Sidebilder
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Ax. 1.

в с 1. If the sides AD, DF, of the

zms ABCD, DBCF, opp. to BC the base, be terminated in the same point D [as in fig. 1], :: each double the ABDC,

Prop. 33. the ABCD= Om DBCF.

Ax. 6. 2. If the sides AD, EF, be not terminated in the same point (figs. 2 and 3), then : ABCD, EBCF are

ŞAD = BC,
EF = BC;

Prop. 33.
AD= EF,
and DE is common to both,
sthe whole or

= whole or remain. DF. remain. AE And ::: EA, AB = FD, DC, ea. to ea.

Prop. 33. ext. LFDC = int. _ EAB,

Prop. 28. base EB = base FC, A EAB=AFDC.

Prop. 4. Take the A EAB from the trapezium ABCF; and from the same trapezium take the a FDC; then the remainders are =; that is,

O" ABCD=OM EBCF. Wherefore parallelograms, &c.

CoR.— Triangles upon the same base, and between the same parallels, are equal to one another. For if the diameters AC, FB, be drawn (figs. 2 and 3), the As ABC, FBC,

• Ax. 2, 3.

Ax. 3.

are the halves of the equal Om ABCD, EBCF, .. the As are equal to one another.

Ax. 7.

PROP. XXXV. THEOR. 36. 1 Eu.

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

Let ABCD, EFGH, be BC, FG, and between the same ||s, AH, BG; then ABCD = OM EFGH.

upon = bases

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Ax. 1.

Join BE, CH,

BC FG, Нур.

EH = FG, Prop. 33.

BC= EH;

str. lines EB, HC, join the extremities of = and || str. lines, they are themselves = and || : that is,

BE is = and || to CH, Prop. 32.

EBCH is a

EBCH= ABCD, , since they are on the same base BC and beProp. 34.

tween the same is.
For the same reason,
EFGH =

EBCH,

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Ax.l.

zms

... ABCD = OMEFGH. Wherefore parallelograms, &c.

Cor. 1.-Triangles upon the equal bases and between the same parallels are equal to each other. For draw the diams. AC, EG; the As ABC, EFG, are the halves of the equal

ABCD, EFGH, and are therefore equal Ax. 7. to cach other.

Cor. 2.-If a parallelogram and a triangle be

upon the same or equal bases, and between the same parallels, the parallelogram is double the triangle. For the Om ABCD is double the A ABC, or double the A EFG, which is its equal, by last Cor.

PROP. XXXVI. THEOR. 39. 1 Eu.

Equal triangles upon the same base, and upon

the same side of it, are between the same parallels.

Let the = As ABC, DBC, be on the same base BC, and on the same side of it; then the As will be between the same ||s.

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Cor.

Ax. 1.

i. e.

For if not, Prop. 30. draw AE || BC, and join EC.

I being on the same

base BC, and beProp.34, Then A ABC = AEBC

tween the same is

BC, AE;
Нур.

but ABC=A DBC,
i. A EBC=A DBC,

less =greater,
which is impossible ;
... AE 8 BC.
In the same manner it may be proved that no
other line but AD || BC,

.. AD || BC.
Wherefore equal triangles upon, &c.

CoR.--In the same way it may be shown,
that equal triangles upon equal bases and to-
wards the same parts, are between the same
parallels

PROP. XXXVII. THEOR. 43. 1 Eu.

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The complements of the parallelograms which

are about the diameter of any parallelo-
gram, are equal to each other.

Let ABCD be a ", AC its diam.; and
EAHK, GKFC, Oms about AC, that is,
through which AC passes. And BEKG,
KHDF, the other omwhich make up the
whole figure ABCD, are called complements.
Then

comp. BEKG = comp. KĀDF.

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с

B G
diam. AC bisects ABCD,

A ABC = A ADC;

AAEK
also
= A AHK,

Prop.33.
AKGC= A KFC,

A AHK + AAx. 2. .A AEK + AKGC

KFC: but the whole 4ABC = whole A ADC; the remains comp.

(remains comp. BEKG

KHDF.

{

Ax. 3.

Wherefore the complements, &c.

PROP. XXXVIII. PROB. 46, 1 Eu.

To describe a square upon a given straight

line.

Let AB be the given str. line; it is required to descr. a square upon AB.

с

D

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А

B

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