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PROPOSITION XXXIX.

THEOR E M.
Equal Triangles constituted upon the same Base,

on the fame Side, are in the fame Parallels.
ET ABC, DBC, be equal Triangles, consti-
tuted

upon the fame Base B C, on the fame Side. I say, they are between the same Parallels. For, let

A D be drawn. I say A D is parallel to B C. 32 of tbis, . For, if it be not parallel, draw * the Right Line

A E thro' the Point A, parallel to BC, and draw EC. + 37 ofibis. Then the Triangle ABC+ is equal to the Triangle

EBC; for it is upon the fame Base B C, and between

the fame Parallels BC, A E. But the Triangle A B C, 1 From Hyp. is I equal to the Triangle D BC. Therefore the

Triangle DBC is also equal to the Triangle E BC, a greater to a lefs, which is impossible. Wherefore A E is not parallel to B C: and by the same way of Reasoning we prove, that no other Line but A D is parallel to BC. Therefore A D is parallel to BC. Wherefore, equal Triangles confiituted upon the same Base, on the same Side, are in the same Parallels ; which was to be demonstrated.

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PROPOSITION XL.

THEOREM.
Equal Triangles constituted upon equal Bases, on

the same Side, are between the same Parallels. LETABC, CDE, be equal Triangles

, constituted upon equal Bases B C, CE. I say, they are between the fame Parallels. For let A D be

drawn. I say, A D is parallel to BE. * 31 of tbis. For, if it be not, let A F be drawn * through A,

parallel to B E, and draw F E. + 38 of rbis. Then the Triangle A B C is equal + to the Triangle

FCE; for they are conftituted upon cqual Bases, and between the same Parallels B E, A F. But the Triangle A B C is equal to the Triangle D CE. There

fore

fore the Triangle D C E shall be equal to the Triangle FCE, the greater to the less, which is impossible. Wherefore A F is not parallel to B E. And in this manner we demonstrate, that no Right Line can be parallel to B E, but A D. Therefore A D is parallel to B E. And so, equal Triangles constituted upon equal Bases, on the fame Side, are between the fame Parallels ; which was to be demonstrated.

PROPOSITION XLI.

THE ORE M.
If a Parallelogram and a Triangle have the same

Base, and are beiween the same Parallels, the
Parallelogram will be double to the Triangle.
E T the Parallelogram A B C D, and the Triangle

EBC, have the same Base, and be between the
fame Parallels, B C, A E. I say, 'the Parallelogram
A B C D is double the Triangle E B C.
For join A C.

Now the Triangle A B C is * equal to the Triangle 37 ofibis. EBC; for they are both constituted upon the fame Base B C, and between the same Parallels B C, A E. But the Parallelogram ABCD is + double the Tri- † 34 of ibis. angle A B C, since the Diameter AC bisects it. Wherefore likewise it shall be I double to the Tri- Ax. 6. angle E BC. If, therefore, a Parallelogram and Triangle have both the same Base,, and are between the same Parallels, the Parallelogram will be double the Triangle; which was to be demonstrated.

PROPOSITION XLII..

THE OR EM.
To constitute a Parallelogran equal to a given

Triangle, in an Angle equal to a given Right

lin'd Angle.

LET the given Triangle be A B C, and the Right

lined Angle D. It is required to constitute a Parallelogram equal to the given Triangle A B C, in a Rght-lind Angle equal to D. D4

Bireet

* 10 of tbis. Birea * B C in E, join A E, and at the Point E, + 23 ofrbis, in the Right Line E C, constitute + an Angle CEF 1 31 of ibis.equal to D. Also draw I AG thro' A, parallel to

E C, and through C the Right Line CG, parallel to
FE.

Now FECG is a Parallelogram: And because *38 of ibis.

BE is equal to EC, the Triangle A B E shall be * equal to the Triangle A EC; for they stand upon equal Bases B E, E C, and are between the same Parallels B C, AG. Wherefore the Triangle A B C is

double to the Triangle A EC. But the Parallelo+ 41 of 1bis, gram FECG is also + double to the Triangle A EC;

for it has the same Base, and is between the same
Parallels. Therefore the Parallelogram F ECG is
equal to the Triangle A B C, and has the Angle
CEF equal to the Angle D. Wherefore, the Pa.
rallelogram F ECG is constituted equal-:o the given
Triangle A B C, in an Angle CEF equal 10 a given
Angle D; which was to be done.
PROPOSITION XLIII.

THEOREM.
In every Parallelogram, the Complements of the

Parallelograms, tbat stand about tbe Diameter,

ore equal between themselves. LETAB CD bea Parallelogram, whose Diameter

; standing about the Diameter BD. Now AK, KC, are called the Complements of the n : I say, the Complement A K is equal to the Complement K C.

For fince A B C D is a Parallelogram, and B D is

the Diameter thereof, the Triangle A B D is * equal 34 of this,

'to the Triangle BD C. Again, because HKFD is a Parallelogram, whose Diameter is DK, the Triangle HD K shall * be equal to the Triangle DFK; and for the same Reason the Triangle KBG is equal to the Triangle K E B. But since the Triangle BEK is equal to the Triangle BGK, and the Triangle HD K to DF K, the Triangle B E K, together with the Triangle HDK, is equal to the Triangle BGK, together with the Triangle D FK. But the whole Triangle A B D is likewise equal to the whole Triangle

BDC.

BDC. Wherefore the Complement remaining,
A K, will be equal to the remaining Complement
KC. Therefore, in every Parallelogram, the Comple-
ments of the Parallelograms that stand about the Diameters

, are equal betweem themselves ; which was to be demonftrated. PROPOSITION XLIV.

PROBLEM.
To apply a Parallelogram to a given Right Line,

equal to a given Triangle, in a given Right

lined Angle. LE ET the Right Line given be A B, the given Tri

angle C, and the given Right-lined Angle D. It is required to the given Right Line A B, to apply a Parallelogram equal to the given Triangle C, in an Angle equal to D.

Make the Parallelogram BEFG equal to the *.42 of tbis.
Triangle C, in the Angle E BG, equal to D. Place
BE in a strait Line with A B, and produce F G to H,
and through A let AH be drawn parallel to either + 32 of ebis.
GB, or FE, and join H B.

Now, because the Right Line HF ralis on the Pa-
rallels A H, EF, the Angles AHF, HFE, are I [ 29 of tbis.
equal to two Right Angles. And so BHF, HFE, are
less than two Right Angles; but Right Lines making
less than two Right Angles, with a third Line, being
infinitely produced, will meet each other. Wherefore . Ax. 12.
HB, FE, produced, will meet each other; which let
be in K, through which draw | K L parallel to E A, || 13 of ibis.
or FH, and produce AH, GB, to the Points L and M.

Therefore HL KF is a Parallelogram, whose Diameter is HK; and A G, ME, are Parallelograms about HK; whereof LB, B F are the Complements. Therefore L B is + equal to BF. But BF is also + 43 of this. equal to the Triangle C. Wherefore likewise LB shall be equal to the Triangle C; and because the Angle GBE is I equal to the Angle A BM, and also 1150f bit. equal to the Angle D, the Angle A BM shall be equal to the Angle D. Theretore, to the given Right Line A B is applied a Parallelogram, equal to the given Triangle C, and the Angle A BM, equal to the given Angle D; which was to be done,

PRO.

1

19 of teza. Blce? * B C in E, join

33 ore in the Right Line EC, co
: ;t of t...equal to D. Also draw i

EC, and through the Ri
FE.

Now FECG is a Para
BE is equal to EC, the T
equal to the Triangle A E(
egoal Bases BE, E C, and ai
relle's BC, AG. Wherefor

ccutie to the Triangle A E
Istogram FECG is also + double

for it has the same Base, and
Parallels. Therefore the Para
equal to the Triangle ABC
CEF equal to the Angle D.
12. clarem FECG is confiitut
Eric ! A B C, in an Angle CI
izgi D; which was to be done.
PROPOSITIO

THEOREM.
I: eety Parallelegram, ibe Curi

Paralelograms, ibat stand abou

are equal beiween theinselves. LE

ETABCD be a Parallelogram,

is DB; and let FH, EG, be ftanding about the Diameter BD. N are called the Complements of the n: 1 piement A K is equal to the Complem

Forince A B C D is a Parallelogran

the Diameter thereof, the Triangle A B 34 og 20.

to the Triangle B D C. Again, becaut a Paralle' gram, whose Diameter is DK geHD K thai] * be equal to the Trian and for the same Reason the Triangle KE to the Triangle K E B. But since the Tria is equal to the Triangle B GK, and the HDK 10 DFK, the Triangle B E K, 10 tre Triangle HDK, is equal to the Triang together with the Triangle D FK. But Triangle A B D is likewise equal to the whole

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