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B C

B C 1. If the sides AD, DF, of the Om ABCD, DBCF, opp. to BC the base, be terminated in the same point D (as in fig. 1],

: each om = double the ABDC, Prop. 33. is the Om ABCD= O DBCF. Ax. 6. 2. If the sides AD, EF, be not terminated in the same point (figs. 2 and 3), then : ABCD, EBCF are oms,

SAD = BC,
EF = BC;

Prop. 33. .. AD= EF,

and DE is common to both, . sthe whole or _.

=whole or remain. DF. Ax. 2, 3. remain. AE And S EA, AB = FD, DC, ea. to ea. Pron 32 ext. LFDC = int. _ EAB,

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

Prop. 4. Take the A EAB from the trapezium ABCF; and from the same trapezium take the Ax. 3. FDC; then the remainders are =; that is,

O" ABCD= D" 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,

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

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PROP. XXXV. THEOR. 36. 1 Eu.

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

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

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..om ABCD= Om EFGH. 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 oms ABCD, EFGH, and are therefore equal Ax. 9. to each 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 Ku.

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 is.

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Cor. .

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

(being on the same

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

tween the same is

BC, AE;
Hyp. but A ABC =ADBC,
Ax. 1. i. A EBC=A DBC,

i. e. 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 towards the same parts, are between the same parallels

PROP. XXXVII. THEOR. 43. 1 Eu. The complements of the parallelograms which are about the diameter of any parallelogram, are equal to each other.

Let ABCD be a OM, 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. KHDF.

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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.

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