will be perpendicular to every other straight line PQ drawn through its foot in the same plane, and thus it will be perpendicular to the plane MN.

Demonstration. Through a point Q, taken at pleasure in PQ, draw the straight line BC in the angle BPC making BQ = QC (242); join AB, AQ, AC.

The base BC being bisected at the point Q, the triangle BPC will give

2

PC + PB = 2PQ+2QC (194).

The triangle BAC will give, in like manner,

AC‍+AB=2AQ+2QT.

If we subtract the first equation from the second, and recollect that the triangles APC, APB, each right-angled at P, give

-2

AC — PC=AP, AB — PB – AP; we shall have

-2

=

AP+AP=2AQ — 2PQ;

or, by taking half of each member,

.

therefore the triangle APQ is right-angled at P (193), and AP is perpendicular to PQ.

326. Scholium. It is evident, then, not only that a straight line may be perpendicular to all those which pass through its foot in the plane, but that this happens, whenever the line in question is perpendicular to two straight lines drawn in the plane; hence the propriety of the definition art. 313.

327. Corollary 1. The perpendicular AP is less than any oblique line AQ; therefore it measures the true distance of a point A from the plane PQ.

328. Corollary 1. Through any given point P in a plane only one perpendicular can be drawn to this plane; for, if there could be two, a plane being supposed to pass through them intersecting the plane MN in PQ, the 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, without a plane two perpendiculars to this plane; for let AP, AQ, be these two perpendiculars, then the triangle APQ would have two right angles APQ, AQP, which is impossible.

Fig. 184.

185.

THEOREM.

329. Oblique lines equally distant from the perpendicular are equal; and of two oblique lines unequally distant from the perpendicular, that which is at the greater distance is the greater.

Demonstration. The angles APB, APC, APD (fig. 184), being right angles, if we suppose the distances PB, PC, PD, equal to each other, the triangles APB, APC, APD, have two sides and the included angle respectively equal, they are consequently equal; therefore the hypothenuses or the oblique lines AB, AC, AD, are equal to each other. Likewise if the distance PE is greater than PD or its equal PB, it is evident that the oblique line AE will be greater than AB or its equal AD.

330. Corollary. All the equal oblique lines AB, AC, AD, &c., terminate in the circumference of a circle BCD described about the foot of the perpendicular P, as a centre; therefore, a point A without a plane being given, to find the point P where the perpendicular A meets this plane, take three points B, C, D, equally distant from the point A, and find the centre of the circle which passes through these points, this centre will be the point Prequired.

331. Scholium. The angle ABP is called the inclination of the oblique line AB to the plane MN. It is manifest that this inclination is the same for all the oblique lines AB, AC, AD, &c., which depart equally from the perpendicular; for all the triangles ABP, ACP, ADP, &c., are equal.

THEOREM.

332. Let AP (fig. 185) be a perpendicular to the plane MN, and BC a line situated in this plane; if from the foot P of the perpendicular a line PD be drawn perpendicular to BC, and AD be joined, AD will be perpendicular to BC.

Demonstration. Take DB = DC, and join PB, PC, AB, AC. Since DB DC, the oblique line PB = PC; and, because PB PC, the oblique lines AB, AC, considered with reference to the perpendicular AP, are equal (329); hence the line AD has two points A and D each equally distant from the extremities B and C; therefore AD is perpendicular to BC (55).

333. Corollary. It will be seen, at the same time, that BC is perpendicular to the plane APD, since BC is perpendicular at the same time to the two straight lines AD and PD.

334. Scholium. The two lines AE, BC, present an example of two lines which do not meet, because they are not situated in the same plane. The least distance of these lines is the straight line PD, which is at the same time perpendicular to the line AP and to the line BC. The distance PD is the shortest; because, if we join two other points, as A and B, we shall have ABAD, AD>PD, and, for a still stronger reason, AB >PD.

The two lines AE, CB, although not situated in the same plane, are considered as making a right angle with each other, because AE and a line drawn through any point in it parallel to BC, would make a right angle with each other. In like manner, the line AB and the line PD, which represent two straight lines not situated in the same plane, are considered as making the same angle with each other, as is made by AB and a line parallel to PD drawn through some point in AB.

THEOREM.

335. If the line AP (fig. 186) is perpendicular to the plane MN, Fig. 186. every line DE parallel to AP will be perpendicular to the same plane.

Demonstration. Let there be a plane passing through the parallels AP, DE, intersecting the plane MN in PD; in the plane MN draw BC perpendicular to PD, and join AD.

According to the corollary of the preceding theorem BC is perpendicular to the plane APDE; consequently the angle BDE is a right angle; but the angle EDP is also a right angle, since AP is perpendicular to PD, and DE is parallel to AP (65); hence the line DE is perpendicular to each of the lines DP, DB; therefore it is perpendicular to the plane MN passing through them (325).

336. Corollary 1. Conversely, if the straight lines AP, DE, are perpendicular to the same plane MN, they will be parallel; for, if they are not, through the point D draw a line parallel to AP; this parallel will be perpendicular to the plane MN, conse quently there would be two perpendiculars to the same plane drawn through the same point, which is impossible (328).

337. Corollary 11. Two lines A and B, parallel to a third C, are parallel to each other; for, let there be a plane perpendicular to the line C, the lines A and B parallel to this perpendicu

Fig. 187.

Fig. 188.

Fig. 189.

Fig. 188.

lar will be perpendicular to the same plane; therefore, by the above corollary, they are parallel to each other.

It is supposed that the three lines are not in the same plane, without which the proposition would already be known (68).

THEOREM.

338. If the straight line AB (fig. 187) is parallel to another straight line CD, drawn in the plane MN, it will be parallel to this plane.

Demonstration. If the line AB, which is in the plane ABCD, should meet the plane MN, this can take place only in some point of the line CD, the common intersection of the two planes; now AB cannot meet CD, because it is parallel to it; consequently it cannot meet the plane MN; therefore it is parallel to this plane (314).

THEOREM.

339. Two planes MN, PQ (fig. 188), perpendicular to the same straight line AB, are parallel to each other.

Demonstration. If they can meet, let O be one of the common points of intersection, and join OA, OB; the line AB, perpendicular to the plane MN, is perpendicular to the straight line OA drawn through its foot in this plane; for the same reason, AB is perpendicular to BO; hence ОA, OB, would be two perpendiculars let fall from the same point O upon the same straight line, which is impossible; consequently the planes MN, PQ, cannot meet; therefore they are parallel.

THEOREM.

340. The intersections EF, GH (fig. 189), of two parallel planes MN, PQ, by a third plane FG, are parallel.

Demonstration. If the lines EF, GH, situated in the same plane, are not parallel, being produced they will meet; consequently the planes MN, PQ, in which they are, would meet; therefore they would not be parallel.

THEOREM.

341. The line AB (fig. 188), perpendicular to the plane MN, is perpendicular to the plane PQ, parallel to the plane MN.

Demonstration. In the plane PQ draw at pleasure the line BC, and through AB, BC, suppose a plane ABC to pass intersecting the plane MN in AD, the intersection AD will be parallel to BC (340); but the line AB, perpendicular to the plane MN, is perpendicular to the straight line AD; consequently it will be perpendicular to its parallel BC; and, since the line AB is perpendicular to every line BC drawn through the foot of it in the plane PQ, it follows that it is perpendicular to the plane PQ.

THEOREM.

342. The parallels EG, FH (fig. 189), comprehended between Fig. 189. two parallel planes MN, PQ, are equal.

Demonstration. Through the parallels EG, FH, suppose a plane EGHF to pass meeting the parallel planes in EF, GH. The intersections EF, GH, are parallel (340) as well as EG, FH; consequently the figure EGHF is a parallelogram; therefore EGFH.

343. Corollary. It follows from this, that two parallel planes are throughout at the same distance from each other; for, if EG, FH, are perpendicular to the two planes MN, PQ, they are parallel to each other (335); therefore they are equal.

THEOREM.

344. If two angles CAE, DBF (fig. 190) not in the same plane, Fig. 190. have their sides parallel and directed the same way, these angles will be equal, and their planes will be parallel.

Demonstration. Take AC = BD, AE = BF, and join CE, DF, AB, CD, EF. Since AC is equal and parallel to BD, the figure ABDC, is a parallelogram (87); therefore CD is equal and parallel to AB. For a similar reason, EF is equal and parallel to AB; consequently CD is also equal and parallel to EF; hence the figure CEFD is a parallelogram, and thus the side CE is equal and parallel to DF; the triangles then CAE, DBF, are equilateral with respect to each other; therefore the angle

CAEDBF.

Again, the plane ACE is parallel to the plane BDF; for, let us suppose the plane parallel to DBF, drawn through the point A, to meet the lines CD, EF, in points different from C and E for example, in G and H; then, according to article 342, the