :: AI or DE: GE; hence PC:: DE: GE, or as DE.OE, which is the measure of the sector DOE, is to GE. ¿OE, which is the measure of the triangle GOE: now this sector is less than the triangle; hence P is less than C; hence the circle is greater than any isoperimetrical polygon.

BOOK. V.

PLANES AND SOLID ANGLES.

Definitions.

313. A straight line is perpendicular to a plane, when it is perpendicular to all the straight lines (326.) which pass through its foot in the plane. Conversely, the plane is perpendicular to the line.

The foot of the perpendicular is the point at which that line meets the plane.

314. A line is parallel to a plane, when it cannot meet that plane, to whatever distance both be produced. Conversely, the plane is parallel to the line.

315. Two planes are parallel to each other, when they cannot meet, to whatever distance both be produced.

316. It will be demonstrated (324.) that the common intersection of two planes which meet each other, is a straight line that granted, the angle or mutual inclination of two planes is the quantity, greater or less, by which they are separated from each other; this quantity is measured by the angle contained between two lines, one in each plane, and both perpendicular to the common intersection at the same point. This angle may be acute, or right, or obtuse.

317. If it is right, the two planes are perpendicular to each other.

318. A solid angle is the angular space included between several planes which meet at the same point.

Thus, the solid angle S, (see the fig. of Art. 364.) is formed by the union of the planes ASB, BSC, CSD, DSA. Three planes at least, are requisite to form a solid angle.

319.

A straight line cannot be partly in a plane, and partly out

of it.

For, by the definition of a plane, when a straight line has two points common with a plane, it lies wholly in that plane.

320.

Scholium. To discover whether a surface is plane, it is necessary to apply a straight line in different ways to that surface, and to observe if it touches the surface throughout its whole extent.

321. Two straight lines, which intersect each other, lie in the same plane, and determine its position.

Let AB, AC be two straight lines which intersect each other in A; a plane may be conceived in which the straight line AB is found; if this plane be turned round AB, until it pass through the point C, then the B line AC, which has two of its points A and C in this plane, lies wholly in it; hence the

position of the plane is determined by the single condition of containing the two straight lines AB, AC.

322. Cor. 1. A triangle ABC, or three points A, B, C

not in a straight line, determine the position of a plane.

323. Cor. 2. Hence also two

parallels AB, CD determine the po- A sition of a plane; for, drawing the

secant EF, the plane of the two

straight lines AE, EF is that of the C parallels AB, CD.

E/

B

D

THEOREM. 3.

324. If two planes cut each other, their common intersection will be a straight line.

For, if among the points common to the two planes, there be three which are not in the same straight line, then the planes, passing each through these three points, must form only one and the same plane; which contradicts the hypothesis.

325. If two straight lines intersect each other, and a third line is perpendicular to both of them at their point of intersection, it will also be perpendicular to all lines drawn through its foot and in the plane of the two first lines, and will, therefore, be perpendicular to the plane of those lines.

Let AP be perpendicular to PB, PC, at the point P, and NM the plane of the lines CP, BD; then will AP be perpendicular to any line of the plane passing through P, and consequently to the plane itself (313.).

Through any point Q in PQ, draw (242.) the straight line BC in the angle BPC, so that BQ=QC; join AB, AQ, AC.

The base BC being divided into two equal parts at the point Q, the triangle BPC (194.) will give

PC2+PB2=2PQ2+2QC2.

The triangle BAC will in like manner give

M

A

B

P

B

N

AC2+AB2=2AQ2+2QC2. ·

Taking the first equation from the second, and observing that the triangles APC, APB, which are both right-angled at P, give AC3-PC2=AP2, and AB2—PB2=AP2; we shall have

AP2+AP=2AQ2-2PQ2.

Therefore, by taking the halves of both, we have AP2= AQ2—PQ2, or AQ2=AP2+PQ3; hence the triangle APQ is right-angled at P; hence AP is perpendicular to PQ.

326. Scholium. Thus it is evident, not only that a straight line may be perpendicular to all the straight lines which pass through its foot in a plane, but that it always must be so, whenever it is perpendicular to two straight lines drawn in the plane; which proves our first Definition to be

accurate.

327. Cor. 1. The perpendicular AP is shorter than any oblique line AQ; therefore it measures the true distance from the point A to the plane MN.

328. Cor. 2. At a given point P on a plane, it is impossible to erect more than one perpendicular to that plane; for if there could be two perpendiculars at the same point P, draw through these two perpendiculars a plane, whose intersection with the plane MN is PQ; then these two perpendiculars would be perpendicular to the line PQ, at the same point, and in the same plane, which is impossible (50.).

It is also impossible to let fall from a given point out of a plane two perpendiculars to that plane; for let AP, AQ, be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible.

THEOREM.

5

329. Oblique lines equally distant from the perpendicular are equal; and, of two oblique lines unequally distant from the perpendicular, the more distant, is the longer.

For the angles APB, APC, APD, being right, if we suppose the distances PB, PC, M PD, to be equal to each other, the triangles APB, APC, APD, will have each an equal angle contained by equal sides; therefore they will be equal; therefore the hypotenuses, or the oblique lines AB, AC,

P

E

B

AD, will be equal to each other. In like manner, if the distance FE is greater than PD or its equal PB, the oblique line AE will evidently be greater than AB, or its equal AD.

330. Cor. All the equal oblique lines AB, AC, AD, &c. terminate in the circumference BCD, described from P the foot of the perpendicular as a centre; therefore a point A

being given out of a plane, the point P at which the perpendicular let fall from A would meet that plane, may be found by marking upon that plane three points B, C, D, equally distant from the point A, and then finding the centre of the circle which passes through these points; this centre will be P, the point sought.

331. Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN; which inclination is evidently equal with respect to all such lines AB, AC, AD, as are equally distant from the perpendicular; for all the triangles ABP, ACP, ADP, &c. are equal to each other.

332. If from a point without a plane, a perpendicular be let fall on the plane, and from the foot of the perpendicular a perpendicular be drawn to any line of the plane, and from the point of intersection a line be drawn to the first point, this latter line will be perpendicular to the line of the plane.

Let AP be perpendicular to the plane NM, PD perpendicular to BC; then will AD also be perpendicular M to BC.

P

C

D

N

Take DB=DC, and join PB, PC, AB,AC. Since DB=DC, the oblique line PB=PC: and with regard to the perpendicular AP, since PB= PC, the oblique line AB÷AC (329.); therefore the line AD has two of its points A and D equally distant from the extremities B and C ; therefore AD is a perpendicular at the middle of BC (55.).

E

333. Cor. It is evident likewise, that BC is perpendicular to the plane APD, since BC is at once perpendicular to the two straight lines AD, PD.

334. Scholium, The two straight lines AE, BC afford an instance of two lines which do not meet, because they are not situated in the same plane. The shortest distance between these lines is the straight line PD, which is at once perpendicular to the line AP and to the line BC. The distance PD is the shortest distance between these two lines; for if we join any other two points, such as A and B, we shall have AB7 AD, AD 7PD; therefore AB 7PD.