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homologous sides are proportional: hence,

OC : OA :: OA : OD;

which was to be proved.

Cor. From the above proportion, we have,

A 02 = OC × OD;

that is, the square of the tangent is equal to the rectangle of the secant and its external segment.

PRACTICAL APPLICATIONS.

PROBLEM I.

To divide a given straight line into parts proportional to given straight lines: also into equal parts.

1o. Let AB be a given straight line, and let it be required to divide it into parts proportional to the lines P, Q, and R.

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draw CI and DF parallel to EB: then will AI, IF, and FB, be proportional to P, Q, and R (P XV., C. 2).

2o. Let AH be a given straight line, and let it be required

to divide it into any number of equal parts, say five.

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then will AH be

KD, LE, and MF, parallel to BH

divided into equal parts at C, D, E, and F (P. XV.,

C. 2).

PROBLEM II.

To construct a fourth proportional to three given straight lines.

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AC: then will DX be the fourth proportional required.

or,

For (P. XV., C.), we have,

DA : ᎠᏴ :: DC: DX;

A : B :: C : DX.

Cor. If DC is made equal to DB, DX will be a third proportional to DA and DB, or to A and B.

PROBLEM III.

To construct a mean proportional between two given straight lines.

Let A and B be the given lines. On an indefinite line, lay off DE equal to A, and EF equal to B; on DF as a diameter describe the semi-circle DGF, and draw EG perpendicular to DF:

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then will EG be the mean proportional required.

For (P. XXIII., C. 2), we have,

DE : EG EG:

EF;

or,

A : EG :: EG : B.

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PROBLEM IV.

To divide a given straight line into two such parts, that the greater part shall be a mean proportional between the whol line and the other part.

Let AB be the given line. At the extremity B, draw BC perpendicular to AB, and make it equal to half of AB. With as a centre, and CB

88 a radius, describe the arc

DBE; draw AC, and produce

D

it till it terminates in the concave arc at E; with A as centre and AD as radius, describe the arc DF: then will AF be the greater part required.

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and, by inversion (B. II., P. V.),

AB : AF :: AF FB.

Scholium. When a straight line is divided so that the greater segment is a mean proportional between the whole line and the less segment, it is said to be divided in extreme and mean ratio.

Since AB and DE are equal, the line AE is divided in extreme and mean ratio at D; for we have, from the first of the above proportions, by substitution,

AE DE :: DE: AD.

PROBLEM V.

Through a given point, in a given angle, to draw a straight line so that the segments between the point and the sides of the angle shall be equal.

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Let BCD be the given angle, and 4 the given point. Through A, draw AE parallel to

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but, FE is equal to EC; hence, EC; hence, FA is equal to AD.

PROBLEM VI.

To construct a triangle equal to a given polygon.

Let ABCDE be the given polygon.

Draw CA; produce EA, and draw BG BG parallel to CA; draw the line CG. Then the triangles BAC and GAC have the common base AC, and because their vertices B and G lie in the

B

A

Ε

same line BG parallel to the base, their altitudes are equal. nd consequently, the triangles are equal: hence, the polygon GCDE is equal to the polygon ABCDE.

Again, draw CE; produce AE and draw DF parallel to CE; draw also CF; then will the triangles FCE and DCE be equal: be equal: hence, the triangle GCF is equal to the polygon GCDE, and consequently, to the given polygon. In like manner, a triangle may be constructel equal to any other given polygon.

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