For, draw AC and DB. The triangles ODB and OAC have the angle O common, and the angles OBD and OCA equal, because each is measured

by half of the arc AD: hence, they are similar, and consequently, their homologous sides are proportional; whence,

that is, the rectangles of each secant and its external seg ment are equal.

If from a point without a circle, a tangent and a secant be drawn, the secant terminating in the concave arc, the tangent will be a mean proportional between the secant and its external segment.

Let ADC be a circle, OC a secant, and OA a tangent: then will

OC : ОА :: OA : OD.

For, draw AD and AC. The triangles OAD and OAC will have the angle common, and the angles OAD and ACD equal, because each is measured by half of the arc AD (B. III., P. XVIII., P. XXI.); the triangles are therefore similar, and consequently, their

A

homologous sides are proportional: hence,

OC : ОА :: OA : OD;

which was to be proved.

Cor. From the above proportion, we have,

A02 = OC × OD;

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

PRACTICAL APPLICATIONS.

3123

PROBLEM L.

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. From one extremity A,

1 F B

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

2°. Let AH be a given straight line, and let it be required to divide it into any number of equal parts, say five.

BH, and from I,
I, K, L, and M, draw the lines IC,
KD, LE, and MF, parallel to BII: then will AH be
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.

AC: then will DX be the fourth proportional required.

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

DA : DB :: DC: DX;

A : B :: C: DX.

OF,

Cor. If DC is made equal to DB, DX thi d proportional to DA and DB, or to A

will be

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:

D

E

B:

AF

then will EG be the mean proportional required,

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

To divide a given straight line into two such parts, that the greater part shall be a mean proportional between the whole 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 C as a centre, and CB 88 a radius, describe the are DBE; draw AC, and produce

it till it terminates in the concave arc at centre and AD as radius, describe the will AF be the greater part required.

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.'