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2o. The polygons will be to each other as the squares of any two homologous sides.

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like parts of the polygons to which they belong, the polygons will be to each other as these triangles; but these triangles, being similar, are to each other as the squares of AB and FG: heuce, the polygons are to each other as the squares of AB and FG, or as the squares of any other two homologous sides; which was to be proved.

Cor. 1. Perimeters of similar polygons are to each other as their homologous diagonals, or as any other homologous lines; and the polygons are to each other as the squares of their homologous diagonals, or as the squares of any other homologous lines.

Cor. 2. If the three sides of a right-angled triangle be made homologous sides of three similar polygons, these polygons will be to each other as the squares of the sides of the triangle. But the square of the hypothenuse is equal to the sum of the squares of the other sides, and consequently, the polygon on the hypothenuse will be equal to the sum of the polygons on the other sides.

PROPOSITION XXVIII. THEOREM.

If two chords intersect in a circle, their segments will be reciprocally proportional.

Let the chords AB and CD intersect at 0: then

will their segments be reciprocally proportional; that is, one segment of the first will be to one segment of the second, as the remaining segment of the second is to the remaining segment of the first.

For, draw CD and BA. Then will the angles ODB and OAC be equal, because each is measured by half of the arc CB (B. III., P. XVIII.). The angles OBD and OCA, will also be equal, because each is measured by

A

half of the arc AD: hence, the triangles OBD and OCA are similar (P. XIX., C.), and consequently, their homologous sides are proportional: hence,

DO : AO :: OB OC;

which was to be proved.

Cor. From the above proportion, we have,

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that is, the rectangle of the segments of one chord is equal to the rectangle of the segments of the other.

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If from a point without a circle, two secants be drawn terminating in the concave arc, they will be reciprocally proportional to their external segments.

Let OB and OC be two secants terminating in the concave arc of the circle BCD: then will

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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,

OB : OC :: OD : OA;

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

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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 gent: then will

be a circle, OC a secant, and OA a tan

OC : OA :: 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

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Cor. From the above proportion, we have,

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

PRACTICAL APPLICATIONS.

PROBLEM I.

To aivide a given line into parts proportional to given lines, also into equal parts.

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

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from the points C C and D,

1 F B

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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 line, and let it be required to divide it into any number of equal parts, say five.

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BH, and from I, K, L, and M, draw the lines IC, KD, LE, and MF, parallel to BH: 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 lines.

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to C; draw AC, and from B draw BX parallel to AC: then will DX be the fourth proportional required. For (P. XV., C.), we have,

or,

DA : DB:: 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.

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