An Encyclopædia of Architecture

twice. Therefore the line A wanting the line B is to the line B as the line C wanting the line D is to the line D.

950. PROP. LXXII. If three lines be proportional, the first is to the third as the square of the first is to the square of the second.

If the line CD is to the line cd as the line cd is to a third line x (fig. 305.), the line CD is to the line as the square of the line CD is to the square of the line cd. Take CF equal to the line x, and draw the perpendicular FB.

Since the line CD is to the line cd as the line cd is to the line CF, the rectangle of the extremes CF, CD, or CL is equal (Prop. 67.) to the rectangle of the means, that is, to the square of cd.

Fig. 305.

Again, the square of CD and the rectangle of the lines CF, CL, being between the same parallels, are to one another (Prop. 58.) as their bases. Therefore CD is to CF, or x, as the square of CD is to the rectangle of CF and CL, or to its equal the square of cd. 951. PROP. LXXIII. If two chords in a circle cut each other, the rectangle of the seg ments of one is equal to the rectangle of the segments of the other Let the two chords AB, CD (fig. 306.) in the circle cut each other in the point F, the rectangle of AF, FB is equal to the rectangle of CF, FD. Draw the two right lines AC, DB. Because in the triangles CAF, BDF the angles at the eircumference A and D are both measured (Prop. 42.) by half the arc CB, they are equal. Because the angles C and B are both measured (Prop. 42.) by half the arc AD, these angles are also equal. And the angles at F are equal, because they are vertical. These two triangles are therefore equiangular, and consequently (Prop. 61.) their sides are proportional. Wherefore the

Fig. 306.

side AF opposite to the angle C is to the side FD opposite to the angle B as the side CF opposite to the angle A is to the side FB opposite to the angle D. Therefore (Prop. 69.) the rectangle of the extremes AF, FB is equal to the rectangle of the means CF, FD.

952. PROP. LXXIV. To find a mean proportional between two given lines.

Let there be two lines A, C (fig. 307.), it is required to find a third line B, such that the line A shall be to the line B as the line B is to the line C.

Place the lines A and C in such manner that they shall form one right line DGL, and bisect this right line in the C. point F. From the point F, as a centre, describe the circumference of a circle DMLN; then, at the point G, where the two lines are joined, raise the perpendicular GM; GM is the mean proportional sought between the lines A and C. Produce MG to N.

Because the chords DL, MN cut each other at the point G, the rectangle of the segments DG, GL is (Prop. 73.) equal to the rectangle of the segments MG, GN.

Because the radius FL is perpendicular to the chord MN, FL (Prop. 38.) bisects MN; therefore GN is equal to GM.

Lastly, because the rectangle of the extremes DG, GL is equal to the rectangle of the means GM, GN, or its equal GM, DG is to GM as GM is to GL. Therefore GM is a mean proportional between DG and GL, that is, between the lines A and C.

953. PROP. LXXV. The bases and altitudes of equal triangles are in reciprocal or inverse ratio.

Let the two triangles ABC, DFG (fig. 308.) be equal; the base AC will be to the base DG, as the perpendicular FM to the perpendicular BL; that is, the bases and altitudes are in reciprocal or inverse ratio.

The triangle ABC (Prop. 54.) is half the product or rectangle of the base AC and the altitude BL. Again, the triangle DFG is (Prop. 54.) half the product or rectangle of the base DG and the altitude FM. The two triangles being equal, the two rectangles, which are double of the triangles, will therefore also be equal.

Again, because the rectangle of the extremes AC, BL is Α L equal to the rectangle of the means DG, FM; AC (Prop. 68.) is to DG as FM is to BL.

Fig. 308.

954. PROP. LXXVI. Triangles the bases and altitudes whereof are in reciprocal or inverse ratio are equal.

In the two triangles ABC, DFG (fig. 309.), if the base AC be to the base DG as the perpendicular FM to the perpendicular BL, the surfaces of the two triangles are equal.

The halves

Because AC is to DG as FM is to BL, the product or rectangle of the extremes AC, BL is (Prop. 67.) equal to the product or rectangle of the means DG, FM. (Corol. to Prop. 27.) of these two rectangles, namely, triangles ABC, DFG, are therefore equal.

955. PROP. LXXVII. Two secants drawn from the same point to a circle are in the inverse ratio of the parts which lie out of the circle.

Let the two secants be CA, CB (fig. 310.); CA is to CB as CD is to CF. Draw the right lines FB, DA.

Fig. 309.

Fig. 310.

In the triangles CDA, CFB the angles A at the circumference A and B, being both measured (Prop. 42.) by half the arc FD, are equal, and the angle C is common to the two triangles. These two triangles are therefore (Prop. 23.) equiangular and (Prop. 61.) have their sides proportional. Wherefore the side CA of the first triangle is to the side CB of the second triangle as the side CD of the first triangle is to the side CF of the second triangle.

956. PROP. LXXVIII. The tangent to a circle is a mean proportional between the secant and the part of the secant which lies out of the circle.

In the circle ABD, CB (fig. 311.) being secant, and CA tangent, CB is to CA as CA is to CD. Draw the right lines AB, AD.

Also

The triangles CAB, CDA have the angle C common to both. the angle B is measured (Prop. 42.) by half the arc AFD; and the angle CAD formed by the tangent AC and the chord AD is measured (Prop. 41.) by half the same are AFD. The two triangles CAB, CDA, having their two angles equal, are (Prop. 23.), equiangular, and consequently (Prop. 61.) have their sides proportional. Hence the side

CB of the greater triangle opposite to the angle CAB is to the side CA of the smaller triangle opposite to the angle D as the side CA of the greater triangle opposite to the angle B is to the side CD of the smaller triangle opposite to the angle A.

COROLLARY. From this proposition is suggested a new method of finding a mean proportional between two given lines.

Fig. 311.

bisect DB; from

Take CB equal to one of the given lines, and CD equal to the other; the point of division, as a centre, describe the circumference DAB; and draw the tangent CA. This tangent is a mean proportional between CB and CD, as appears from the proposition. 957. PROP. LXXIX.

ratio.

To cut a given line in extreme and mean

Let it be required to divide the line CA (fig. 312.) in extreme and mean ratio; that is, to divide it in such a manner that the whole line shall be to the greater part as the greater part is to the less.

At the extremity A of the line CA raise a perpendicular AG equal to half the line CA; from the point G, as a centre, with the radius GA, describe the circumference ADB; draw the line CB through the centre, and take CF equal to CD; the line CA will be divided at the point F in extreme and mean ratio.

Because (Prop. 78.) CB is to CA as CA is to CD, by division, (Prop. 71.) CB wanting CA or its equal DB is to CA, as CA wanting

Ꭺ

Fig. 312.

CD or its equal CF is to CD; that is, CD or CF is to CA, as FA is to CD or CF; or, inversely, CA is to CF as CF is to FA, or the line AC is cut in extreme and mean ratio.

SIMILAR FIGURES.

958. DEFINITIONS. 1. Figures are similar which are composed of an equal number of physical points disposed in the same manner. Thus, the figures ABCDF, abcdf (fig. 313.) are similar, if every point of the first figure has its corresponding point placed in the same manner in the second. Hence it follows, that if the first figure is, for example, three times greater than the second, the points of which it is composed are three times greater than those of the second figure.

2. In similar figures, those lines are said to be homologous
which are composed of an equal number of corresponding points.

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, ca, CF, cf; CA is to CF 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, cf. If, for instance, the line CA is
composed of 40 equal points, and the line CF of 30, the B
line ca will necessarily be composed of 40 points, and the line
ef of 30; 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.

Fig. 314.

The circumference DCB (fig. 315.) is to the radius AB as the circumference dcb 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 propor tional. Therefore the circumference DCB is to the radius AB as the circumference dcb 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.) homologous sides CD, cd form the squares B, b. square B is to the square b.

Upon the

Fig. 315.

The surface A is to the surface a as the

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 b 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 sur- L face A is to the surface a as the square B to the square b.

Fig. 316.

COROLLARY. 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 formed upon the other sides.

962. PROP. 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 dcb as the square formed upon the radius AB to the square formed upon the radius ab.

Fig. 317.

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 have 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.

Fig. 318.

If the triangle ABC were not similar to the triangle abc, 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 abc,

will also (Prop. 84.) be equiangular to abc; which is impossible, since the triangle ABC 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 AD is to the line AF (fig. 319,), the square 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; Awith 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

-D

Fig. 319.

have their sides about equal angles proportional; they are therefore (Prop. 693.) equiangular, and consequently (Prop. 85.) similar whence they are to one another (Prop. 82.) as the squares of their homologous sides. If, then, the triangle BAD be a third part of the triangle CAF, the square of the side AB will be a third part of the square of the side 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 number of similar.

Let the similar figures be ABCDF, abcdf, and draw the homologous lines CA, ca, CF, cf; 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, fcd are also, for B the same reason, similar. Wherefore the similar figures ABCDF, abcdf are divided into an equal number of similar triangles.

967. PROP. LXXXVIII. Similar figures are equiangular.

Fig. 320.

The similar figures ABCDF, abcdf (see fig. preced. Prop.) have their angles equal. Draw the homologous lines CA, ca, CF, cf. The triangles BCA, bea are similar, and consequently equiangular. Therefore the angle B is equal to the angle b, the angle BAC to the angle bac, and the angle BCA to the angle bca. The triangles ACF, acf, FCD, fed are also equiangular, because they are similar. Therefore all the angles of the similar figures ABCDF, abcdf are equal.

968. PROP. LXXXIX. Equiangular figures the sides of which are proportional are similar.

If the figures ABCDF, abcdf (fig. 321.) have their angles equal and their sides proportional, 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. lines CA, ca are therefore (Prop. 80.) proportional.

The

Fig. 321.

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.

PLANES.

969. DEFINITIONS. 1. A plane is a surface, such that if a right line applied to it touches it in two points it will touch it in every other point. The surface of a fluid at rest, or of a well-polished table, may be considered as a plane.

2. A right line is perpendicular to a plane if it make right angles with all lines which can be drawn from any point in that plane. Thus BA (fig. 322.) is perpendicular to the plane MLGFPN, because it makes right angles with the lines AM, M AL, AG, &c. drawn from the point A.

3. Let AB (fig. 323.) be the common intersection of two planes.

Fig. 322.

If two right lines LM, FG be drawn, in these two planes, perpendicular to the line AB, these will form four angles at the point C, which are called the inclinations of the two planes, or the angles formed by the two planes. 4. If the line AB (fig. 324.) revolves about itself, without changing its place, the line AC, which makes an acute angle with AB, will

M
Fig. 323.

Fig. 324.

describe in the revolution a concave surface LAC; and the line AD, which makes 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 describe in the revolution a surface which will be neither concave nor convex, but plane: and the line AB will be perpendicular to the plane MLG FPN, because it will make right angles with the lines AM, AL, AG, &c. drawn from the point A in that plane.

6. Two planes are parallel when all perpendiculars drawn from 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.

JD G

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. Upon 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.

Fig. 326.

The distance of one point from another is measured by a right 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.

973. PROP. XCIII. If three points, not in a right line, are common 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 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 the same plane. The point D, for example, is common to both planes. lines AB, CD.

Because the right line AB touches the two planes in the points A and B, it will touch 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 point 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.

974. PROP. XCIV. If a right line be perpendicular to two right lines which cut each other, it will be perpendicular to the plane of these right lines.

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