Demonstr. 1. Join BD, and AE; in the two triangles ACE, and BCD, the angle DBE, and DAE, are equal as angles at the circumference of a circle, standing in the same segment; for the same reason, also, the angles at D, and E, are equal; the angles at C, are equal, as vertical angles within, and as identical without, the circle; therefore, these triangles are similar, and their Homologous sides are proportional, or the following proportion results, viz:

CA: CE=CB : CD;

CA: CB-CE: CD;

But these lines are the segments reciprocally of the two sections of the lines crossing each other in C, between the point of intersection and the circumference of the circle, therefore these lines are cut reciprocally proportional; as was to be demonstrated.

II. From the point F, where the tangent from C. touches the circumference of the circle draw FB, and FE; the angle CFE, between the tangent CF, and the chord FE, is equal to the angle FBE, in the alternate segment; the angle at C, is common to the two triangles CFE, and CFB, therefore these triangles are similar, and their homologous sides are proportional: that is,

CE: CF CF: CB; and by the fore

going proportion also CD: CF-CF: CA;

that is: the tangent CF, is a mean proportional between the

two segments of a line drawn from the same point C, without

the circle, that cuts the circumference in two points.

§ 72. PROBLEM. To divide a given straight line in extreme and mean ratio; that is: so that the whole line shall be to the greater part, as this greater part to the smaller part.

APPLIC.

The given straight line AB, (fig. 78,) shall be divided in extreme and mean ratio; that is: so that the following proportion result, viz:

AB AF-AF : FB.

Solution. At the point B, erect a perpendicular upon AB, and make BC, equal to the half of AB; with BC, as radius, from C, as centre, describe the circle EBD; join AC, and protract it to the intersection of the circumference in E; the AE, will be cut by this circumference in the two points D, and E; from AB, cut off the AF, equal to AD; the point F, will divide the given straight line AB, in extreme and mean ratio; that is, it will give the proportion:

AB: AF-AF: FB.

Proof. By the foregoing Theorem, second part, was demonstrated that AB, as tangent from A, to the circumference of the circle at B, is a mean proportional between the two seg ments AE, and AD; thence here

AE: AB-AB: AD;

and by the differences according to the principles of proportions, AE-AB: AB-AB-AD: AD;

and because AB=2CB=DE, and also AD=AF, and ABAD-AB-AF-FB, by the construction; this proportion

that is: the AB, is divided in the point F, so as to make AF, a mean proportional to the whole AB, and the other part of it FB, as has been required.

$73. PROBLEM. To determine the side of a regular decagon or a rectilineal figure of ten sides inscribed in the circle; and also the side of the regular pentagon, or of five equal sides.

APPLIC. In the circle AFDBG, (fig. 79,) the centre of which is C, it is required to describe the side of the regular decagon, and of the regular pentagon.

Solution. Divide the radius CB, of the circle in extreme and mean ratio in the point E, the greater part of it, with CE, lay off the chord BD, it will be the side of the regular deca

gon inscribed in the circle; and the double of the arc which it intersects, or BDF, will have for its chord the side of the regular pentagon.

Proof. The radius CB, being divided in E, so as to make the proportion CB: CE-CE: EB; and by construction BD -CE; join CD, and ED; the two triangles CBD, and BDE, have the angle at B, common; and because of BD=CE, they have the sides about these angles proportional, that is:

CB: BD=BD : EB,

therefore, these triangles are similar, and the angle EDB, is equal to the angle C, at the centre. The triangle BCD, is isosceles, therefore, also, the triangle EBD; the angles BED, EBD, and CDB, are equal, and the side ED, is equal to BD, thence also to CE; therefore, also, the triangle CED, is isosceles, and the angle ECD=CDE. The angle BED, is outward angle to the triangle CED, therefore, equal to the sum of the two inward and opposite angles ECD, and CDE, together, therefore, it is double the angle C, at the centre; and because of BED=CBD-CDB, also, the angle CBD, or CDB, is double the angle C, at the centre; therefore, the angle C, is one-fifth part of two right angles; the chord BD, therefore, subtends an angle at the centre equal to the fifth part of the semi-circumference; that is, the tenth part of the whole circumference; it is, therefore, the chord of the regular decagon inscribed in the circle. Laying of the chord BD, from D, to F; the chord BF, will be the side of the regular pentagon, because the arc BE, is double the arc BD, and therefore contained only five times in the whole circumference.

Corol. The chord of the pentagon can evidently also be determined by drawing from A, the other end of the diameter, the AF, parallel to CD; the arc intercepted will again be the BF, double of BD, because it subtends at the circumference an angle equal to half the angle at the centre.

Scholium. The difference between the arc of the decagon, and that of the hexagon, (determined by § 56:) is the arc of the regular poligon of fifteen sides, inscribed in the same circle:

1

1

for the arc of the decagon being and that of the hexagon of

10

arc of the poligon of 15 equal sides. In like manner, the difference between the arc of the pentagon and that of the hexagon, is

equal to the arc of the poligon of 30 equal sides; that is: their

difference gives

ᄒᄒ

1 30

or an arc that will be contained 30

times in the circumference of the circle. The difference of the arcs of the square and the hexagon, is the arc of the regular

1 1
4

2 1 24 12

; the difference

dodecagon for the difference: 1-6

between the arc of the square and that of the pentagon will be,

20

1. or the arc of the regular poligon of 20 sides.

CHAPTER VI.

Application of Arithmetic to Plane Geometry.

§ 74. PROBLEM. To determine one of the sides of a triangle by the two others, and the segment of the one of them made by a perpendicular from its opposite angle.

APPLIC. The side AC, in the triangle ABC, (fig. 80,) shall be determined by the sides AB, BC, and the segment BD, made by the perpendicular CD, from the angle C, opposite to AB, whether the triangle be acute or obtuse angled.

Determination I. If the triangle is acute angled, then by the § 36, the perpendicular CD, will be determined equally by the two following expressions, or what might be called equations, as in Algebra, as they are to be used as such; viz: CD2 = CA2 AD2 == CB2 DB2;

and because AD — AB — BD; the equation becomes

or

CA2 — (AB — BD)2 = CB2 — DB2 ;
CA2-AB2-BD2+2ABX BD CB-DB2 ;

=

=

that is, CA-AB2 + 2AB × BD CB2; thence CA — CB2 + AB2 - 2AB × BD; which was required to determine.

II. If the triangle is obtuse angled, the simple consideration of CD, falling outside of the triangle, thence being added to AB, instead of subtracted, as above, to give the value of AD, or that AD=AB+ BD, as by the figure, will change the sign in the term in which BD, is a factor, and therefore give the result immediately.

CA2 CB2 + AB2 + 2AB × BD;

which was required.

$75. PROBLEM. The three sides of a triangle being given to determine the perpendicular upon one of the sides from the opposite angle; and the content of the triangle.

APPLIC. In the triangle ABC, (fig. 80,) having given the three sides, AB, AC, and CB, to determine the perpendicular CD, and the content of the triangle.

Determination. By the foregoing proposition (in the first case) has been determined

thence if the above value of BD2, is substituted:

reducing the right hand parts to a common denominator, ob