equal to the Bafe DB; and the Triangle ABC is equal to the Triangle B CD. Wherefore, the Didmeter BC bifects the Parallelogram ACDB; which was to be demonftrated:

PROPOSITION XXXV.

THEOREM.

Parallelograms conftituted upon the fame Base, and between the fame Parallels, are equal bétween themselves.

LE

ET ABCD, EBCF, be Parallelogram's conftituted upon the fame Base BC, and between the fame Parallels A F and BC. I fay, the Parallelogram ABCD is equal to the Parallelogram EBC F.

For, because ABCD is a Parallelogram, AD is equal to BC; and for the fame Reafon E F is equal * 34 of this. to BC; wherefore A D fhall be † equal to EF; but† 4x. 1. DE is common. Therefore the whole A E is equal † Ax. 2. to the whole DF. But A B is equal to DC; wherefore EA, AB, the two Sides of the Triangle A BE, are equal to the two Sides F D, DC, each to each; and the Angle FDC* equal to the Angle E A B, the 29 of tbit. outward one to the inward one. Therefore the Bafe EB is equal to the Bafe CF, and the Triangle E A B+ 4 of this. to the Triangle FDC. 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 GBC, which is common, be added, the Parallelogram A B C D will be equal to the Parallelogram EBCF. Therefore, Parallelograms conftituted upon the fame Bafe, and between the fame Parallels, are equal between themselves ; which was to be demonftrated.

PRO

* Hyp•

PROPOSITION XXXVI.

THEOREM.

Parallelograms conftituted upon equal Bafes, and between the fame Parallels, are equal between themselves.

LET

ET the Parallelograms A BCD, EFGH, be conftituted upon the equal Bafes BC, FG, and between the fame Parallels AH, BG. I fay, the Parallelogram ABCD is equal to the Parallelogram EFGH.

For join BE, CH. Then because BC is equal to F G, and FG to EH; BC will be likewife equal to EH; and they are parallel, and BE, CH, join them. But two Right Lines joining Right Lines, which are equal and parallel, towards the fame Parts, +33 of this are + equal and parallel: Wherefore E BCH is a Pa35 of this rallelogram, and is equal to the Parallelogram

ABCD; for it has the fame Base BC, and is conftituted between the fame Parallels B C, A H. For the fame Reason, the Parallelogram EF G H is equal to the fame Parallelogram EBCH. Therefore the Parallelogram ABCD fhall be equal to the Parallelogram EFGH. And fo Parallelograms conflituted upon equal Bafes, and between the fame Parallels, are equal between themfelves; which was to be demonftrated.

PROPOSITION XXXVII.

THEOREM.

Triangles conftituted upon the fame Bafe, and between the fame Parallels, are equal between themselves.

LE

ET the Triangles ABC, DB C, be conftituted upon the fame Bafe B C, and between the fame Parallels A D, BC. I fay, the Triangle ABC is equal to the Triangle D BC.

For produce AD both Ways to the Points E and 31 of this. F; and through B draw B E parallel to C A; and through C, CF, parallel to BD.

Where

*

Wherefore both EBCA, DBCF, are Parallelograms; and the Parallelogram E BCA is equal to 35 of this. the Parallelogram DBCF; for they ftand upon the fame Base B C, and between the fame Parallels BC,

EF. But the Triangle ABC is + one half of the Pa-† 34 of this. rallelogram EBC A, because the Diameter A B bifects it; and the Triangle DBC is one half of the Parallelogram D BCF, for the Diameter DC bifects it. But Things that are the Halves of equal Things, are ‡‡ Ax. 7. equal between themfelves. Therefore the Triangle ABC is equal to the Triangle D BC. Wherefore, Triangles conflituted upon the fame Bafe, and between the fame Parallels, are equal between themselves; which was to be demonftrated.

PROPOSITION XXXVIII.

THEOREM.

Triangles conftituted upon equal Bafes, and between the fame Parallels, are equal between themselves.

LET the Triangles ABC, DCE, be conffituted upon the equal Bafes BC, CE, and between the fame Parallels BE, AD. I fay, the Triangle A B C is equal to the Triangle D C E.

For, produce AD both Ways to the Points, G,

H; through B draw * BG parallel to CA; and * 31 of this. through E, EH, parallel to DC.

Wherefore both G BCA, DCEH, are Paralle

lograms; and the Parallelogram G BCA is + equal † 36 of this, to the Parallelogram DCEH: For they ftand upon equal Bafes, BC, CE, and between the fame Parallels BE, GH. But the Triangle ABC is one half 34 of this. of the Parallelogram G B C A, for the Diameter A B bifects it; and the Triangle DCE is one half of the Parallelogram DCEH, for the Diameter DE bifects it. But Things that are the Halves of equal Things, are equal between themfelves. Therefore* Ax. 7. the Triangle ABC is equal to the Triangle DCE, Wherefore, Triangles conftituted upon equal Bafes, and between the fame Parallels, are equal between themselves; which was to be demonftrated. D 3

PRO

32 of this.

PROPOSITION XXXIX.

THEOREM.

Equal Triangles conftituted upon the fame Bafe, on the fame Side, are in the fame Parallels.

L

ET ABC, DB C, be equal Triangles, conftituted upon the fame Bafe B C, on the fame Side. I fay they are between the fame Parallels. For, let AD be drawn. I fay, AD is parallel to B C.

For, if it be not parallel, draw the Right Line A E thro' the Point A, parallel to BC, and draw EC, +37 of this. Then the Triangle ABC + is equal to the Triangle EBC; for it is upon the fame Bafe BC, and between the fame Parallels BC, A E. But the Triangle ABC From Hyp. is equal to the Triangle DBC. Therefore the Triangle DBC is also equal to the Triangle EBC, a greater to a lefs, which is impoffible. Wherefore AE is not parallel to B C: And by the fame Way of Reasoning we prove, that no other Line but AD is parallel to BC. Therefore A D is parallel to BC. Wherefore, equal Triangles conftituted upon the fame Bafe, on the fame Side, are in the fame Parallels; which was to be demonftrated.

31 of this.

་

PROPOSITION XL.

THEOREM.

Equal Triangles conftituted upon equal Bafes, on the fame Side, are between the fame Parallels.

LET ABC, CDE, be equal Triangles, conftituted upon equal Bafes BC, CE. I fay, they are between the fame Parallels. For, let AD be drawn. I fay, AD is parallel to BE.

For, if it be not, let AF be drawn through A, parallel to B E, and draw F E.

Then the Triangle ABC is +equal to the Triangle 38 of this FCE; for they are conftituted upon equal Bafes, and

between the fame Parallels B E, AF. But the Triangle ABC is equal to the Triangle DCE. There

fore