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PROP. XXXV.-THEOREM.

Parallelograms upon the same base, and between the same parallels, are equal to each other.

(References-Prop. 1. 4, 29, 34; ax. 1, 3, 6.)

Let the parallelograms ABCD, EBCF, be on the same base BC, and between the same parallels AF, BC.

Then the parallelogram ABCD shall be equal to the parallelogram EBCF.

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If the sides AD, DF, of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D,

it is plain that each of the parallelograms ABCD, DBCF, is double of the triangle BDC; (1. 34)

and therefore the parallelograms ABCD, DBCF, are equal to one another. (ax. 6.)

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But if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point;

then, because ABCD is a parallelogram,

therefore AD is equal to BC; (I. 34)

for the same reason EF is equal to BC;

wherefore AD is equal to EF; (ax. 1)

and DE is common;

therefore the whole, or remainder* AE, is equal to the whole, or remainder DF; (ax. 2 or 3)

*The words 'whole or remainder' are used to suit both figures (2 and 3), since, in figure 2, DE is added to AD and EF, which therefore gives the whole AE; and in figure 3, DE is taken away from AD and EF, which therefore leaves the remainder AE.

and AB is equal to DC. (1. 34.)

Wherefore, in the triangles EAB, FDC,

because the two EA, AB, are equal to the two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB, (1. 29)

therefore the base EB is equal to the base FC,

and the triangle EAB equal to the triangle FDC. (I. 4.)

Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB;

the remainders are therefore equal, (ax. 3)

that is, the parallelogram ABCD is equal to the parallelogram EBCF.

Therefore, parallelograms upon the same base, &c.

Q. E. D.

PROP. XXXVI.-THEOREM.

Parallelograms upon equal bases and between the same parallels are equal to one another.

(References-Prop. 1. 33, 34, 35; ax. 1.)

Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG.

Then the parallelogram ABCD shall be equal to the parallelogram EFGH.

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From B to E, draw the straight line BE, and from C to H, the straight line CH.

DEMONSTRATION

Because BC is equal to FG, (hyp.) and FG to EH, (1. 34)

therefore BC is equal to EH; (ax. 1)

and they are parallels, and joined towards the same parts by the straight lines BE, CH; (hyp.)

but 'straight lines which join equal and parallel straight lines towards

the same parts, are themselves equal and parallel ;' (1. 33)

therefore EB, CH, are both equal and parallel;

wherefore EBCH is a parallelogram. (1. 34, def.)

Then because the parallelograms ABCD, EBCH, are upon the same base BC,

and between the same parallels BC, AH;

therefore the parallelogram ABCD is equal to the parallelogram EBCH; (1.35)

for the like reason, the parallelogram EFGH is equal to the same EBCH;

therefore the parallelogram ABCD is equal to the parallelogram EFGH. (ax. 1.)

Wherefore, parallelograms, &c.

Q. E. D.

PROP. XXXVII.-THEOREM.

Triangles upon the same base, and between the same parallels, are equal to one another.

(References-Prop. 1. 31, 34, 35; ax. 7.)

Let the triangles ABC, DBC, be on the same base BC, and between the same parallels AD, BC.

Then the triangle ABC shall be equal to the triangle DBC.

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Produce AD both ways to the points E, F,

and through B draw BE parallel to CA, and through C draw CF parallel to BD. (I. 31.)

DEMONSTRATION

Then, each of the figures EBCA, DBCF, is a parallelogram; (1. 34, def.)

and because the parallelograms EBCA, DBCF, are upon the same base BC,

and between the same parallels BC, EF,

therefore the parallelogram EBCA is equal to the parallelogram DBCF; (1. 35)

and because the diameter AB bisects the parallelogram EBCA;

therefore the triangle ABC is the half of EBCA; (1. 34)

and because the diameter DC bisects the parallelogram DBCF,

therefore the triangle DBC is the half of DBCF;

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but things which are halves of the same are equal to one another,' (ax. 7)

therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles, &c.

Q. E. D.

PROP. XXXVIII.—THEOREM.

Triangles upon equal bases and between the same parallels are equal to one another.

(References-Prop. 1. 31, 34, 36; ax. 7.)

Let the triangles ABC, DEF, be on the equal bases BC, EF, and between the same parallels AD, BF.

Then the triangle ABC shall be equal to the triangle DEF.

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Produce AD both ways to the points G, H,

and through B draw BG parallel to CA, and through F draw FH parallel to ED. (L. 31.)

DEMONSTRATION

Then, each of the figures GBCA, DEFH, is a parallelogram; (1 34 def.)

and because they are upon equal bases BC, EF,

and between the same parallels BF, GH,

therefore these parallelograms are equal to one another. (1.36.) And because the diameter AB bisects the parallelograms GBCA, therefore the triangle ABC is the half of GBCA; (1. 34)

and because the diameter DF bisects the parallelogram DEFH,

therefore the triangle DEF is the half of DEFH;

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but things which are halves of the same are equal to one another,' (ax 7.)

therefore the triangle ABC is equal to the triangle DEF. Wherefore, triangles, &c.

Q. E. D.

PROP. XXXIX.- THEOREM.

Equal triangles upon the same base and on the same side of it, are between the same parallels.

(References Prop. I. 31, 37; ax. 1.)

Let the equal triangles ABC, DBC, be upon the same base BC, and on the same side of it.

Then they shall be between the same parallels.

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For, if it be not, through the point A draw AE parallel to BC, (I. 31) and join EC.

DEMONSTRATION

Then, because the triangles ABC, EBC, are on the same base BC, and are assumed to be between the same parallels BC, AE;

therefore the triangle ABC must be equal to the triangle EBC; (1. 37) but the triangle ABC is equal to the triangle DBC; (hyp.)

therefore the triangle DBC must be equal to the triangle EBC, (ax. 1) the greater to the less, which is impossible;

therefore AE is not parallel to BC.

In the same manner, it can be demonstrated that no other line but AD is parallel to BC;

therefore AD is parallel to BC.

Wherefore, equal triangles upon, &c.

Q.E.D.

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Equal triangles upon equal bases in the same straight line and towards the same parts are between the same parallel.

(References Prop. 1. 31, 38; ax. 1.)

Let the equal triangles ABC, DEF, be upon the equal bases BC, EF, in the same straight line BF, and towards the same parts.

Then the triangles ABC, DEF, shall be between the same parallels.

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