Sidebilder
PDF
ePub

a 2 Cor.

15. 1.

Book I. all thé angles of these triangles are equal to twice as many

right angles as there are triangles, that is, as there are sides of the figure ; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles ; that isa, together with four right angles. Therefore, twice as many right angles as the figure has sides are equal to all the angles of the figure, together with four right angles, that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four.

Cor. 2. All the exterior angles of any rectilineal figure are together equal to four right angles.

Because every interior angle ABC, with its adjacent ex

terior ABD, are equal b 13. 1. to two right anglesb; therefore all the inte

АТ
rior, together with all
the exterior angles of
the figure, are equal
to twice as many right

D
B

c
angles as there are
sides of the figure;
that is, by the forego-
ing corollary, they are equal to all the interior angles of the
figure, together with four right angles; therefore all the ex-
terior angles are equal to four right angles.

PROP. XXXIII. THEOR.

THE straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and

A

B
parallel straight lines, and joined
towards the same parts by the
straight lines AC, BD; AC,
BD are also equal and parallel.

Join BC; and because AB is
parallel to CD, and BC meets
them, the alternate angles ABC, BCD are equala; and be-

a 29. 1.

căuse AB is equal to CD, and BC common to the two tri Book I. angles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the tri- b 4. 1. angle ABC to the triangle BCD, and the other angles to the other anglesb, each to each, to which the equal sides are opposite; therefore the angle ACB is equal to the angle CBD; and because 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 c 27. 1. be equal to it. Therefore, straight lines, &c. Q. E. D.

PROP. XXXIV. THEOR.

THE opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it 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 bisects it. Because AB is parallel to CD,

A

B and BC meets them, the alternate angles ABC, BCD are equala to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal

c

D 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

a 29. 1.

Book I. angles; therefore their other sides are equal, each to each,

and the third angle of the one to the third angle of the b 26. 1. otherb, viz. the side AB to the side CD, and 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 a parallelogram are equal to one another. Also, its diameter bisects it; 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; c 4.1. therefore the triangle ABC is equal to the triangle BCD,

and the diameter BC divides the parallelogram ACDB into two equal parts. Therefore, &c. 'Q. E. Ď,

[ocr errors][merged small]

PARALLELOGRAMS upon the same base, and between 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 pagures.

rallelogram ABCD is equal to the parallelogram EBCF.
If the sides AD, DF of the

D
parallelograms ABCD,DBCF,

F opposite to the base BC, be terminated in the same point D;

it is plain that each of the paa 34. 1. rallelograms is doublea of the triangle BDC; and they are

B В 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 b 1. Ax. equala to BC; for the same reason EF is equal to BC: wherec 2. or 3.

fore AD is equalb to EF; and DE is common ; therefore Ax. the whole, or the remainder AE is equal to the whole, or

[ocr errors]

37

the remainder DF; and AB is also equal to DC; therefore

Book I.

[merged small][merged small][merged small][merged small][ocr errors]

the two EA, AB are equal to the two FD, DC, each to each; but the exterior angle FDC is equald to the interior EAB; d 29. 1. wherefore the base EB is equal to the base FC, and the triangle EABe to the triangle FDC. Take the triangle e 4. 1. FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders will then be equall, that is, the parallelogram ABCD is equal to the f 3. Ax. parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q. E. D.

PROP. XXXVI. THEOR.

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

Let ABCD, EFGH be parallelograms upon

A
D C

H
equal bases BC, FG,
and between the same
parallels AH, BG; the
parallelogram ABCD is
equal to EFGH.
Join BE, CH; and B

c because BC is equal to FG, and FG toa EH, BC is equal to EH; and they are pa- a 34. 1. rallels, and joined towards the same parts by the straight lines BE, CH; therefore EB, CH are both equal and parallelo, b 33. 1.

[ocr errors]

Book I. and EBCH is a parallelogram ; and it is equal to ABCD:

but the parallelogram EFGH is equal to the same EBCH; c 35. 1. therefore also the parallelogram ABCD is equal to EFGH.

Wherefore, parallelograms, &c. Q. E. D.

PROP. XXXVII. THEOR.

TRIANGLES

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

a 31. 1.

Let the triangles ABC, DBC be upon the same base BC, and between the same pa

3
rallels AD, BC; the tri-

А A D
angle ABC is equal to the
triangle DBC.

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

B C draw CF parallel to BD: then each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal to DBCF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects it; and the triangle DBC is the half of the parallelogram DBCF; therefore the triangle ABC is equal to the triangle DBC, Wherefore, triangles, &c. Q. E. D.

b 35. 1.

c 34. 1.

d 7. Ax.

PROP. XXXVIII. THEOR.

.

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 BF, AD; the triangle ABC is equal to the triangle DEF.

« ForrigeFortsett »