46. I

† 43. I.

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; which was to be demonftrated.

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 contained under the whole Line, and the faid Segment, together with the Square, made of the other Segment.

ET the Right Line A B be any how cut in the Point C. I fay the Squares of AB, BC, to gether, are equal to double the Rectangle contained under AB, BC, together with the Square, made of AC.

For let the Square of AB be described, viz. ADE B, 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 fhall be equal to the whole Rectangle CE, and fo 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, fhall be double to the Rectangle AF. But double the Rectangle under AB, 1 Cor. 4. 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 con❤ tained under AB, BC. And if HF, which is common, being the Square of AC, be added to both; then the Gnomon KLM, and the Squares CF, HF, are equal to double the Rectangle contained

§ A Figure is faid to be conftructed, 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 likewife a double Figure is faid to be conftructed, when two Right Lines parallel to the Sides, make four Parallelograms about the Diameter, and four Complements.

under

under AB, BC, together with the Square of AC. But the Gnomon KLM, together with the Squares CF, HF, are equal to ADE B, and CF, viz. the Squares of AB, BC. Therefore the Squares of AB, BC, are together equal to double the Rectangle contained under AB, BC, together with the Square of AC. Therefore, 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. contained under the whole Line, and the faid Segment, together with the Square, made of the other Segments which was to be demonftrated.

PROPOSITION VIII.

THEOREM.

If a Right Line be any how cut into two Parts, four times the Rectangle, contained under the whole Line, and one of the Parts, together with the Square of the other Part, is equal to the Square of the Line, compounded of the whole Line, and the first Part taken as one Line.

ET the Right Line AB be cut any how in C. I fay four times the Rectangle contained under AB, BC, together with the Square of AC, is equal to the Square of AB, and BC taken as one Line.

For let the Right Line A B be produced to D, so that BD be equal to BC, defcribe the Square AE FD, on AD, and conftruct the double Figure.

*

Now fince CB is equal to BD, and alfo to Hyp GK, and BD is equal to KN: GK fhall be t. 34. I likewife equal to KN; by the fame Reasoning, PR is equal to RO. And fince CB is equal to BD, and GK to KN, the Rectangle CK will be ‡ 36. 1. equal to the Rectangle BN, and the Rectangle GR to the Rectangle RN. But CK is equal to RN;* 43. T for they are the Complements of the Parallelogram CO. Therefore BN is equal to GR, and the four Squares BN, KC, GR, RN, are equal to each other; and fo they are together Quadruple CK. Again, because CB is equal to BD, and BD to BK, that is, equal to CG; and the faid CB is equal

E 2

alfo

+ Cor. 4. of this.

allo to GK, that is, to GP; therefore CG fhall be equal to GP. But PR is equal to RO; therefore the Rectangle AG fhall be equal to the Rectangle MP, and the Rectangle PL equal to RF. But MP is equal to PL; for they are the Complements of the Parallelogram ML. Wherefore AG is equal alfo to RF. Therefore the four Parallelograms AG, MP, PL, RF, are equal to each other, and accordingly they are together Quadruple of AG. But it has been proved that the four Squares CK, BN, GR, RN, are Quadruple of CK. Therefore the four Rectangles, and the four Squares, making up the Gnomon STY, are together Quadruple of AK; and because A K is a Rectangle contained under AB, and BC, for BK is equal to BC; four times the Rectangle under AB, BC, will be Quadruple of A K. But the Gnomon STY has been proved to be Quadruple of AK. And fo four times the Rectangle contained under AB, BC, is equal to the Gnomon STY. And if XH, being equal to † the Square of AC, which is common, be added to both: Then four times the Rectangle contained under AB, BC, together with the Square of A C, is equal to the Gnomon STY, and the Square XH, But the Gnomon STY and HX, make AEFD, the whole Square of AD. Therefore four times the Rectangle contained under AB, BC, together with the Square of AC, is equal to the Square of AD, that is, of A B and B C taken as one Line. Wherefore, if a Right Line be any how cut into two Parts, four times the Rectangle contained under the whole Line, and one of the Parts, together with the Square of the other Part, is equal to the Square of the Line, compounded of the whole Line, and the firft Part taken as one Line; which was to be demonftrated.

PRO

PROPOSITION IX.

THEOREM.

If a Right Line be any how cut into two equal, and two unequal Parts; then the Squares of the unequal Parts together, are double to the Square of the half Line, and the Square of the intermediate Part.

ET any Right Line AB be cut unequally in D, and equally in C. I fay the Squares of AD, DB, together, are double to the Squares of AC and CD together.

For let* CE be drawn from the Point C at Right * 11. 1. Angles to AB, which make equal to AC, or CB, and join EA, EB. Alfo thro' D let + DF be drawn † 31. 1. parallel to CE, and FG thro' F parallel to AB, and draw AF.

*

Now because AC is equal to CE, the Angle EAC will be equal to the Angle AEC; and fince the ‡ 5. 1. Angle at C is a Right one, the other Angles AEC, EAC, together, fhall make one Right Angle, and 3. Cor. are equal to each other: And fo AEC, EAC, are each 32. I. equal to half a Right Angle. For the fame Reason are also CEB, EBC, each of them half Right Angles. Therefore the whole Angle AEB is a Right Angle. And fince the Angle GEF is half a Right one, and EGF is a Right Angle; for it is † equal to the † 29. 1. inward and oppofite Angle ECB, the other Angle EFG will be alfo equal to half a Right one. Therefore the Angle GEF is equal to the Angle EF G. And fo the Side EG is equal to the Side GF. Again, ‡ 6. 1. because the Angle at B is half a Right one, and FDB is a Right one, because equal to the inward and oppofite Angle ECB, the other Angle BFD will be half a Right Angle. Therefore the Angle at B is equal to the Angle BFD; and fo the Side DF is equal to the Side DB. And because AC is equal to CE, the Square of AC will be equal to the Square of CE Therefore the Squares of AC, CE, together, are double to the Square of AC; but the Square of EA is equal to the Squares of AC, CE, together, finee 4 47. x« ACE

E 3

+ 47. I,

ACE is a Right Angle. Therefore the Square of EA is double to the Square of AC. Again, because EG is equal to GF, and the Square of E G is equal to the Square of GF: Therefore the Squares of EG, GF, together, are double to the Square of GF. But the Square of EF is † equal to the Squares of EG, GF, Therefore the Square of EF is double the Square of GF: But GF is equal to CD; and fo the Square of EF double to the Square of CD. But the Square of AE is likewife double to the Square of AC. Wherefore the Squares of AE, and EF, are double to the Squares of AC and CD. But the Square of AF is equal to the Squares of AE and EF; because the Angle AEF is a Right Angle, and confequently the Square of AF is double to the Squares of AC, and CD. But the Squares of AD, DF, are equal to the Square of AF: For the Angle at D is a Right Angle. Therefore the Squares of AD, and DF, together, fhall be double to the Squares of AC and CD together. But DF is equal to DB. Therefore the Squares of AD, and DB, together, will be double to the Squares of AC and CD, together. Wherefore, if a Right Line be any how cut into two equal, and two unequal Parts, then the Squares of the unequal Parts together, are double to the Square of the half Line, and the Square of the intermediate Part; which was to be demonftrated.

PROPOSITION X.

THEOREM.

If a Right Line be cut into two equal Parts, and to it be directly added another; the Square made on [the Line compounded of] the whole Line, and the added one, together with the Square of the added Line, fhall be double to the Square of the balf Line, and the Square of [that Line which is compounded of] the half, and the added Line.

L

ET the Right Line AB be bisected in C, and bany ftraight Line BD added directly thereto. I fay the Squares of AD, DB, together, are double to the Squares of AC, CD, together,