Any two homologous triangles are like parts of the polygons to which they belong.

For, the homologous triangles being similar, we have,

hence, ABC: FGH :: ACD : FHI ::

ADE : FIK.

Whence, by composition (B. II., P. X.),

ABC : FGH :: ACD + ABC + ADE : FHI+FGH + FIK; ABC : FGH :: ABCDE : FGHIK.

that is,

Cor. 2. If two polygons are made up of similar triangles, similarly placed, the polygons themselves are similar.

The perimeters of similar polygons are to each other as any two homologous sides; and and the polygons are to each other as the squares of any turo homologous sides.

1°. Let ABCDE and FGHIK be similar polygons: then their perimeters are to each other as any two homologous sides.

For, any two homologous sides, as AB and FG, are like parts of the perimeters to which they belong: hence (B. II., P. IX.), the perimeters of the polygons are

B

to each other as AB to FG, or as any other two homologous sides; which was to be proved.

2°. The polygons are to each other as the squares of any two homologous sides.

which they belong, the polygons are to each other as these triangles; but these triangles, being similar, are to each other as the squares of AB and FG: hence, 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 are made made homologous sides of three similar polygons, these polygons are 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 are reciprocally proportional.

Let the chords AB and CD intersect at O: then are

their segments 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 CA and BD. ODB and OAC are equal,

Then the angles

because each is

measured by half of the arc CB (B. III., P. XVIII.). The angles OBD and OCA are also equal, because each is measured by half of the arc AD: hence, the triangles OBD

B

and OCA are similar (P. XVIII., 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,

DO XOC = AO X OB;

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

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

Let OB and OC be two secants terminating in the

concave arc of the circle BCD: then

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 are AD: hence, they are similar, and consequently, their homologous sides are proportional; whence,

OB : OC :: OD : OA;

which was to be proved.

Cor. From the above proportion, we have,

OB X OA OC XOD;

B

D

A

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

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

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

OC : OA :: OA: OD.

For, draw AD and AC. The triangles OAD and OAC have the angle O 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 homologous sides are proportional hence,

A

AO2 = OCXOD;

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.

From one extremity A, draw the indefinite line AG, making any angle with AB; lay off AC equal to P, CD equal to Q, and DE equal to R; draw EB, and from the points C and D, draw Cl and DF parallel to EB: then

PH

RH

A

F B

Al, IF, and FB, are proportional to P, Q, and R (P. XV.,

C. 2).