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PROPOSITION X. THEOREM.

329. Straight lines drawn through the same point intercept proportional segments upon two parallels.

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Let the two parallels AE and A'E' cut the straight Lines OA, OB, OC, OD, and OE.

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Proof. Since A'E' is to AE, the pairs of A OAB and OA'B', OBC and OB'C', etc., are mutually equiangular and

similar,

AB OB

and

A'B' OB'

BC OB
B'C' OB

(homologous sides of similar ▲ are proportional).

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REMARK. A condensed form of writing the above is

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§ 319

Ax. 1

Q. E. D.

where a parenthesis about a ratio signifies that this ratio is used to prove the equality of the ratios immediately preceding and following it.

PROPOSITION XI. THEOREM.

330. CONVERSELY: If three or more non-parallel straight lines intercept proportional segments upon two parallels, they pass through a common point.

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Let AB, CD, EF, cut the parallels AE and BF so that AC: BD CE : DF.

To prove that AB, CD, EF prolonged meet in a point.

Proof. Prolong AB and CD until they meet in O.

Join OE.

If we designate by F the point where OE cuts BF, we shall have by $ 329,

But by hypothesis

AC: BD CE: DF'.

AC: BDCE: DF

These proportions have the first three terms equal, each to each; therefore the fourth terms are equal; that is,

DFDF

.. F' coincides with F

.. EF prolonged passes through O.

.. AB, CD, and EF prolonged meet in the point O.

Q. E. D.

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331. If two polygons are composed of the same number of triangles, similar each to each, and similarly placed, the polygons are similar.

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In the two polygons ABCDE and A'B'C'D'E', let the triangles AEB, BEC, CED be similar respectively to the triangles A'E'B', B'E'C', C'E'D'.

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In like manner we may prove BCD = Z B'C'D', etc. Hence the two polygons are mutually equiangular.

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(the homologous sides of similar are proportional). Hence the homologous sides of the polygons are proportional. Therefore the polygons are similar, § 319 (having their homologous equal, and their homologous sides proportional).

Q. E D.

PROPOSITION XIII. THEOREM.

332. If two polygons are similar, they are composed of the same number of triangles, similar each to each, and similarly placed.

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Let the polygons ABCDE and A'B'C'D'E' be similar. From two homologous vertices, as E and E', draw diagonals EB, EC, and E'B', E'C'.

To prove

A EAB, EBC, ECD

similar respectively to ▲ E'A'B', E'B'C', E'C'D'.

Proof. In the ▲ EAB and E'A'B',

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(being homologous sides of similar polygons).
..A EAB and E'A'B' are similar,

(having an of the one equal to an ▲ of the other, and the including sides

§ 326

proportional).

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(having an of the one equal to an ▲ of the other, and the including sides

proportional).

In like manner we may prove ▲ ECD and E'C'D' similar.

PROPOSITION XIV. THEOREM.

Q. E. D.

333. The perimeters of two similar polygons have the same ratio as any two homologous sides.

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Let the two similar polygons be ABCDE and A'B'C'D'E', and let P and P' represent their perimeters.

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AB: A'B' BC: B'C' CD: C'D', etc.,

=

=

(the homologous sides of similar polygons are proportional).

=

§ 319

.. AB+ BC, etc. : A'B' + B'C', etc. AB: A'B', § 303 (in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent).

That is,

P: P

AB: A'B'.

Q. E D.

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