An Encyclopaedia of Architecture, Historical, Theoretical, & Practical

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Let the two parallelograms ABCD, FGLM (fig. 291.) be between the same parallek, BL, AM, the surface of the parallelogram ABCD contains the surface of the parallelogram FGLM as many times exactly as the base AD contains the base FM. Suppose, for example, that the base AD is triple of the base FM; in this case the surface ABCD will also be triple of the surface FGLM.

Divide the base AD into three parts, each of which is equal to the base FM, and draw from the points of division the lines NP, RS parallel to the side AB.

Fig. 291.

The parallelograms ABPN, FGLM being between the same parallels and having equal bases, the parallelogram ABPN is (Prop. 30.) equal to the parallelogram FGLM. For the same reason, the parallelograms NPSR, RSCD are also equal to the parallelogram FGLM. The parallelogram ABCD is therefore composed of three parallelograms, each of which is equal to the parallelogram FGLM. Consequently the parallelogram ABCD is triple of the parallelogram FGLM.

937. PROP. LIX. Triangles which are between the same parallels are to one another as their buses.

M F

Let the two triangles ABC, DFG (fig. 292.) be between the same parallels LF, AG, the surface of the triangle ABC contains the surface of the triangle DFG as many times as the base AC contains the base DG. Suppose, for example, that the base AC is triple of the base DG, in this case the surface ABC will be triple of the surface DFG.

Divide the base AC into three equal parts, AN, NR, RC, each of which is equal to the base DG, and draw the right lines BN, BR.

Fig. 292.

The triangles ABN, DFG being between the same parallels and having equal bases, the triangle ABN is (Prop. 31.) equal to the triangle DFG. For the same reason, the triangles NBR, RBC are each equal to the triangle DFG. The triangle ABC is therefore composed of three triangles, each of which is equal to the triangle DFG. Wherefore the triangle ABC is triple of the triangle DFG.

938. PROP. LX. If a line be drawn in a triangle parallel to one of its sides, it will cut the other two sides proportionally.

In the triangle BAC (fig. 293.), if the line DF be parallel to the side BC, it will cut the other two sides in such manner that the segment AD will be to the segment DB as the segment AF is to the segment FC. Suppose, for instance, the segment AD to be triple of the segment DB, the segment AF will be triple of the segment FC. Draw the diagonals

DC, FB.

The triangles AFD, DFB are between the same parallels, as will be easily conceived by supposing a line drawn through the point F parallel to the side AB. These two triangles are therefore to one another (Prop. 59.) as their bases; and since the base AD is triple of the base DB, the triangle AFD will be triple of the triangle DFB.

Again, the triangles BFD, FDC are between the same parallels DF, BC, and upon the same base DF. These two triangles are therefore (Prop. 31.) equal; and since the triangle AFD is triple of the triangle DFB, it will also be triple of the triangle FDC. Lastly, the triangles ADF, FDC are between the same parallels, as will be easily conceived by supposing a line drawn through the point D parallel to the side AC. These two triangles are therefore to one another (Prop. 59.) as their bases; and since the triangle ADF is triple of the triangle FDC, the base AF will be triple of the base FC.

939. PROP. LXI. Equiangular triangles have their homologous sides proportional. In the two triangles ABC, CDF (fig. 294.), if the angle A be equal to the angle C, the angle B equal to the angle D, and the angle C equal to the angle F; the side AC, for example, opposite to the angle B is to the side CF opposite to the angle D as the side AB opposite to the angle C is to the side CD opposite to the angle F. Place the two triangles so that the sides AC, CF shall form one right line, and produce the sides AB, FD till they meet in G.

Fig. 291.

The interior and exterior angles GAF, DCF being equal, the A lines GA, DC are (Prop. 20.) parallel. In like manner, the alternate angies GFA, BCA on the same sides being equal, the lines GF, BC are (Prop. 20.) parallel. Wherefore the quadrilateral figure BGDC is a parallelogram, and consequently its opposite sides are equal. In the triangle GAF the line BC, being parallel to the side

FG, cuts (Prop. 60.) the other two sides proportionally; that is, AC is to CF as AB is to BG, or its equal CD.

940. PROP. LXII. Triangles the sides of which are proportional are equiangular.

In the two triangles BAC, FDG (fig. 295.), if the side AB is to the side DF as the side BC is to the side

FG and as the side AC to the side DG, these two triangles have their angles equal.

Let the side AB be supposed triple of the side DF; the side AC must be triple of the side DG, and the side BC triple of the side FG.

If the triangle FDG be not equiangular with the triangle BAC, another triangle may be formed equiangular B with it; for example, FLG. But this is impossible;

Fig. 295.

I D

for if the two triangles BAC, FLG were equiangular, their sides would be (Prop. 61.) proportional; and BC being triple of FG, AB would be triple of LF. But AB is triple of DF; whence LF would be equal to DF. For the same reason, LG would be equal to DG. Thus, the two triangles FLG, FDG, having their three sides equal, would be (Prop. 5.) identical; which is absurd, since their angles are unequal.

941. PROP. LXIII. Triangles which have an angle in one equal to an angle in the other, and the sides about these angles proportional, are equiangular.

If in the two triangles BAC, NMP (fig. 296.) the angle A be equal to the angle M. and the side AB be to the side MN as the side AC is to the side MP, the two triangles are equiangular.

If AB be triple of MN, AC must be triple of MP. Now, if the angle MNP, for example, is not equal to the angle ABC, another angle may be made, as MNR, which shall be equal to it. But this is impossible; for the two triangles BAC, NMR, having two angles equal, would be equiangular, and consequently (Prop. 61.) would have their sides proportional; wherefore, AB being triple of MN, AC would be triple of MR, which cannot be, since AC is triple of MP.

Fig. 296.

942. PROP. LXIV. A right line which bisects any angle of a triangle divides the side opposite to the bisected angle into two segments, which are proportional to the two other sides. In the triangle BAC, let the angle BAC be bisected by the right line AD, making the angler equal to the angle s. The segment BD is to the segment

DC as the side BA to the side AC.

Produce the side BA, and draw CF parallel to DA.

The lines DA, CF being parallel, the interior and exterior angles -, Fare (Prop. 19.) equal, and the alternate angles s, C are (Prop. 17.) also equal. And since the angle r is equal to the angle s, the angle F will also be equal to the angle C; and consequently the side AF is equal to the side AC.

Fig. 297.

In the triangle BFC, the line AD being parallel to the side FC; BD (Prop. 60.) will be to DC as BA is to AF, or its equal AC. 943. PROP. LXV. To find a fourth proportional to three given lines. Let the three lines be A, B, C (fig. 298.), it is required to find a fourth line D, such that the line A shall be to the line B as the line C is to A the line D.

Form any angle RFG, make FM equal to the line A, MG equal to the line B, and FN equal to the line D C; draw the right line MN, and through the point G draw GL parallel to MN; NL will be the fourth proportional required.

In the triangle FLG the line NM, being parallel to F

the side LG, cuts the other two sides (Prop. 60.) propor

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tionally. Wherefore FM is to MG as FN is to NL; that is, A is to B as C is to D. 944. PROP. LXVI. To find a third proportional to two given lines.

Let the two lines be A, B (fig. 299.), it is required to find a third line C, such that the line A shall be to the line B as the line B is to the line C.

Form any angle LFG, make FM equal to the line A, MG equal to the line B, and FN equal to the line B; draw the right line MN, and through the point G draw GL parallel to MN; NL will be the third proportional required.

Fig. 299.

In the triangle FLG the line NM, being parallel to the side LG, cuts the other two

sides (Prop. 60.) proportionally. Wherefore FM is to MG as FN is to NL; that is, is to B as B is to C.

945. PROP. LXVII. If four lines be proportional, the rectangle or product of the extrem is equal to the rectangle or product of the means.

Let the line A be to the line B as the line C is to the line D (fig. 300.); the rectang formed by the lines A and D is equal to the rectangles formed A by the lines B and C

Let the four lines meet in a common point, forming at that C point four right angles; and draw the lines parallel to them to complete the rectangles x, y, z.

If the line A be triple of the line B, the line C will be triple of the line D.

The rectangles or parallelograms x, z being between the same parallels, are to one another as their bases. Since the base A is triple of the base B, th rectangle is triple of the rectangle z. In like manner, the rectangles or parallelogram y, z, being between the same parallels, are to one another as their bases: since the bas C is triple of the base D, the rectangle y is therefore triple of the rectangle z. Where fore, the rectangle a being triple of the rectangle z, and the rectangle y being triple of the same rectangle z, these two rectangles x and y are equal to one another.

946. PROP. LXVIII. Four lines which have the rectangle or product of the extremes equa to the rectangle or product of the means are proportional.

Let the four lines A, B, C, D (fig. 301.) be such that the rectangle of A and D is equal to the rectangle of B and C, the line A will be to the line B as the line C to the line D.

Let the four lines meet in a common point, forming at that point four right angles, and complete the rectangles x, y, z.

If the line A be triple of the line B, the line C will be triple of the line D.

The rectangles 2 and 2, being between the same parallels, are to one another as their bases: since the base A is triple of

C y

Fig. 301.

the base B, the rectangle x will be therefore triple of the rectangle z.

And the rectangle

y is, by supposition, equal to the rectangle r; the rectangle y is therefore also triple of the rectangle z.

But the rectangles y, z, being between the same parallels, are to one another as their bases. Hence, since the rectangle y is triple of the rectangle z, the base C is also triple of the base D.

947. PROP. LXIX. If four lines be proportional, they are also proportional alternately. If the line A is to the line B as the line C to the line D (fig. 302.), A they will be in proportion alternately; that is, the line A will be to the line C as the line B to the line D.

Because the line A is to the line B as the line C is to the line D, C. the rectangle of the extremes A and D is equal to the rectangle of the Dmeans B and C; whence it follows (Prop. 68.) that the line A is to the line C as the line B is to the line D.

Fig. 302

Otherwise, Suppose the line A to be triple of the line B, the line C will be triple of the line D. Hence, instead of saying A is to B as C to D, we may say three times B is to B as three times D is to D. Now it is manifest that three times B is to three times D as B is to D. Therefore the line (which is equal to three times B) is to the line C (which

is equal to three times D) as the line B is to the line D.

948. PROP. LXX. If four lines be proportional, they will be proportional by composi

tion.

Let the line A be to the line B as the line C is to the line D (fig. 303.), they will be proportional by composition; that is, the line A joined to the line B will be to the line B as the line C joined to the line D is to the line D.

If the line A contain the line B. for example, three times, and the line C contain the line D three times; the line A joined to the line B will contain the line B four times, and the line C joined to the line D will D contain the line D four times. Therefore the line A joined to the line B is to the line B as the line C joined to the line D is to the line D. 949. PROP. LXXI. If four lines be proportional, they will be also proportional by division.

Fig. 303.

If the line A is to the line B as the line C is to the line D (fig. 304.), they will be proportional by division; that is, 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. If the line A contain the line B, for example, three times, and the line C contain the line D three times, the line A wanting the line B will contain the line B only twice; and the line C wanting the line D will also contain the line r

Fig. 304.

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

Draw

C F

L B

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 r, 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 segments 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. 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 Bare 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 Cpoint F. From the point F, as a centre, describe the cir-. cumference 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.

Fig. 307.

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 GI,, that is, between the lines A and C.

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

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 tases 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 A equal to the rectangle of the means DG, FM; AC (Prɔp. 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.

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. The halve (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.

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

L C

Fig. 309.

Fig. 310.

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.

Take CB equal to one of the given lines, and CD equal to the other; bisect DB; from 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.

Fig. 312.

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 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 A 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.

Fig. 313.

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