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therefore the area of a trapezoid is equal to the product of its altitude and the line joining the middle points of the sides. which are not parallel.

THEOREM VI.

16. A line drawn parallel to one side of a triangle divides the other sides proportionally.

In the triangle ABC let DE be drawn parallel to BC; then

AE: ECAD:DB

Draw DC and BE; the triangles A DE and ED C, having the same vertex D, have the same altitude; therefore (13)

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B

D

E

And the triangles A D E and DE B, having the same vertex E, have the same altitude; therefore (13)

ADE: DEBAD: DB

But the triangles EDC and D E B are equivalent (11), since they have the same base D E and the same altitude, viz., the perpendicular distance between the two parallels DE and BC. Therefore (Pn. 11) AE: ECAD: DB

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18. A line dividing two sides of a triangle proportionally is parallel to the third side of the triangle.

In the triangle ABC if DE divides A B and AC so that AE: EC: AD: DB, then D E is parallel to B C.

For if DE is not parallel to BC, through D draw DF parallel to BC; then (16)

AD: DB-AF: FC

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or (Pn. 15)

AF: AE FC: EC

But this proportion is absurd; for AF is less than AE, while FC is greater than EC; therefore D E is parallel to B C.

19. Definition. Similar Polygons are those which are mutually equiangular, and have their homologous sides, that is, the sides including the corresponding angles, proportional.

THEOREM VIII.

20. Two triangles mutually equiangular are similar.

In the two triangles A B C, DEF, let the angle AD, BE, and CF; then the triangles are similar.

As the triangles are equiangular, we have only to prove

B.

G

A

H

C

E

D

F

the homologous sides proportional. Cut off A G and A H equal respectively to DE and DF, and join GH; the triangle AG H is equal to DEF (I. 40), and the angle AGHE; but EB; therefore A GHB, and GH is parallel to BC (I. 18); and (17)

or

AB: AGAC: AH
AB: DE AC: DF.

In like manner it may be proved that

Ꭺ Ᏼ : Ꭰ Ꭼ BC: EF

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AC: DF

21. Corollary. Two triangles whose sides are equally inclined to each other are similar. For if one of the triangles is

turned through an angle equal to the angle of inclination of the sides, the sides of the triangles become respectively parallel; they are therefore equiangular (I. 12) and similar (20).

THEOREM IX.

22. The altitudes of two similar triangles are proportional to the homologous sides.

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ABG, DEH are equiangular (I. 35), and similar (20); therefore BG:EH=AB:DE=AC:DF = BC: E F

THEOREM X.

23. Two triangles having an angle of the one equal to an angle of the other, and the sides including these angles proportional, are similar.

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spectively equal to D E and D F, and join G H; the triangle

AGH

or

DEF, and the angle A GHE (I. 40).

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that is, the sides A B, A C are divided proportionally by the

line GH; therefore G H is parallel to BC (18), and the angle AGH = B (I. 18); but the angle AGH = E; therefore BE, and the two triangles are mutually equiangular and therefore similar (20).

THEOREM XI.

24. In a right triangle the perpendicular drawn from the vertex of the right angle to the hypothenuse divides the triangle into two triangles similar to the whole triangle and to each other.

In the right triangle ABC if BD is drawn from the vertex B of the right angle to the hypothenuse A C, the two triangles ABD, BCD are similar to 4 ABC and to each other.

B

D

The two right triangles A B D and A B C have the acute angle A common; they are therefore equiangular (I. 35), and similar (20). The two right triangles ABC and BCD have the acute angle C common; therefore they are equiangular and similar. The two triangles A B D and B C D, being each similar to ABC, are similar to each other.

25. Cor. 1. Since A B C and A B D are similar triangles AC: ABAB: AD

And since ABC and BCD are similar

AC:CB=CB:CD

that is, in a right triangle either side about the right angle is a mean proportional between the whole hypothenuse and the segment adjacent to that side cut off by the perpendicular drawn from the vertex of the right angle to the hypothenuse.

26. Cor. 2. As ABD and BCD are similar triangles

AD: DB DB: DC

that is, in a right triangle the perpendicular from the vertex of the right angle is a mean proportional between the segments of the base.

THEOREM XII.

27. The square described on the hypothenuse of a right angle is equivalent to the sum of the squares described upon the other two sides.

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that is, the square BIKC is equivalent to the rectangle DCEL. In the same way the square A G H B can be proved equivalent to the rectangle A DLF: therefore the sum of the two rectangles, that is, the square A CEF is equivalent to the sum of the squares BIKC and AGHB; or

28. Corollary. Since

and

A C2 = A B2 + B C2

AC2 = AB2+BC2

BC-AC-AB2

-BCVAC2 — A B2
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