PROPOSITION II.

THEOREM.

If a Right Line be any how divided, the Rectangles contained under the whole Line, and each of the Segments, or Parts, are equal to the Square of the whole Line.

ET the Right Line A B be any how divided in the Point C. I fay, the Rectangle contained under AB, BC, together with that contained under AB and AC, is equal to the Square made on AB.

For let the Square ADE B be defcribed * on A B,* 46. г. and thro' C let CF be drawn parallel to AD or BE. Therefore AE is equal to the Rectangles AF and CE. But AE is a Square described upon AB; and AF is the Rectangle contained under BA, A C; for it is contained under D A and A C, whereof AD is equal to AB; and the Rectangle CE is contained under AB, BC, fince BE is equal to AB. Wherefore the Rectangle under AB and AC, together with the Rectangle under AB and B C, is equal to the Square of AB. Therefore, if a Right Line be any how divided, the Rectangles contained under the whole Line, and each of the Segments, or Parts, are equal to the Square of the whole Line.

PROPOSITION III.

THEOREM.

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 firft-mentioned Part.

ET the Right Line A B be any how cut in the Point C. I fay, the Rectangle under A B and BC is equal to the Rectangle under AC and BC, together with the Square described on BC.

For

46. I.

1

For defcribe the Square CDEB upon BC; ↑ 31. 1. produce ED to F; and let AF be drawn + thro' A, parallel to CD or BE.

* 46. I. † 31. 1.

$29. I.

5. I.

† 6.1. $34 I

Then the Rectangle AE fhall be equal to the two Rectangles AD, CE: And the Rectangle A E is that contained under AB and BC; for it is contained under A B and B E, 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 described upon B C. 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 Rectan= gle 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 firft-mentioned Part; which was to be demonftrated.

PROPOSITION IV.

THEOREM.

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 Rettangle contained under the Segments.

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

*

Then because CF is parallel to AD, and BD falls upon them, the outward Angle BGC fhall 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 fo the Side BC 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 CGKB

is equilateral. I fay, it is alfo Right-angled; for becaufe 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 alfo is a Right Angle, and the oppofite Angles GCB, CGK, GKB, fhall be Right Angles. Therefore CGKB is a Rectangle. But it has been proved to be equilateral. Therefore CGKB is a Square defcribed upon BC. For the fame Reafon HF is allo 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 AG is *equal to the Rectangle GE, and AG is that which * 43 *. is contained under AC and CB, for GC is equal to CB: GE fhall 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, CB. 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 how 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 manifeft, that the Parallelograms which stand about the Diameter of a Square, are likewife Squares.

PRO

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.

LET any Right Line A B be cut into two equal

Parts in C, and into two unequal Parts in D. I fay the Rectangle contained under AD, DB, together with the Square of CD, is equal to the Square of B C.

Fort defcribe CEF B, the Square of BC, draw BE, and thro' D draw* DHG, parallel to CE, or BF, and thro' H draw KLO, parallel to CB, or EF, and AK thro' A, parallel to CL, or BO.

*

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 AL, because AC is equal to CB; therefore AL is equal to DF, and adding CH, which is common, the whole AH fhall be equal to FD, DL, together. But AH is the Rectangle contained under AD, DB; for DH is equal to DB, and FD, DL, is the Gnomon MNX; therefore MNX is equal to the Rectangle contained 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 contained under AD, DB, together with the Square of CD; but the Gnomon MNX, and LG, make up the whole Square CEF B, viz. the Square of CB. Therefore the Rectangle under AD, 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 of the intermediate Distance, is equal to the Square made of half the Line; which was to be demonstrated,

PRO

PROPOSITION VI

THEOREM.

If a Right Line be divided into two equal Parts, and another Right Line be added directly to the fame, the Rectangle contained 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.

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

For defcribe CEFD, the Square of CD, and * 46. 1. join DE; draw + BHG thro' B, parallel to CE, † 31. 1. or DF, and KLM thro' H, parallel to AD, or EF, as alfo AK thro' A, parallel to CL, or DM.

Then because AC is equal to CB, the Rectangle AL fhall be equal to the Rectangle CH, but CH* 36. t. is equal to HF. Therefore AL will be equal to ‡ 43. I. HF; and adding CM, which is common to both, then the whole Rectangle AM, is equal to the Gnomon NXO. But AM is that Rectangle which

is contained under AD, DB, for DM is equal to * Cor. 4. DB; therefore the Gnomon NXO is equal to of this. the Rectangle under AD, and DB. And adding LG, which is common, viz. † the Square of CB; † Cor. 4. and then the Rectangle under AD, DB, together of this. with the Square of BC, is equal to the Gnomon NXO with LG. But the Gnomon NX O, 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 contained under the Line, compounded of the subole and added Line, (taken as one Line,) and the added

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