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* 46. 1. +41.1.

#29. 1.

* 5. 1.

† 6. 1.

Then the Rectangle AE shall be equal to the two Rectangles AD, CE: And the Rectangle A E is that contained under A B and BC; for it is contained under AB and BE, whereof BE is equal to BC: And the Rectangle AD is that contained under AC and CB, fince DC is equal to CB: And DB is a Square defcrib'd upon BC. Wherefore the Rectangle under AB and BC is equal to the Rectangle under AC and CB, together with the Square defcribed upon BC. Therefore if a Right Line be any how cut, the Rectangle contained under the whole Line, and one of its Parts, is equal to the Rectangle contained under the two Parts together, with the Square of the first-mention'd Part; which was to be demonftrated

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PROPOSITION IV.

THEOREM.

Right Line be any how cut, cut, the Square which is made on the whole Line will be equal to the Squares made on the Segments thereof, together with twice the Rectangle contained under the Segments."

LET the Right Line AB be any how cut in C. I fay, the Square made on AB is equal to the Squares of A C, CB, together, with twice the Rectangle contained under AC, CB.

For * defcribe the Square ADEB upon AB, join BD, and thro C draw CGF parallel to AD or BE; and also thro' G draw HK parallel to AB or DE.

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Then because CF is parallel to AD, and BD falls upon them, the outward Angle BGC shall be + equal to the inward and oppofite Angle ADB: But the Angle ADB is * equal to the Angle ABD, fince the Side BA is equal to the Side A D. Wherefore the Angle CGB is equal to the Angle GBC; and so the Side B C equal to the Side CG; but likewise the Side CB is equal to the Side GK, and the Side CG to BK: Therefore GK is equal to KB, and C G K B is equilateral. I say, it is also Right-angled; for because CG is parallel to BK, and CB falls on them, ... the Angles KBC, GCB, are equal to two Right Angles. But KBC is a Right Angle. Wherefore GBC

+ 34. 1.

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GBC also is a Right Angle, and the oppofite Angles CGK, GKB, shall be Right Angles. Therefore CGKB is a Rectangle. But it has been proved to be equilateral. Therefore CGKB is a Square described upon BC. For the fame Reason HF is also a Square made upon HG, that is equal to the Square of AC. Wherefore HF and CK are the Squares of AC and CB. And because the Rectangle A Gis * equal to the Rectangle GE, and AG is that which * 43 is contained under AC and CB, for GC is equal to CB: GE shall be equal to the Rectangle under AC, and CB. Wherefore the Rectangles AG, GE, are equal to twice the Rectangle contained under AC, CB; and HF, CK, are the Squares of AC, Св. Therefore the four Figures HF, CK, AG, GE, are equal to the Squares of AC and CB, with twice the Rectangle contained under AC and CB. But HF, CK, AG, GE, make up the whole Square of AB, viz. ADEB. Therefore the Square of AB is equal to the Squares of AC, CB, together with twice the Rectangle contained under AC, CB. Wherefore, if a Right Line be any bow cut, the Square which is made on the whole Line, will be equal to the Squares made on the Segments thereof, together with twice the Rectangle contained under the Segments; which was to be demonftrated.

Coroll. Hence it is manifest, that the Parallelograms which stand about the Diameter of a Square, are likewife Squares.

PROPOSITION V.

THEOREM.

If a Right Line be cut into two equal Parts, and into
two unequal ones; the Rectangle under the unequal
Parts, together with the Square that is made of the
intermediate Distance, is equal to the Square made
of half the Line.

TET any Right Line AB be cut into two equal equal
Parts in C, and into two unequal Parts in D. I

fay the Rectangle contain'd under AD, DB, toge

E

ther

† 46. 1.

ther with the Square of CD, is equal to the Square of CB,

For + describe CEFB, the Square of BC, draw BE, and thro' D draw * DHG, parallel to CE, or *31.1. BF, and thro' H draw KLG, parallel to CB, or EF, and AK thro' A, parallel to CL, or BO.

#43.1.

143, 1.

Now the Complement CH is † equal to the Complement HF. Add DO, which is common to both of them, and the whole CO, is equal to the whole DF; but CO is † equal to A L, because AC is equal to CB; therefore AL is equal to DF, and adding CH, which is common, the whole AH shall be equal to FD, DL, together. But AH is the Rectangle contain'd under AD, DB; for DH is + equal * Cor. 4. to DB, * and FD, DL, is the Gnomon MNX; of this. therefore MNX is equal to the Rectangle contain'd under AD, DB, and if LG, being common, and * equal to the Square of CD be added; then the Gnomon MNX, and LG, are equal to the Rectangle contain'd under AD, DB, together with the Square of CD; but the Gnomon MNX, and LG, make up the whole Square CE F B, viz. the Square of CB. Therefore the Rectangle under A D, DB, together with the Square of CD, is equal to the Square of CB. Wherefore, if a Right Line be cut into two equal Parts, and into two unequal ones; the Rectangle under the unequal Parts, together with the Square that is made the intermediate Distance, is equal to the Square made of half the Line; which was to be demonstrated.

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PRO

PROPOSITION VI.

THEOREM.

If a Right Line be divided into two equal Parts, and another Right Line be added directly to the same, the Rectangle contain'd under the Line, compounded of the whole and added Line, (taken as one Line,) and the added Line, together with the Square of half the Line, is equal to the Square of the Line compounded of half the Line, and the added Line taken as one Line.

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LET the Right Line AB be bisected in the Point
C, and BD added directly thereto. I say the
Rectangle under AD, and DB, together with the
Square of BC, is equal the Square of CD.

* 46. 1.

For describe * CE FD, the Square of CD, and † 36.1. join DE; draw + BHG thro' B, parallel to CE, or DF, and KLM thro' H, parallel to AD, or EF, as also AK thro' A, parallel to CL, or DM.

Then because A C is equal to CB, the Rectangle # 43.1. A L shall be equal to the Rectangle CH, but CH is ‡ equal to H F. Therefore AL will be equal to HF; and adding CM, which is common to both, then the whole Rectangle A M, is equal to the Gnomon NXO. But AM is that Rectangle which is * Cor. 4, contain'd under A D, DB, for DM is * equal to DB; of this. therefore the Gnomon NXO is equal to the Rectangle under A D, and DB. And adding LG, which + Cor.4. is common, viz. f the Square of CB; and then the of this. Rectangle under AD, DB, together with the Square of BC, is equal to the Gnomon NXO with LG. But the Gnomon NXO, and LG, together, make up the Figure CEFD, that is the Square of CD. Therefore the Rectangle under AD, and DB, together with the Square of BC, is equal to the Square of CD. Therefore, if a Right Line be divided into two equal Parts, and another Right Line be added directly to the fame, the Rectangle contain'd under the Line, compounded of the whole and added Line, (taken as one Line,) and the added Line, together with the Square

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of

* 46. 1.

† 43.1.

of half the Line, is equal to the Square of the Line compounded of half the Line, and the added Line taken as one Line; which was to be demonstrated.

PROPOSITION VII.

THEOREM.

If a Right Line be any how cut, the Square of the whole Line, together with the Square of one of the Segments, is equal to double the Rectangle contain'd under the whole Line, and the Said Segment, together with the Square, made of the other Segment.

LET the Right Line AB be any how cut in the Point C. I say the Squares of AB, BC, together, are equal to double the Rectangle contain'd under AB, BC, together with the Square, made of AC.

For let the Square of A B be * defcrib'd, viz. A ADEB, and conftruct the Figure.

Then because the Rectangle AG, is equal to the Rectangle GE. If CF, which is common, be added to both, the whole Rectangle AF shall be equal to the whole Rectangle CE, and so the Rectangles AF, CE, are double to the Rectangle AF; but AF, CE, make up the Gnomon KLM, and the Square CF. Therefore the Gnomon KLM, together with the Square CF, shall be double to the Rectangle AF. But double the Rectangle under A B, BC, is double the Rectangle AF, for BF is † equal to BC. Therefore the Gnomon KLM, and the Square CF, are equal to twice the Rectangle contain'd under A B, BC. And if HF, which is com

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mon, being the Square of A C, be added to both;

then

A Figure is faid to be constructed, when Lines, drawn in a Parallelogram parallel to the Sides thereof, cut the Diameter in one Point, and make two Parallelograms about the Diameter, and two Complements. So likewise a double Figure is faid to be constructed, when two Right Lines parallel to the Sides, make four Parallelograms about the Diameter, and four Complements.

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