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THE

PROPOSITION XLIII. THEOREM XXXII.

THEOREM XXXII.
H E complements (AF, FD,) of the parallelograms (HG, EI,) about

,
the diagonal (BC) of any parallelogram (AD), are equal to one another.
Hypothesis.

TheGs. 1. AD is a Pgr, whose diagonal is BC. Tbe Pgrs AF, FD, which are the II. HG, EI, are the Pgrs aboựt the complements of the Pgrs HG, EI, diagonal.

are = to one another, DEMONSTRATION.

BECAUSE

ECAUSE AD is a Pgr, whose diagonal is BC (Hyp. 1.). 1. This diagonal divides the Pgr into two equal

parts.

P. 34. B.1. 2. Consequently, the ACAB is = to the ABDC.

Likewise, El being a Pgr, whose diagonal is BF (Hyp. 2.). 3. It divides also the Pgr into two equal parts.

P. 34. B. 1. 4. Wherefore the FEB is = to the BIF.

In fine, HG is a Pgr, whose diagonal is FC (Hyp. 2.). 5. Which consequently divides it into two equal parts.

P. 34. B. 1. 6. Consequently, the A CHF is = to the A FGC.

Since then the AFEB is = to the ABIF ( Arg. 4.), & the A CHF

= to the FGC ( Arg. 6.).
7. The A FEB, together with the ACHF is = to the A BIF, together

with the A FGC.
But the whole A CAB, BDC, being = to one another (Arg. 2.); if
there be taken away from both, the A FED + CHF, & the A BIF

+ FGC, which are equal ( Arg. 7.).
8. The remaining Pgrs AF, FD, which are the complements of the Pgrs
HG, EI, will be also = to one another.

Which was to be demonstrated.

Ax. 2.

Ax. 3:

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UPON

PO N a given straight line (AB), to make a parallelogram (BC) which shall be equal to a given triangle (T), and have one of its angles as (BAC) equal to a given rectilineal angle (M). Given

Sought 1. The straight line AB.

A Pgr made upon a firaght line AB II. The AT.

= io the AT, having one of its y III. The rectilineal V M.

BAC = to the given V M.

Resolution.
1. Produce the straight line AB indefinitely.

Pos. 2. 2. Take AL = to one of the sides of the given AT.

P. 3. B. 1. 3. Make the A AKL = to the given AT.

P. 22. B. 1. 4. Describe the Pgr EH= to the A AKL, having an V HAE = to the given VM.

P. 42. B. 1. 5. Thro' the point B, draw a straight line BF plle to EA or GH. P. 31. B. 1. 6. Produce GH indefinitely, as also GE, until it meets BF in F. Pof. 2. 9. Thro' the points F & A, draw the straight line FA, which Pof. 1,

when produced will meet GH produced, somewhere in I. 8. Thro’ the point I, draw the straight line ID plle to HB or GF. P. 31. B. 1. 9. Produce FB, EA, until they meet ID in the points D&C. Pof. 2.

DEMONSTRATION.

, ,

ECAUSE in the figure DG the opposite sides GI, FD, & GF, ID, are plle (Ref. 5. 6. 8. & 9.). !. The figure DG is a Pgr.

D. 35. B. 1.

Ax. 1.

Again, the opposite sides EA, FB, & EF, AB; also HI, AC, &

HA, IC, of the figures EB, HC, being plle (Ref. 5. 6. 8. & 9.). 2. Those figures EB, HC, are Pgrs.

D. 35. B. 1. But the straight line FI is the diagonal of the Pgr DG (Ref. 7.), &

EB, HC, are Pgrs about this diagonal ( Arg. 2, & Ref. 7.). 3. Consequently, the Pgrs BC, EH, which are the compliments, are = to one another.

P.43. B. I. But the Pgr EH is = to the A AKL (Ref. 4.), & the given A Tis =

to the fame A AKL (Ref. 3.). 4. From whence it follows, that the Pgr EH is = to the given AT.

The Pgr EH being therefore = to the given AT (Ărg. 4.), & this

fame Pgr EH being = to the Pgr BC (Arg. 3.). 5. The Pgr BC is = to the given AT.

Ax. 1. Moreover, because the V HAE, BAC, are vertically opposite. 6. Those V are = to one another.

P. 15. B. 1. Wherefore, V HAE being = to the given V M (Ref. 4.). 7. The VBAC is also = to this given V M.

Ax. 1. 8. Therefore, upon the given straight line AB, there has been made a Pgr

BC = to the given ÅT (Arg. 5.), & which has an V BAC = to the given VM ( Arg: 7.).

Which was to be done.

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Тоа

O describe a parallelogram (AF), equal to a re&ilineal figure (IH); and having an angle (n) equal to a given rectilineal angle (N). Given

Sought 1. A rectilineal figure IH.

The construction of a Pgr=io the realilineal II. A rectilineal y N.

figure IH, & having an yn=toagiven VN..

Resolution.
1. Draw the diagonal GK.

Pos. 1.
2. Upon an indefinite straight line AP, make the Pgr AE = to
the AGHK, having an Vn= to the given V N.

P. 42. B. 1.
3. Upon the side BE of the Pgr AE, make the Pgr DF = to
the ACIK; having an Vr=to the given VN.

P. 44. B. 1. DEMONSTRATION. Because

ECAUSE V N is = to each of the V n&r (Ref. 2 & 3.).. 1. The Vnis = to the V r.

If the V m be added to both. 2. The Vn+m will be = to the Vrt m.

Ax, 2. But because the sides AD, BE, are plles (Res. 2,) cut by the same

straight line AB. 3. The two interior Vn+m, are = to iwo L.

P. 29. B. 1. 4. Consequently, the adjacent Vot m, which are = to them

(Arg. 2.), are also = to two La.
The straight lines AB, BC, which meet on the opposite sides of the
line BE at the point B, making with this straight line Be the sum of

the adjacent Vrtm = to two L ( Arg. 4.). 5. Those straight lines AB, BC, form but one & the same straight line AC. P. 14. B. 1.

Moreover, the straight lines DE, AC, being two plles (Ref. 2.) cut by the same straight line BE.

Ax. 1.

Ax. I.

Ax. i.

6. The alternate V r&s, are = to one another.

P. 29. B. 1.. And if the V ube added to both. 7. The Vrtu, will be = to Vstu.

Ex. 2. But because the sides EF, BC, are iwo plles (Res. 3.) cut by the fame

straight line BE. 8. The interior Votu, are = to two L.

P. 29. B. 1. 9. From whence it follows, that the adjacent Vstu, which are = to

them ( Arg. 7.), are also = to two L.
The Itraight lines DE, EF, which meet on the opposite sides of the
line BE at the point E, making with this straight line BE, the

fuin of the adjacent Vstu= to two L. (Arg. 9.).
10. Those straight lines DE, EF, form but one and the saine straight
line DF.

P. 14. B. 1. But since the ftraight lines AD, BE, & BE, CF, are the opposite sides of the Pgrs AE, BF, (Ref. 2 & 3.).

P. 34. B. 1. 11. The straight line AD is = & plle to BE, & BE is = & plle to CF.

SP. 30. B. 1. 12. Consequently, AD is = & plle to to CF.

Ax. 1. Moreover, those == and plle straight lines AD, CF, are joined by the straight lines AC, DF, ( Arg. 5 & 10.).

S P. 33. B. 1. 13. Confequently, the figure AF is a Pgr.

ID. 35. B.1. And because the Pgr BF is = to the A GIK (Ref. 3.), the Pgr

AE is = to the AGHK, & Vn= to the given V Ň (Ref. 2.).
14. The whole Pgr AF is = to the rectilineal figure IH; & has an Vn
= to the given V N.

Ax. 2.
Which was to be demonstrated.

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