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959. Prop. LXXX. In similar figures the homologous sides are proportional.
Let the similar figures be ABCDF, abcdf (fig. 314.), and the homologous lines CA, C, CF, cf ; CA is to ČF as ca to cf.
Since the lines CA, ca are homologous, they are composed of an equal number of corresponding points; as are also the homologous lines CF, ef. If, for instance, the line CA is composed of 40 equal points, and the line CF of 30, the Bx line ca will necessarily be composed of 40 points, and the line çf of 50; and it is manifest that 40 is to 30 as 40 to 30. Therefore CA is to CF as ca to cf. 960. Prop. LXXXI. The circumferences of circles are as their radii.
The circumference DCB (fig. 315.) is to the radius AB as the circumference deb is to the radius ab.
All circles are similar figures, that is, are composed of an equal number of points disposed in the same manner. They have therefore (Prop. 80.) their homologous lines proportional. Therefore the circumference DCB is to the radius AB as the circumference deb is to the radius ab.
961. Prop. LXXXII. Similar figures are to each other as the squares of their homologous sides.
Let the two similar figures be A, a (fig. 316.) Upon the homologous sides CD, cd form the squares B, b. The surface A is to the surface a as the square B is to the square b.
Since the figures A, a are similar, they are composed of an equal number of corresponding points; and since the homologous sides CD, cd are composed of an equal number of points, the squares drawn upon these lines B, b are also composed of an equal number of points.
If it be supposed that the surface A is composed of 1000 points and the square B of 400 points, the surface A will be also composed of 1000 points and the square 6 of 400. Now it is manifest that 1000 is to 400 as 1000 to 400. Wherefore the surface A is to the square B as the surface a is to the square b; and, alternately (Prop. 69.), the surface A is to the surface a as the square B to the square b.
COROLLABY. It follows that if any three similar figures be formed upon the three sides of a right-angled triangle, the figure upon the hypothenuse will be equal to the other two taken together; for these three figures will be as the squares of their sides; therefore, since the square of the hypothenuse is equal to the two squares of the other sides, the figure formed upon the hypothenuse will also be equal to the two other similar figures formes upon the other sides.
962. Peop. LXXXIII. Circles are to each other as the squares of their radii. Let two circles DCB, dcb (fig. 317.) be drawn.
The surface contained within the circumference DCB is to the surface contained within the circumference dch as the square formed upon the radius AB to the square formed upon the radius ab.
The two circles, being similar figures, are composed of an equal number of corresponding points, and the radii AB, ab being composed of an equal number of points, the squares of these radii will also be composed of an equal number of points. Suppose, for example, that the greater circle DCB is composed of 800 points, and the square of the greater radius AB of 300 points, the smaller circle dcb will also be composed of 800 points, and the square of the smaller radius of 300. Now it is manifest that 800 is to 300 as 800 to 300. Therefore the greater circle DCB is to the square of its radius AB as the smaller circle dcb is to the square of its radius ab; and, alternately, the greater circle is to the lesser circle as the greater square is to the lesser square.
963. Prop. LXXXIV. Similar triangles are equiangular.
If the two triangles ABC, abc (fig. 318.) be composed of an equal number of points disposed in the same manner, they are equiangular.
For, since the triangles ABC, abc are similar figures, they hare their sides (Prop. 80.) proportional; they are therefore (Prop. 62.) equiangular.
964. Prop. LXXXV. Equiangular triangles are similar
If the triangles ABC, abc are equiangular, they are also similar. See fig. 318. If the triangle ABC were not similar to the triangle abc,
Fig. 318. another triangle might be formed upon the line AC; for example, ADC, which should be similar to the triangle abc. Now, the triangle ADC, being similar to the triangle ubr,
will also (Prop. 84.) be equiangular to abc ; which is impossible, since the triangle AB is supposed equiangular to abc.
965. Prop. LXXXVI. If four lines are proportional, their squares are also proportional
If the line AB be to the line AC as the line A D is to the line AF (fig. 319,), the squa: of the line AB will be to the square of the line AC A As the square of the line AD is to the square of the A line AF.
With the lines AB and AD form an angle BAD; A with the lines AC and AF form another angle CAF equal to the angle BAD, and draw the right lines BD, CF.
Because AB is to AC as AD to AF, and the contained angles are equal, the two triangles BAD, CAF have their sides about equal angles proportional; they are therefore (Prop. 65.) equiangula and consequently (Prop. 85.) similar : whence they are to one another (Prop. 82.) a the squares of their homologous sides. If, then, the triangle BAD be a third part ! the triangle CAF, the square of the side A B will be a third part of the square of the sid AC, and the square of the side AD will be a third part of the square of the side AF Wherefore these four squares will be proportional.
966. Prop. LXXXVII. Similar rectilineal figures may be divided into an equal numbı of similar triangles.
Let the similar figures be ABCDF, abcdf, and draw the homologous lines CA, ca, CF, c) these two figures will be divided into an equal number of similar triangles.
The triangles BCA, bca (fig. 320.), being composed of an equal number of corresponding points, are similar. The triangles ACF, acf and the triangles FCD, fed are also, for B the same reason, similar. Wherefore the similar figures ABCDF, abcdf are divided into an equal number of similar
Fig. 320. triangles.
967. Prop. LXXXVIII. Similar figures are equiangular.
The similar figures ABCDF, abcdf (see fig. preced. Prop.) have their angles equal Draw the homologous lines CA, ca, CF,cf. The triangles BCA, bca are similar, and con sequently equiangular. Therefore the angle B is equal to the angle b, the angle BAC t the angle bac, and the angle BCA to the angle bca. The triangles ACF, acf, FCD, fc are also equiangular, because they are similar. Therefore all the angles of the simila figures ABCDF, abcdf are equal.
968. Prop. LXXXIX. Equiangular figures the sides of which are proportional ar similar.
If the figures ABCDF, abcdf (fig. 321.) have their angles equal and their sides propor tional, they are similar. Draw the right lines CA, ca, CF, cf.
The triangles CBA, cba, have two sides proportional and the contained angle equal ; they are therefore (Prop. 63.) equiangular, and consequently (Prop. 85.) similar. The lines CA, ca are therefore (Prop. 80.) proportional.
The triangles CAF, caf have two sides proportional and the contained angle equal; for if from the equal angles BAF, baf be taken the equal angles BAC, bac, there will remain the equal angles CAF caf. These two triangles are therefore equiangular, and consequently similar. In the same manner it may be proved that the triangles CFD, cfd are similar.
The two figures ABCDF, abcdf are then composed of an equal number of similar triangles that is, they are composed of an equal number of points disposed in the same manner, or are similar.
969. Definitions. - 1. A plane is a surface, such that if a right line applied to it
touch it in two points it will touch it in every other point.
be considered as a plane.
angles with all lines which can be drawn from any point in
AL, AG, &c. drawn from the point A.
If two right lines LM, FG be drawn, in these two planes, perpendicular to the line
formed by the two planes.
revolve about itself, with-
an obtuse angle with AB, will describe in the revolution a convex surface MAD. 5. But the line AF (fig. Defin. 2.), which makes a right angle with AB, will de
scribe in the revolution a surface which will be neither con-
one to the other are equal. See fig. 325., wherein AB, CD
are equal between the surfaces LM, FG. 970. Prop. XC. A perpendicular is the shortest line which can be drawn from any point to a plane.
Fig. 325. From the point B (fig. 326.), let the right line BA be drawn perpendicular to the plane DF; any other line, as BC, will be longer than the line BA. L'pon the plane draw the right line AC.
Because the line BA is perpendicular to the plane DF, the angle BAC is a right angle. The square of BC is therefore (Prop. 32.) equal to the squares of BA and AC taken together. Consequently the square of BC is greater than the square of BA, and the line BC longer than the line BA.
971. Prop. XCI. A perpendicular measures the distance of any point from a plane. The distance of one point from another is measured by a right
c line, because it is the shortest line which can be drawn from one point to another. So the distance from a point to a line is measured by a perpendicular, because this line is the shortest which can be drawn from the point to the line. In like manner, the distance from a point to a plane must be measured by a perpendicular drawn from that point to the plane, because this is the shortest line which can be drawn from the point to the plane.
972. Prop. XCII. The common intersection of two planes is a right line.
Let the two planes ALBMA, AFBGA (fig. 327.) intersect each other; the line which is common to both is a right line. Draw a right line from the point A to the point B.
Because the right line AB touches the two planes in the points A and B, it will touch them (Defin. 1.) in all other points ; this line therefore, is common to the two planes. Wherefore the common intersection of the two planes is a right line.
978. Prop. XCIII. If three points, not in a right line, are com. non to two planes, these two planes are one and the same plane. Let two planes be supposed to be placed upon one another, in such
Fig. 327. manner that the three points A, B, C shall be common to the two planes ; all their other points will also be common, and the two planes will be one and the same plane. The point D, for example, is common to both planes. Draw the right lines AB, CD
Because the right line AB (fig. 328.) touches the two planes in the points A and B, it will tonch them (Defin. 1.) in every other point; it will therefore touch them in the point F. The point F is therefore common to the two planes.
Again, because the right line CD touches the two planes in the points C and F, it will touch them in the poiut D; therefore the point D is common to the two planes. The same may be shown concerning every other point. Wherefore the two planes coincide in all points, or are one and the same plane.
274. Prop. XCIV. If a riyht line be perpendicular to two right lines which cut each other, it will be perpendiculur to the plare of these right lines.
Let the line AB (fig. 329.) make right angles with the lines AC, AD, it will be perpei dicular to the plane which passes through these lines.
If the line A B were not perpendicular to the FDCG, another plane might be made to pass through the point A, to which the AB would be perpendicular. But this is impossible ; for, since the angles BAC, BAD are right angles, this other plane (Defin. 2.) must pass through the points (, D; it would therefore (Prop. 99.) be the same with the plane FDCG, since these two planes would have three common points + A, C, D.
975. Prop. XCV. From a given point in a plane to raise a perpendicular to that plane.
Let it be required to raise a perpendicular from the point A (fiy. 330.) in the plane LA
Form a rectangle CDFG, divide it into two rectangles, having a common section AB, and place these rectangles upon the plane LM in D such a manner that the bases of the two rectangles AC, AG shall be in the plane LM, and form any angle with each other; the line AB shall be perpendicular to the plane LM, The line A B makes right angles with the two lines AC, AG, which, by supposition, are in the plane LM; it is therefore (Prop. 94.) perpendicular to the plane LM.
976. Prop. XCVI. If two planes cut each other at right angles, and a right line be drawn in one of the planes perpendicular to thei common intersection, it will be perpendicular to the other plane.
Let the two planes AFBG, ALBM (fig. 331.), cut each other at right angles; if th line LC be perpendicular to their common intersection, it is also perpendicular to the plane AFBG. Draw CG perpendicular to AB.
Because the lines CL, CG are perpendicular to the common in. tersection AB, the angle LCG (Defin. 3.) is the angle of inclination of the two planes. Since the two planes cut each other perpendicularly, the angle of inclination LCG is therefore a right angle.
And because the line LC is perpendicular to the two lines CA, CG in the plane ABFG, it is (Prop. 94.) perpendicular to the plane AFBG. 977. Prop. XCVII. If one plane meet another plane, it makes
Fig. 331. angles with that other plane, which are together equal to two right angles.
Let the plane ALBM (fig. 332.) meet the plane AFBG; these planes will make with each other two angles, which will together be equal to two right angles. Through any point C draw the lines FG, LM perpendicular to the line AB. The line CL makes with the line FG two angles together equal to two right angles. But these two-angles are (Defin. 3.) the angles of inclination of the two planes. Therefore the two planes make angles with each other, which are together equal to two right angles.
COROLLARY. It may be demonstrated in the same manner that planes which intersect each other have their vertical angles equal, that parallel planes have their alternate angles equal, &c. 978. Prop. XCVIII. If two planes be parallel to each other,
Fig. 332. a right line, which is perpendicular to one of the planes, will be also perpendicular to the other. Let the two planes LM, FG (fig. 333.) be parallel. If the line BA
-CM be perpendicular to the plane FG, it will also be perpendicular to the plane LM. From any point C in the plane LM draw CD perpendicular to the plane FG, and draw BC, AD.
Because the lines BA, CD are perpendicular to the plane FG, the angles A, D are right angles.
Because the planes LM, FG are parallel, the perpendiculars AB, F DC (Defin. 6.) are equal; whence it follows that the lines BC, AD are parallel.
The line BA, being at right angles to the line AD, will also (Prop. 19.) be at right angles to the parallel line BC. The line BA is therefore perpendicular to the line BC.
In the same manner it may be demonstrated that the line BA is at right angles to all other lines which can be drawn from the point B in the plane LM. Wherefore (Defin. 2.) the line BA is perpendicular to the plane LM.
SOLIDS. 979. DEFINITIONS. – 1. A solid, as we have before observed, is that which has length,
breadth, and thickness. 2. A polyhedron is a solid terminated by plane surfaces. 3. A prism is a solid terminated by two identica plane bases
parallel to each other, and by surfaces which are parallelo.
grams. (Fig. 394.) 4. A parallelopiped is a prism the bases of which are parallelo
grams. (Fig. 395.) 5. A cube is a solid terminated by six square surfaces : a die,
for example, is a cube. (Fig. 336.) 6. If right lines be raised from every point in the perimeter of
Fig. 334. any rectilineal figure, and meet in one common point, these lines together with the rectilineal figure inclose a solid which
is called a pyramid. (Fig. 337.)
planes, which are equal and parallel cir-
in the circumference of a circle, and meet in one common point, these lines together with the circle inclose a solid, which
is called a cone. (Fig. 339.) 9. A semicircle revolving about its diame
ter forms a solid, which is called a
sphere. (Fig. 340.)
dicular be let fall upon the opposite
Fig. 341. 11. Solids are said to be equal, if they inclose an equal space: thus a cone and a pyramid
are equal solids if the space inclosed within the cone be equal to the space inclosed
within the pyramid. 12. Similar solids are such as consist of an equal number of physical points disposed in
the same manner. Thus (in the fig. Defin. 10.) the larger pyramid ACD and the smaller pyramid Acd are similar solids if every point in the larger pyramid has a point corresponding to it in the sınaller pyramid. A hundred musket balls, and the same number of cannon balls, disposed in the same manner, form two similar solids.
980. Prop. XCIX. The solid content of a cube is equal to the product of one of its sides trcice multiplied by itself.
Let the lines AB, XD (fig. 342.) be equal. Let the line AD, drawn perpendicular to AB, be supposed to move through the whole length of AB; when it arrives at BC, and coincides with it, it will have formed the square DABC, and will have been multiplied by the line AB.
Next let the line AF be drawn equal to AD, and perpendicular to the plane DABC; suppose the plane DABC to move perpendicularly through dk the whole length of the line AF; when it arrives at the plane MFGL, and coincides with it, it will have formed the cube AFLC, and will have been multiplied by the line AF.
llence it appears, that to form the cube AFLC, it is necessary, first, to multiply the side AD by the side AB equal to AD; and then to multiply the product, that is, the square of AC, by the side AF equal to AD; that is, it is necessary to multiply AD by AD, and to multiply the product again by AD
981. Prop. C. Similar solids have their homologous lines proportional.
Let the two solids A, a ( fig. 343.) be similar ; and let their bomologous lines be AB, ab, BG, bg; AB will be to BG at ab to bg.
Because the solids A, a are similar, every point in the solid A bas a point corresponding to it, and disposed in the same