In the same manner all the other lines drawn as prescribed (as CD), pass through F.

SCHOLIŪM. This is the foundation of Dr. Brook Taylor's method of putting a point in perspective.

PROPOSITION XXVI. Let there be a point E on one side of the plane MN, and any number

of parallel lines AB, A'B', etc., on the other; and let EF be drawn parallel to them meeting the plane MN in F: then lines drawn from points B, C, B', C", etc., in the lines AB, A' B' to the point Ě, all cut the plane MN in points b, c, b', d' in the lines AF, A'F; and lines drawn to every point in AB, A'B' etc., cut MN in points lying between A and F, A' and F, etc.


For AB, EF being parallel, they are in one plane; and hence EC, EB, etc., are in that plane. Whence they cut the plane MN. Their intersections with the plane MN are therefore in the line AF, in which ABEF intersects MN. In the same manner for the points B', C' in the line A'B'.

SCHOLIUM. This is expressed in perspective by saying that parallel lines AB, A'B', etc.," have the same vanishing point.”

PROPOSITION XXVII. If there be three parallel planes MN, PQ, RS, and any equal lines

AB, A'B' be taken in one of them, RS, and from any points E, E' in MN lines EA, EB, EA, E' B' be drawn to meet P Q in ab, a'b': then ab, a'b' will be equal.

For since the four straight lines EA, EB, E'A', E'B' are cut by the three parallel planes MN, PQ, RS, they are divided proportionally (Chap. I., 22). That is,

EA: EB:: E'A': E'B':: Ea: Eb::E'a': E'b'. Also, since the planes AEB, A'E'B' are cut by the parallel planes, the lines AB, A'B' are respectively parallel to ab, a'b'; and the triangles AEB, A'E'B' are respectively similar to aEb, a'E'V.

Wherefore AE : AB :: aE : ab.

A'E': A'B'::a'E': a'b'.
But AE ; aE :: A'E': a'E', and hence

ab : a'b' :: AB : A'B'. But by hypothesis AB = A'B', and hence ab = a'b'.

Of course, if E and E' be made to coincide, the same equality of ab to a'b' holds good.

This proposition is the fundamental one of Gravesandes's method of perspective. This method, and that of Dr. Brook Taylor, have been advantageously combined by the late Peter Nicholson in his work on the subject.


PROPOSITION I. From any point in a given straight line innumerable perpendiculars

may be drawn to that line. Let B be a given point in the given line AB; innumerable perpendiculars to AB may be drawn from B.

For through AB innumerable planes may be drawn, as AP, AQ, AR, AS, etc. In each of these, one perpendicular can be drawn, as BP, BQ, BŘ, BS, S etc. Whence the conclusion follows with respect to B.

In the same manner from any other se point b, the innumerable perpendiculars bp, bq, br, bs. etc. may be drawn in the same planes. The conclusion now follows universally.

PROPOSITION II. If any number of lines be drawn in a plane through the same point,

and if at that point another straight line be drawn perpendicular to two of them, it will be perpendicular to all the others.

In the plane MN let any number of lines AB, AC, AD, etc., pass through the point A; and let the line EF be perpendicular to any two of them AB, AC at A; then it will be perpendicular to AD, etc.


For, draw any line in the plane to cut AB, AC, AD in B, C, D. Make EA, AF equal; and join EB, EC, ED, and FB, FC, FD.

Then because EA, AB are equal to FA, AB and the angles EAB, FAB equal (being right angles), EB EE is equal to FB (Euc. 1. 4).

Similarly, EC is equal to FC.

Wherefore the two sides EB, BC are equal to the two FB, BC, and the base EC to the base FC; and therefore the angle EBC to the angle FBC (Euc, 1. 8).

Again, since EB, BD are equal to FB, BD and the angle EBD to the angle FBD, the bases ED, DF are also equal (Euc. 1. 4).

And, lastly, since EA, AD are equal to FA, AD, and the base ED to the base FD, the angle EAD is equal to the angle FAD; and they are adjacent angles, and hence right angles; that is, the line EAF is perpendicular to AD.

In the same manner it is proved to be perpendicular to any, and therefore to every other line drawn in the plane MN through the point A.

SCHOLIUM. It is on account of this theorem that a line perpendicular to a plane is often defined as “ that which is perpendicular to any line in the plane drawn through the point of intersection.”

As a definition, this is open to the same objection that has been often made against some of Euclid's other definitions—that it involves superfluous conditions. Perpendicularity to two of the lines completely defines the perpendicular; perpendicularity to all is a demonstrated property, and should be quoted as such.

When, however, a line is predicated to be perpendicular to a plane, or a plane to a line, we are at full liberty to quote this property, without naming the specific lines through which the plane is drawn.

PROPOSITION III. All lines perpendicular to the same line at the same point in it are

situated in the same plane. Let any number of lines BC, BD, BF, etc. be perpendicular to AB at the point B; they will all lie in the same plane.

For, through any two of them BC, BF let a plane MN be drawn; and if any one of the perpendiculars (as BD) be not in that plane, it will be either above or below it. Let it be above, as in the diagram ; and through AB and BD draw the plane AE cutting MN in BE.

Then because AB is perpendicular to BC, BF, it is perpendicular to BE (Prop. 11.), and by hypothesis ABD is a right angle: whence the angles. ABD, ABE in the same plane AE are equal to one another, the less to the greater, which is impossible.

Whence BD does not lie above the plane MN; and in the same way it does not lie below it. It is therefore in the plane MN.

The same demonstration applies to every other line perpendicular to AB at B; and hence the proposition is true.

PROPOSITION IV. From the same point there can be drawn only one perpendicular

to a plane, whether that point be in the plane or without it. (1.) Let A be a point in the plane MN: there can only be drawn from A one perpendicular to the plane MN.

For, if possible, let there be two, viz., AB, AC, and through them draw the plane PQ, cutting MN in AQ.

Then since BA is perpendicular to the plane MN, BAQ is a right angle (Prop. 11.); and since CA is perpendicular to MN, CAQ is a right angle. The angles BAQ, CAQ in the same plane PQ are therefore equal ; the less to the greater, which is impossible. Whence from A situated in the plane MN only one perpendicular can be drawn.

(2.) Let A be a point without the P plane MN, only one line can be drawn from A perpendicular to MN.

For, if possible, let there be two, viz., AB, AC, and through them draw a plane PQ, cutting MN in BCQ.

Then since AB, AC are perpendicular to the plane MN, the angles ABC, ACB of the triangle ABC are two right angles (Prop. 11.); which is impossible (Ěuc. i. 17). Whence both lines cannot be perpendicular to the plane MN.

Whence, whether the point be in the plane or without it, only one line through the point can be perpendicular to the plane.

PROPOSITION V. Through the same point only one plane can be drawn perpendicular to

a line, whether that point be in the line or without it. (1.) Let A be a point in the line BC: only one plane prpendicular to BC can be drawn through A.

For, if possible, let there be two planes MN, MP drawn through A perpendicular to BC; and let AŇ be their intersection; and draw any other plane BQ through BC, and not passing through AM, cutting MN and MP in AD and AE.

Then since BC is perpendicular to the planes MN, MP, the angles BAD, BAE are right angles (Prop. 11.); and they

are in one plane BQ, which is impossible. Whence MN and MP cannot both be perpendicular to BC.

(2.) Let A be without the line BC; only one plane can be drawn through A perpendicular to BC.

For, if possible, let there be two, MN, MP, the common section of which is AM; and let a plane be drawn through BC and A, cutting MN, MP in AD, AE. Then since EA is in the plane MP perpendicular to CB, the angle AEB is a right angle; and since DA is in the plane MN perpendicular to CB, the angle BDA is a right angle. But AD, AE, and BC are in the same plane. Wherefore the interior angle ADE of the triangle ADE is equal to the exterior; which is impossible (Euc. 1. 16.). Wherefore only one plane through A can be perpendicular to BC.

PROPOSITION VI. (1.) If two planes be perpendicular to a line, they are parallel to

one another; and (2.) If one of two parallel planes be perpendicular to a line, the other

is also perpendicular to it.

(1.) Let the planes MN, PQ be per- M pendicular to AB: they are parallel to one another.

For, through AB draw any two planes AD, AF, cutting MN in AC and AE, and PQ in BD and BF.

Then since AB is perpendicular to MN, the angles BAC, BAE are right angles (Prop. 11.); and since AB is perpendicular to PQ, the angles ABD, ABF are right angles.

Then, since in the same plane AD, the lines AC, BD make right angles with AB, they are parallel. Similarly AE, BF are parallel. Wherefore the two lines CA, AE which meet in A being parallel to the two DB, BF which meet in B, the planes MN, PQ which contain them are parallel (Prop. VII. Chap. 1.).

(2.) Let MN, PQ be parallel planes, and one of them MN be perpendicular to the line AB: then the other, PQ, will also be perpendicular to AB.

For draw any two planes AD, AF through AB cutting the planes in AC, AE and BD, BF as before.

Then since the planes MN, PQ are parallel, the line AC is parallel to BD and AE to BF. Whence the two angles EAB, ABF are together equal to two right angles; and one of them EAB is a right angle (Prop. 11.); and consequently the other, ABF is a right angle, or BF is perpendicular to AB. In the same manner BD is perpendicular to AB; and hence (Def. 5) the plane PQ through BD, BF is perpendicular to AB. VOL. II.


« ForrigeFortsett »