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BOOK SECOND.

DEFINITIONS.

1. In a right-angled triangle, the side opposite the right-angle is called the hypothenuse; and the other two sides are called the legs, and sometimes the base and perpendicular.

2. The base of any rectilineal figure is the side on which the figure is supposed to stand.

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posite side produced, considered as the base; thus, CD is the altitude of the triangle ABC.

4. The altitude of a parallelogram is the perpendicular between two opposite sides considered as bases; thus, FG is the altitude of the parallelogram ABCD.

5. The altitude of a trapezoid is the perpendicular drawn between its two parallel sides; thus, FG is the altitude of the trapezoid ABCD.

A

A

D F

DF

6. A rectangle is said to be con- D tained by its adjacent sides; thus, the rectangle ABCD is contained by the sides AB, CD. For brevity,

it is often referred to as the rectangle BA.AD.

B

THEOREM.

PROPOSITION I.

Two parallelograms having the same base and same altitude, are equivalent.

Let the two parallelograms ABCD, ABFG, have the same base AB, and the same altitude; then will they be equivalent.

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Since they have the same altitude, their upper bases will be in the same line GC, parallel to the common base AB. We have BC equal to AD, and BF equal to AG; we also have DC equal GF, each being equal to AB. If from the whole line GC we take GF, then will remain FC, and if from the same line GC we take DC, then will remain GD equal to FC. If each of these equal lines be taken from the whole line GC, there will remain the line GD. in the one case, equal to the line FC in the other, (Ax. III.) Therefore the three sides of the triangle ADG are equal to the three sides of the triangle BCF, and consequently they are equal, (B. I, Prop. vIII.) If each of these equal triangles be taken from the whole space ABCG, there will remain the

parallelogram ABCD in the one case, equal to the parallelogram ABFG in the other, (Ax. III.)

Cor. 1. Parallelograms having the same base, and being situated between the same parallels, are equivalent; for their altitudes will be perpendiculars between the two parallels, which are all equal by the definition of parallels.

Cor. 2. Parallelograms having equal bases and altitudes are equivalent; for they may be so applied as to have their equal bases coincide, and then by this proposition they will be equivalent.

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PROPOSITION II.

THEOREM. Two triangles having the same base and same altitude, are equivalent.

Let the two triangles ABC, ABD have the same base AB, and the same altitude; then will they be equivalent.

D

A

B

their vertices will be Draw AF parallel to

Since they have the same altitude, the line CD which joins parallel to the common base AB. BC, and BG parallel to AD; thus forming the two parallelograms ABCF and ABGD, which are equivalent. (B. II, Prop. 1.) The triangle ABC is one half of the parallelogram ABCF, and the triangle ABD is one half of the parallelogram ABGD, (B. I, Prop. xxvII;) therefore the triangle ABC is equivalent to the triangle ABD.

Cor. 1. Triangles having the same base, and situated

between the same parallels, are equivalent; for the altitude is the perpendicular between the two parallels, which is everywhere equal.

Cor. 2. Triangles having equal bases and the same altitude, or being situated between the same parallels, are equivalent.

(44.) PROBLEM. To find a triangle that shall be equivalent to a given polygon.

Let ABCDF be the given polygon. Draw the diagonal CF, cutting off the triangle CDF: through the point D, draw DG parallel to CF, and meeting DF produced; draw CG, and the polygon ABCDF will be equivalent to ABCG, which has one side less than the original polygon.

B

H

A

F G

For, the triangles CDF, CGF have the common base CF; and they have the same altitude, since their vertices D and G are situated in the line DG which is parallel to the base CF; therefore these triangles are equivalent, (B. II, Prop. 11.) To each add the figure ABCF, and we shall have the polygon ABCDF equivalent to the polygon ABCG.

For a similar reason, the triangle CAH may be substituted for the equivalent triangle CAB, and thus the pentagon ABCDF will be changed into an equivalent triangle HCG.

It is obvious the same process may be applied to a polygon having any number of sides; each step lessening the number of sides by one, until we finally arrive at an equivalent triangle.

PROPOSITION III.

THEOREM. If a parallelogram and a triangle have the same base and the same altitude, the triangle will be half the parallelogram.

Let the parallelogram ABCD, and the triangle ABF, have the same base AB, and the same altitude; then will the triangle be half the parallelogram.

B

For, draw the diagonal BD of the parallelogram, dividing it into two equal parts, (B. I, Prop. xxvII.) Now, since the triangles ABF, ABD have the same base and the same altitude, they are equivalent, (B. II, Prop. II;) and since ABD is half the parallelogram ABCD, it follows that ABF is also half the same parallelogram.

Cor. A triangle is half the parallelogram having the same base, and being situated between the same parallels; for, being between the same parallels, they must have the same altitude, and by this proposition the triangle must be half the parallelogram.

PROPOSITION IV.

THEOREM. A trapezoid is equivalent to half a parallelogram, whose base is the sum of those two sides, and its altitude the perpendicular distance between them.

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the sum of the two parallel sides; also produce DC, and let the three lines CK, HB, GF be parallel to AD;

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