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31 each to each, to which the equal fides are opposite. therefore the Book !. angle ACB is equal to the angle CBD. and because the straight line m BC meets the two straight lines AC, BD and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD, c. 27. 1. and it was fhewn to be equal to it, therefore straight lines &c. Q. E. D.

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PROP. XXXIV. THEOR.

а

2. 29.

THE opposite sides and angles of parallelograms are

equal to one another, and the diameter bisects them, that is, divides them into two equal parts.

N. B. A Parallelogram is a four sided figure of which the opposite fides are parallel. and the diameter is the fraight line joining two of its opposite angles.

Let ABCD be a parallelogram, of which BC is a diameter. the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it.

Becaufe AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to

А

B one another, and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal o to one another. wherefore the two triangles ABC, C

D CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their cqual angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other 5, viz. the side AB to the side CD, and AC to BD, and the b. 26.1. angle BAC equal to the angle BDC. and because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB; the whole angle ABD is equal to the whole angle ACD, and the angle BAC has been shewn to be equal to the angle BDC; therefore the opposite sides and angles of parallelograms are equal to one another. also, their diameter bisects them. for, AB being equal to CD, and BC common; the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC is equal to the angle BCD;

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gures.

Book I. therefore the triangle ABC is equal to the triangle BCT, and the

diameter BC divides the parallelogram ACDB into two cqual parts. C. 4. 1. Q. E. D.

PROP. XXXV. THEOR. See N. PARALLELOGRAMS upon the same base and between

the same parallels, are equal to one another,

Let the parallelograms ABCD, EBCF be upon the same base BC and 3d Fi-and between the same parallels AF, BC. the pa, allelogram ABCD

shall be equal to the parallelogram EBCF.

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 A D F of the parallelograms is double of the triangle BDC; and they are therefore equal to one another.

But if the sides AD, EF opposite to the base BC of the parallelograms ABCD, B С EBCF be not terminated in the same point; then because ABCD is a parallelogram, AD is equal“ to

BC; for the same reason, EF is equal to BC; wherefore AD is 6. 1. Ax. equal 6 to EF; and DE is common; therefore the whole, or the c. 2. or 3. remainder, AE is equal to the whole, or the remainder DF; AB

also is equal to DC; and the two EA, AB are therefore equal to

А DE F A E D F

2. 34. I.

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B

B C the two FD, DC, each to each; and the exterior angle FDC is ed. 29. 1. qual d to the interior EAB; therefore the base EB is equal to the

base FC, and the triangle EAB equal o to the triangle FDC. take the triangle FDC from the trapezium ABCF, and from the same

trapezium take the triangle EAB ; the remainders therefore are f. 3. Ax. equalf, that is, the parallelogram ABCD is equal to the paralle

logram EBCF. therefore parallelograms upon the fame base &c. Q. E. D.

PROP.

6. 4. 1.

33 PROP. XXXVI. THEOR.

Book I. PARALLELOGRAMS upon equal bases and between the

same parallels, are equal to one another. Let ABCD, EFGH be parallelograms upon equal bases BC, FG, and between the same

A D E H parallels AH, BG; the párallelogram ABCD is equal to EFGH.

Join BE, CH; and becáuse BC is equal to FG, and FG to a EH, BC is

B

С F equal to EH; and they

a. 34. ##

G are parallels, and joined towards the fame parts by the straight lines BE, CH. but straight lines which join equal and parallel straight lines towards the fame parts, are themselves equal and parallel b; there. 6. 33. ti fore EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD, because it is upon the fame base c. 35. i. BC, and between the fame parallels BC, AD. for the like reason the parallelogram EFGH is equal to the same EBCH. therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D.

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PROP. XXXVII. THEOR.
RIANGLÈS upon the same base, and between the

same parallels, are equal to one another.
Let the triangles ABC, DBC be upon the famie base BC and be-
tween the fame parallels AD,

E A D F BC. the triangle ABC is equal to the triangle DBC.

Produce AD both ways to the points E, F, and thro' B

a. 31. t. draw a BE parallel to CA; and thro' C draw CF parallel

B

С to BD. therefore each of the figures EBCA, DBCr is a parallelogram; and EBCA is equal to b. 35.11 DBCF, because they are upon the fame bafe BC, and between the fame parallels BC, EF; and the triangle ABC is the half of the pa

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rallelogram

A

Book I. ralfelogram EBCA, because the diameter AB bisects e it; and the

triangle DBC is the half of the parallelogram DBCF, because the C. 34, 1. diameter DC bisects it. but the halves of equal things are equal d; d. 7. Ax. therefore the triangle ABC is equal to the triangle DBC. Where

fore triangles, &c. Q. E. D.

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'RIANGLEs upon equal bases, and between the same

parallels, are equal to one another.

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Let the triangles ABC, DEF Le upon equal bases BC, EF, and between the fame parallels BF, AD. the triangle ABC is equal to the triangle DEF.

Produce AD both ways to the points G, H, and thro'B draw.BG X. 31. 1. parallel to CA, and thro' F draw FH parallel to ED. then cach of the figures GBCA,

G А. D H. DEFH is a parallelo

gram; and they are 8. 36. 1. equal b to one ano

b
ther, because they
are upon equal bales
BC, EF and between

B CE F the fame parallels BF, €. 34. 1. GH; and the triangle ABC is the halfo of the parallelogram GBCA,

because the diameter AB bisects it; and the triangle DEF is the

half of the parallelogram DEFH, because the diameter DF bisects d. 7. Ax. it. but the halves of equal things are equal d; therefore the tri

angle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E, D.

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EQUAL triangles upon the fame base, and upon the

fame side of it, are between the same parallels.

Let the equal triangles ABC, DBC be upon the fame base BC, and upon the same fide of it; they are between the same parallels.

Join AD; AD is parallel to BC; for if it is not, thro' the point 2.31. 5. A draw? AE parallel to BC, and join EC, the triangle ABC is

.

equal

2

3$ equat b to the triangle EBC, because it is upon the same base BC, Book I.

b and between the same parallels BC, AE. A

D but the triangle ABC is equal to the tri

b; 37: angle BDC; therefore also the triangle

E BDC is equal to the triangle EBC, the greater to the less, which is impossible. therefore AE is not parallel io BC. in the same manner it can be B

с demonstrated that no other line but AD'is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upor, &c. C. E. D.

PROP. XL. THEOR.

EQUAL triangles upon equal bases, and towards the

same parts, are between the same parallels.

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Let the equal triangles ABC, DEF be upon equal bases BC, EF; and towards the fame parts;

А D they are between the same parallels. Join AD; AD is parallel

G to EC. for if it is not, thro' A draw a AG parallel to BF, and join GF. the triangle

B CE Е. ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and between the fame parallels BF, AG, but the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible. therefore AG is not parallel to BF. and in the same manner it can be demonstrated that there is no other parallel to it but AD, AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D.

F B. 38. 84

PROP. XLI. THEOR.
IF a parallelogram and triangle be upon the fame base,

and between the same parallels; the parallelogram shall be double of the triangle.

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