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But similar figures which have their homologous sides also equal, are. altogether equal.

(2.) Let the pyramid SABCDE (S being the vertex, and ABCDE the base), be cut by a plane abcde parallel to the base: then the figure abcde will be similar to the base ABCDE.

For, as in the preceding case, ab, bc, cd, de, ea are respectively parallel to AB, BC, CD, DE, EA; and the angles abc, bed, etc., equal to the angles ABC, BCD, etc. The figure abcde is therefore equiangular with the base.

Again, the triangles aSb, ASB are similar, since ab is parallel to AB; and similarly with respect to the other pairs. Whence

ab: AB::bS: BS :: bc: BC, or ab: bc:: AB: BC;

that is the sides about the equal angles abc, ABC are proportional.

In the same manner it may be proved for the sides about the other equal

angles. The figure abcde is hence similar to ABCDE.

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Scholium. The cutting plane may be below the base, between the base and vertex, or beyond the vertex. It is analogous to the three cases of Euc. VI. 2. In the last case, however, the figure will have a reversed position, as may be seen in the figure.

PROPOSITION XXIV.

If a cylinder or cone be cut by a plane parallel to the base, the section will be a circle; and in the cylinder this circle will be also equal to the base.

(1.) Let ABC be the base of a cylinder, and E its centre; let the cylinder be cut by a plane parallel to the base in abc: then abc is a circle.

For let Ee be the intercepted portion of the axis, and Aa, Bb, Cc, etc., the intercepted portions of the edges. All these, by the definition of the cylinder, are parallel; and being between parallel planes, they are equal (Prop. xv.); and each equal to Ee.

Also, since the plane A ae E cuts the parallel planes in AE, ae, these lines are parallel (Prop. 1.); and hence A ae E is a parallelogram, having ae AE. In like manner in all the corresponding cases, be BE, ce = CE, etc. But AE CE, etc.; and hence ae = ce, etc., and e is the centre of a circle, the

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BE

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be

radius of which is equal to that of the base. The section, therefore, is a circle equal to the base.

(2.) Let S be the vertex of a cone, of which

the base is ABC and axis SE; and let it be cut by a plane in abc; then abc is a circle.

For draw the planes SEA, SEB, etc., cutting ABC in AE, BE, etc., and abc in ae, be, etc.

Then the parallel planes ABC, abc, being cut by SEA, the lines AE, ae are parallel (Prop. 1 ). In the same manner BE, be are parallel, and

so on.

Wherefore the triangle Sae is similar to SAE, Sbe to SBE, etc.; and hence

AE

ae: AE::eS: ES :: be: BE, or

ae: eb:: AE: EB.

But E is the centre of the circle ABC, and hence EB; and therefore ae = eb. In the same manner ae = ec, etc.

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Consequently, the points a, b, c, etc., being in one plane, and all equidistant from a point e in it, they are in the circumference of a circle, of which e is the centre.

SCHOLIUM. The same remark applies to the different positions of the cutting plane in respect of the coue that was made respecting the pyramid.

PROPOSITION XXV.

If from two fixed points E, P any parallel straight lines be drawn, viz., EA and PB, EC and PD, etc., to meet the fixed plane MN in A and B, C and D, etc.: then AB, CD, etc., will always pass through the same point in the plane, viz., the point F in which EP

meets it.

For since EA, PB are parallel, they are in one plane. Also the points E, P being in that plane, the line EP is in that plane. The point F being in the line EP is also in that plane; and hence the intersection AB of the plane EABP with MN contains the point F. That is, the line AB passes through F.

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

SCHOLIUM. 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 E, all cut the plane MN in points b, c, b', c' 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.

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F

*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, E'A', E'B' be drawn to meet PQ in ab, a'b': then ab, a'b' will be equal.

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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'b'.

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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.

CHAPTER II.

PERPENDICULARITY.

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, BR, BS, 8 etc. Whence the conclusion follows with respect to B.

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

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PROPOSITION II.

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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, 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 E 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

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therefore the angle EBC to the angle FBC (Euc. 1. 8).

D.

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

F

M

A

(Prop. II.), 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.

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