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Proposition 19. Theorem.

316. Two triangles which have their sides parallel or perpendicular, each to each, are similar.

Hyp. In the As ABC, A'B'C', let AB, AC, BC be || or I, respectively, to A'B', A'C', B'C'.

To prove A ABC similar to
A A'B'C'.
Proof. Since the sides are 11 or

A
I, each to each, the included Z8

B' are equal or supplementary (80, 83).

... three hypotheses may be made:

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B

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(1) A + A' = 2 rt. Zs, B+B' = 2 rt. Zs, C + C'= 2 rt. Zs. (2) A = A', B+B' = 2 rt. Zs, C+C' = 2 rt. Zs. (3) A = A',

B = B', ..C= C'. The first and second hypotheses are inadmissible, since the sum of the Zs of the two As cannot exceed 4 rt. Ls (97). Therefore the third is the only admissible hypothesis.

... the As ABC, A'B'C' are similar,
being mutually equiangular (309).

Q.E.D.

317. Sch. The homologous sides of the two triangles are either the parallel or the perpendicular sides.

EXERCISE.

If the sides of the triangle in Prop. 15 are AB = 16, BC = 10, and AC = 8, find the lengths of the segments BD and CD.

=

To prove

B

Proposition 20. Theorem. 318. If in any triangle a parallel be drawn to the base, all lines from the vertex will divide the base and its parallel proportionally.

Hyp. Let ABC be a A, B'C' || to
BC, and AD, AE two lines intersect-
ing B'C' at D'E'.
BD DE EC

D'/ E'
B'D' D'E EO
Proof. Since B'C' is || to BC, the

D E C
AS AB'D', ABD are similar; also the
AS AD'E' and ADE; and the As AE'C' and AEC. (311)

AD BD AD DE

and
AD B'D' AD = D'E' >
their homologous sides being proportional (307).

BD DE
B'D' D'E

(Ax. 1) Ᏼ '

''
In the same way it may be shown that

DE EC
D'E E'C'

BD DE ЕС
B'D' D'E E'C'.

Q.E.D. 319. COR. If BD = DE EC, we shall have B'D' = D'E' = E'C'. Therefore, if the lines from the vertex

. divide the base into equal parts, they will also divide the parallel into equal parts.

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EXERCISES.

1. If BD = 8, B'D' = 6, DE = 10, and EC = 3, find the lengths of D'E' and E'C'. 2. If AD = 20, AD' = 16, AE

15, BD = 7,

and DE = 5, find B'D', D'E', and AE'.

Proposition 21. Theorem.

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

Hyp. Let the As ABC, ACD, ADE E of the polygon ABCDE be similar respectively to the As A'B'C', A'C'D',

A A'D'E' of the polygon A'B'C'D'E', and similarly placed.

B
To prove the polygons are similar.

E'
Proof. Since the homologous Z8 of
similar As are equal,

(307)

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In like manner, Z CDE = _ C'D'E'; etc.

.:. the polygons are mutually equiangular.

Since similar As have their homologous sides proportional,

(307)

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... the homologous sides of the polygons are proportional.
.. the polygons are similar,
being mutually equiangular and having their homologous sides
proportional (307).

Q.E.D.

Proposition 22. Theorem.

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321. Conversely, two similar polygons may be divided into the same number of triangles, similar each to each, and similarly placed.

Hyp. Let ABCDE, A'B'C'D'E' be two similar polygons divided into As by the diagonals AC, AD, A'C', A'D' drawn

A from the homologous Zs A and A'.

B To prove As ABC, ACD, ADE sim

E' ilar respectively to As A'B'C', A'C'D', A'D'E'.

c Proof. Since the polygons are similar, AB BC

B' ...B= LB', and

(307) A'B' B'C'

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.:. ABC is similar to A A'B'C'.

(314)

ACB = L A'C'B'.

(307)

But
_ BCD = _ B'C'D'.

(307) ... remaining – ACD = remaining _ A'C'D'.

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In the same way

it

may be shown that As ADE, A'D'E' are similar.

Q.E.D. COR. The homologous diagonals of two similar polygons are proportional to the homologous sides.

Proposition 23. Theorem.

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322. The perimeters of two similar polygons are to each other as any two homologous sides.

Hyp. Let ABCDE, A'B'C'D'E' be two similar polygons; denote their perimeters by P and P'. То prove P:P' = AB : A'B'.

B

E' Proof. Since the polygons are similar,

A

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1. From the ends of a side of a triangle any two straight lines are drawn to meet the other sides in P, Q ; also from the same ends two lines parallel to the former are drawn to meet the sides produced in P', Q': show that PQ is parallel to P'Q'.

Let ABC be the A; BP, CQ the lines. AC: AP = AQ': AB, etc.

2. ABC is a triangle, AD any line drawn from A to a point D in BC; a line is drawn from B bisecting AD in E and cutting AC in F: prove that BF is to BE as 2 CF is to AC. Draw EG || to AC meeting BC in G. DE = EA, .. AC = 2 EG, etc.

3. If in Prop. 20, AD = 24, AD' = 20, AE = 16, BD = 8, and DE= 6, find B'D', D'E', and AE'.

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