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fore the angle ACB is equal to the angle CBD; and be- Book I. cause the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD; and it was shown to e 27. 1. be equal to it. Therefore, straight lines, &c. Q.E. D.

PROP. XXXIV. THEOR. 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 sides are parallel ; and the diameter is the straight line joining two of its opposite angles. .

Let ACDB 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 bise: ts it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one anothera; and

* 29. 1. because AC is parallel to BD, and BC meets them, the alter

D pate angles ACB, CBD, are equala to one another; wherefore the two triangles ABC, 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 equal 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”, viz. the side AB to the side CD, and b 26. . AC to BD, and the 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 shown 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; therefore the triangle

Book I. ABC is equalc to the triangle BCD, and the diameter BC

n divides the parallelogram ACDB into two equal parts.

* 4. 1. 2. E. D.

PROP. XXXV. THEOR. Sve N. PARALLELOGRAMS upon the same base, and be

tween the same parallels, are equal to one another. See the 2d. Let the parallelograms ABCD, EBCF be upon the same and 3d fi- base BC, and between the same parallels AF, BC; the pasures.

rallelogram ABCD shall be equal to the parallelogram
EBCF.
· If the sides AD, DF of the pa- A
rallelograms ABCD, DBCF, op-
posite to the base BC, be termi-
nated in the same point D; it is

plain that each of the parallelo* 34. 1. grams is doublea 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, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD

is equala to BC; for the same reason EF is equal to BC; 01 Ax. wherefore AD is equalb to EF; and DE is common; c 2 or 3 therefore the whole, or the 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

A D E F A E DF

Ax

BČ..

the two FD, DC, each to each; and the exterior angle 29. 1. FDC is equald to the interior EAB, therefore the base EB • 4. 1. is equal to the base FC, and the triangle EAB equale to

the triangle FDC. Take the triangle FDC from the tra.

pezium ABCF, and from the same trapezium take the trif3 Ax. angle EAB : the remainders therefore are equalf, that is, the

parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E. D. Book I. · PROP. XXXVI. THEOR. PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another. Let ABCD, EFGH, be

DE parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram ABCD is, equal to EFGH.

'Join BE, CH; and R because BC is equal to FG, and FG toa EH, BC is equal • 34. 1. to EH; and they are parallels, and joined towards the same parts by the straight lines BE, CH. But straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallell; therefore EB,.33. 1. CH, are both equal and parallel, and EBCH is a parallelogram; and it is equal to ABCD, because it is upon the < 35. 1. same base BC, and between the same parallels BC, AD: For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD iş equal to EFGH. Wherefore parallelograms, &c. Q. E.D.

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PROP. XXXVII. THEOR. TRIANGLES upon the same base, and between the same parallels, are equal to one another.

Let the triangles ABC, DBC, be upon the same base BC, and between the same f parallels AD, BC: The triangle ABC is equal to the triangle DBC.

Produce AD both ways to the points E, F, and through B drawa BE pa

. *31.1. rallel to CA; and through C draw CF parallel to BD; therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equalb. 35. 1. to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram, EBCA, because the

Book I: diameter AB bisects it; and the triangle DBC is the half

of the parallelogram DBCF, because the diameter DC bic 34. 1. Ax: sects it: but the halves of equal things are equald; there.

fore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

PROP. XXXVIII. THEOR.

A

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

Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BÈ,AD: The triangle ABC is equal to the triangle DEF.

Produce AD both ways to the points G, H, and through * 31. 1. B draw BG parallela to CA, and through F draw FH pa

rallel to ED: Then
each of the figures
GBCA, DEFH, is
a parallelogram;

and they are equal • 36. 1. tobone another, be

cause they are upon
equal bases BC,

C E EF, and between the same parallels BF, GH; and the 341. triangle ABC is the halfc of the parallelogram GBCA, · because the diameter AB bisects it; and the triangle DEF

is the half of the parallelogram DEFH, because the dia

meter DF bisects it: But the halves of equal things are 27 As. equald; therefore the triangle ABC is equal to the triangle

DEF. Wherefore triangles, &c. Q.E.D.

PROP. XXXIX. THEOR.

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

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

Join AD; AD is parallel to BC; for, if it is not, through ..31. 1. the point A drawa AE parallel to BC, and join EC: The

Book I.

37. 1.

triangle ABC is equalb to the trian-
gle EBC, becaụse it is upon the
same base BC, and between the same
parallels BC, AE: But the triangle
ABC is equal to the triangle BDC,
therefore also the triangle BDC ; is
equal to the triangle EBC, the
greater to the less, which is impos-
sible: 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; AD is therefore parallel to it. Where-
føre equal triangles upon, &c. Q. E. D.

PROP. XL. THEOR. EQUAL triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels.

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

Join AD; AD is parallel to BC: For, if it is not, through A drawa b

C E F .31. 1. AG parallel to BF, and join GF: The triangle ABC is. . equal to the triangle GEF, because they are upon equal" 38. 1. bases BC, EF, and between the same parallels BF, AG : But the triangle ABC is equal to the triangle DEF; there, fore 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.

PROP. XLI. THEOR. If a parallelogram and triangle be upon the same base, and between the same parallels ; the paral. lelogram shall be double of the triangle. .

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