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a. 2. Cor. 19. I.

Book 1. to each of its angles. And, by the preceeding proposition, all

the 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 is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides.

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 exterior b 13. 1. ABD, is equal b to two right

angles; therefore all the interi-
or, together with all the exterior
angles of the figure, are equal
to twice as many right angles as
there are sides of the figure; that
is, by the foregoing corollary, D B
they are equal to all the inte-
rior angles of the figure, toge.
ther with four right angles; therefore all the exterior angles are
equal to four right angles.

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TH
"HE straight lines which join the extremities of two

equal and parallel straight lines, towards the fame
parts, are also themselves equal and parallel.
Let AB, CD be equal and pa-

B
rallel 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 pa-
rallel to CD, and BC meets them,

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D the alternate angles ABC, BCD are equal“; and becaule AB is equal to CD, and BC common to the two triangles ABC, DCB, the two fides AB, BC are e. qual to the two DC, CB; and the angle ABC is equal to the angle ECD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles

2 29. I.

b 4. I.

to

to the other angles , each to each, to which the equal fides are
cpposite ; therefore the angle ACB is equal to the angle CBD; M
and because the straight line BC meets the two straight lines b 4. 1.
AC, BD, and makes the alternate angles ACB, CBD equal to
ene another, AC is parallel to BD; and it was shown to be c 27. 1,
cqual to it. Therefore straight lines, &c. Q. E. D.

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THE
"HE opposite sides and angles of parallelograms are

equal to one another, and the diameter bisects them, that is, divides them in 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 bifects it. Because AB is parallel to CD,

A

B and BC meets them, the alternate angles ABC, BCD are equal · to one another; and be

a 29. 1 caule AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal a 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 angles; there. fore their other fides shall be equal, each to each, and the third angle of the one to the third angle of the other b, viz. b 26, I. the fide 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 parallelograms are equal to one another ; also, their diameter bisects them; for

l 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

C4

Book 1. is equal to the angle BCD; therefore the triangle ABC is e mqual to the triangle BCD, and the diameter BC divides the

parallelogram ACDB into two equal parts. Q. E. D.

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See N.

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

a

a 34. I.

See the ad Let the parallelograms ABCD, EBCF, be upon the same and 3d fi- base BC, and between the fame parallels AF, BC; the parallegures.

logram ABCD Mall be equal to the parallelogram EBCF.

If the fides AD, DF of the pa.
rallelograms ABCD, DBCF opposite A D F
to the base BC be terminated in the
same point D; it is plain that each
of the parallelograms is double a of
the triangle BDC; and they are there-
fore equal to one another.

But, if the fides AD, EF, opposite B
to the base BC of the parallelograms
ABCD, EBCF, be not terminated in the same point; then, be-
cause ABCD is a parallelogram, AD is equal a to BC; for the
fame reason EF is equal to BC; wherefore AD is equal o to
EF; and DE is common; therefore the whole, or the remain-
der, AE is equal to the whole, or the remainder DF; AB al-
fo is equal to DC; and the two EA, AB are therefore equal to

А. DE F A E DF

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b 1. Ax.

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c 2. or 3. Ax.

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d 29. 1.

e 4: I.

B

B C
the two FD, DC, each to each; and the exterior angle FDC
is equal to the interior EAB; therefore the base EB is equal
to the base FC, and the triangle EAB equal to the triangle
FDC; take the triangle FDC from the trapezium ABCF, and
from the same trapezium take the triangle EAB; the remain-
ders therefore are equals, that is, the parallelogram ABCD is
equal to the parallelogram EBCF. Therefore parallelograms
upon the same base, &c. Q. E. D.

1

PRO P.

f 3. Ax.

Book I.

PROP. XXXVI.

THEOR.

P

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

A

Let ABCDEFGH be

DE parallelograms upon e

H qual bases BC, FG, and between the same parallels AH, BG; the paral. lelogram ABCD is equal to EFGH.

Join BE, CH; and because BC is equal to

B

C F G FG, and FG to * EH, BC is equal to EH; and they are pa- a 34. I. rallels, 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 parallel 5; therefore EB, CH are both equal and parallel, and 5 33. . EBCH is a parallelogram; and it is equal to ABCD, because c 35.1. it is upon the same bafe BC, and between the fame parallels BC, AD: For the like reason, the parallelogram EFGÀ 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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TRA

RIANGLES upon the same base, and between the

fame parallels, are equal to one another.

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

E А D F
AD, BC: The triangle ABC
is equal to the triangle
DBC.

Produce AD both ways to the points E, F, and thro' Bdraw - BE parallel to CA;

a 31. 8, and thro' C draw CF paral

B Icl to BD: Therefore each

C of the figures EBCA, DBCF is a parallelogram ; and EBCA is equal o to DBCF, because they are upon the same base BC, and b 35. I. between the same parallels BC, EF; and the triangle ABC is

the

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Book 1. the half of the parallelogram EBCA, because the diameter AB

bisects Cit; and the triangle DBC is the half of the parallelo

gram DBCF, because the diameter DC bifects it. But the d 7. Ax.

halves of equal things are equald; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

C 34. I.

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a 31. I.

PRO P. XXXVII. THEOR. TRIANGLEs upon equal bases, and between the fame

parallels, are equal to one another. Let the triangles ABC, DEF be 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 through B draw BG parallela to CA, and through F draw FH parallel to ED: Then each of

G A D Η.
the figures GBCA,
DEF# is a paralle-
logram; and they
are equal to one an-
other, because they
are upon equal bases
BC,EFand between
the same parallels

B CE
C

F
BF, GH; and the triangle ABC is the half of the parallelogram
GBCA, because the diameter AB bisects it; and the triangle
DEF is the half of the parallelogram DEFH, because the di-
ameter DF bisects it : But the halves of equal things are equald;
therefore the triangle ABC is equal to the triangle ĎEF. Where.
fore triangles, &c. Q. E. D.

b 36. I.

E

C 34. 1.

dy. Ax.

PRO P. XXXIX. THE O R.
QUAL triangles upon the fame base, and upon the

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

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

angle

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A 33.

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