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equal to the Base DB; and the Triangle A B C is equal to the Triangle BCD. Wherefore, the Diameter B C bifects the Porallelogram ACDB; which was to be demonstrated,

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THEOREM
Parallelograms constituted upon the same Base,

and between the same Parallels, are equal be-
tween themselves.

, , be
ftituted upon the fame Base B C, and between
the same Parallels A F and BC. I say, the Paral.
lelogram A B C D is equal to the Parallelogram,
EBCF.

For, because A B C D is a Parallelogram, A D is * equal to BC; and for the fame Reason E F is equal - 34 of ibis. to BC; wherefore A D fall be + equal to EF; but + Ax. 1. DE is common. Therefore the whole A E is I equal | Ax. 2. to the whole D F. But A B is equal to DC; wherefore E A, A B, the two Sides of the Triangle A BE, are equal to the two Sides FD, DC, each to each and the Angle FDC * equal to the Angle E A B, the * 29 of ibis. outward one to the inward one. Therefore the Base EB is + equal to the Base CF, and the Triangle E A B + 4 of ibis. 10 the Triangle FD C. If the common Triangle DGE be taken from both, there will remain the Ax. 3, Trapezium ABGD, equal to the Trapezium FCGE; and if the Triangle G B C, which is common, be added, the Parallelogram ABCD will be equal to the Parallelogram E B CF. Therefore, Parallelograms constituted upon the fame Base, and between the same Parallels, are equal between themselves ; which was to be demonstrated.

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Hyp.

PROPOSITION XXXVI.

THEOR E M.
Parallelograms constituted upon equal Bases, and

between the same Parallels, are equal between
themselves.
E T the Parallelograms ABCD, EFGH, be

constituted upon the equal Bases B C, FG, and between the same Parallels A H, BG. I say, the Parallelogram A B C D is equal to the Parallelogram EFG H.

For join B E, CH. Then because BC* is equal to F G, and F G to EH; B C will be likewise equal to E H; and they are parallel, and B E, C H, join them. But two Right Lines joining Right Lines,

which are equal and parallel, towards the fame Parts, + 33 of tb.si are + equal' and parallel : Wherefore E BCH is a 1 35 of this. Parallelogram, and is equal to the Parallelogram

ABCD; for it has the same Base B C, and is constituted between the fame Parallels BC, A H. For the same Reason, the Parallelogram E F G H is equal to the same Parallelogram EĎ CH. Therefore the Pa. rallelogram A B C D shall be equal to the Parallelogram E F G H. And fo Parallelograms constituted upon equal Bases, and between the same Parallels, are equal between themselves; which was to be demonstrated.

PROPOSITION XXXVII.

THEOREM.
Triangles constituted upon the same Base, and

between the same Parallels, are equal between

themselves.
LE A

upon the same Base B C, and between the same Parallels A D BC. I say, the Triangle ABC is equal to the Triangle D BC.

For produce A D both Ways to the Points E and *gr of tbis. F; and through B draw * BÉ parallel to C A; and through C, C F, parallel to B D.

Where

Wherefore both EBCA, DBCF, are Parallelograms; and the Parallelogram E BC A is * equal to 35 of Ibis, the Parallelogram DBCF; for they stand upon the same Base BC, and between the same Parallels B C, EF. But the Triangle A BC is tone half of the Paral. 1 34 of tbit. lelogram, EBCA, because the Diameter A B bisects it; and the Triangle D B C is one half of the ParalleJogram D BCF, for the Diameter D C bisects it. But Things that are the Halves of equal Things, are 1 1 Ax. 9. equal between themselves. Therefore the Triangle A B C is equal to the Triangle D BC. Wherefore, Triangles constituted upon the same Base, and between the same Parallels, are equal beiween themselves; which was to be demonstrated.

PROPOSITION XXXVIII.

THEOREM.
Triangles constituted upon equal Bases, and be-

tween the same Parallels, are equal between
themselves.
E T the Triangles A BC, DCE, be constituted

upon the equal Bases B C, CE, and between the same Parallels B E, A D I say, the Triangle ABC is equal to the Triangle DC E.

For, produce A D both Ways to the Points, G, H; through B draw * BG parallel to CA; and 31 of ibis. through E, E H, parallel to D C.

Wherefore both GBCA, DCEH, 'are Parallelograms; and the Parallelogram GBC A is + equal + 36 of tbis. to the Parallelogram DCEH: For they stand upon equal Bases, BC, CE, and between the fame Paral. leis BE, G H. But the Triangle A B C is f one half t 34 of tbis. of the Parallelogram GBCA, for the Diameter AB bisects it, and the Triangle DCEI is one half of the Parallelogram DCEÅ, for the Diameter D Ebifects it. But Things that are the Halves of equal Things, are * equal between themselves. Therefore * Ax. 7. the Triangle A B C is equal to the Triangle DCE. Wherefore, Triangles constituted upon equal Bases, and between the same Parallels, are equal between themselves; which was to be demonstrated.

PRO

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