are equal to the double Rectangle contained under A B, and B C, together with the Square of A C. But the Gnomon K L M, together with the Squares CF, HF, are equal to A DEB, and C F, viz. the Squares of A B, BC. Therefore the Squares of A B and B C, are together equal to double the Rectangle contained under A B,, and B C, together with the Square of A C. 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 Segment; which was to be demonftrated.

PR POPSITION VIII.

THEOREM.

If a Right Line be any bow 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.

L

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

For, let the Right Line A B be produced to D, fo that B D be equal to BC; defcribe the Square A EFD, on A D, and conftruct the double Figure.

*

Now, fince CB is equal to BD, and alfo to †, GK, and BD is equal to K N; therefore G K fhall be likewife equal to KN: By the fame Reasoning, PR is equal to RO. And fince C B is equal to BD, and G K to K N, the Rectangle CK will be equal to the Rectangle B N, and the Rectangle G R to the Rectangle RN. But CK is equal to RN; for they are the Complements of the Parallelogram C O. Therefore BN is equal to G R, and the four Squares BN, CK, GR, R N, are equal to each other; and fo they are together quadruple CK. Again, because CB is equal to B D, and B D to B K, that is, equal

to

to CG; and the faid C B is equal alfo to G K, that is, to GP; therefore C G fhall be equal to G P. But PR is equal to RO; therefore the Rectangle A G fhall be equal to the Rectangle M P, and the Rectangle P L equal to RF. But MP is equal to P L; for they are the Complements of the Parallelogram M L. Wherefore A G is equal alfo to R F. Therefore the four Parallelograms AG, MP, PL, R F, are equal to each other, and accordingly they are together quadruple of A G. But it has been proved that the four Squares C K, 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 A B, and BC, for B K is equal to BC; therefore four times the Rectangle under A B, and BC; will be quadruple of A K. But the Gnomon STY has been proved to be quadruple of A K. And fo four Times the Rectangle contained under AB and B C, is equal to the Gnomon S T Y. And if X H, being equal to + the Square of A C, which is + Cor. 4. of common to be added to both; then four Times the tbis. Rectangle contained under A B, and B C, together with the Square of A C, is equal to the Gnomon STY, and the Square X H. But the Gnomon STY and X H make A EFD, the whole Square of A D. Therefore four Times the Rectangle contained under A B and BC, together with the Square of A C, is equal to the Square of A D, that is of A B and BC 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 de monftrated.

II. I.

† 31. 1.

$5.1.

* Cor. 3. 32. I.

† 29. I.

6.1.

† 47. I.

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.

LET any Right Line A B be cut unequally in D,
and equally in C. I fay, the Squares of A D,
D B, together, are double to the Squares of A C and
CD together.

For, let C E be drawn from the Point C at Right Angles to A B, which make equal to AC, or CB; and join E A, E B. Alfo through D let + D F be drawn parallel to C E, and F G through F parallel to A B, and draw A F.

Now, because AC is equal to CE, the Angle EAC will be equal to the Angle E A C; and fince the Angle at C is a Right one, the other Angles, AEC, EAC, together, fhall * make one Right Angle, and are equal to each other: And fo A EC, EAC, are each equal to half a Right Angle. For the fame Reafons are alfo CE B, EBČ, each of them half a Right Angle. Therefore the whole Angle A E B is a Right Angle. And fince the Angle GEF is half a Right one, and EGF is a Right Angle; for it is t equal to the inward and oppofite Angle E CB; the other Angle E F G will be alfo equal to half a Right Therefore the Angle G E F is equal to the Angle EFG. And fo the Side E G is equal to the Side G F. Again, because the Angle at B is half a Right one, and FDB is a Right one, because equal to the inward and oppofite Angle E C B, 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 D F is equal to the Side D B. And because A C is equal to C E, the Square of A C will be equal to the Square of CE. Therefore the Squares of A ̊C, and CE together, are double to the Square of A C; but the Square of E A is † equal to the Squares of

one.

A C,

A C, and CE, together, fince ACE is a Right Angle. Therefore the Square of EA is double to the Square of A C. Again, becaufe E G is equal to GF, and the Square of EG is equal to the Square of GF; therefore the Squares of EG, GF, together, are double to the Square of G F. But the Square of EF is equal to the Squares of EG, GF. Therefore † 47. 1. the Square of EF is double the Square of G F: But GF is equal to CD; and fo the Square of E F is double to the Square of CD. But the Square of A E is likewife double to the Square of A C. Wherefore the Squares of A E and E F, are double to the Squares of AC, and CD. But the Square of A F is + equal to the Squares of AE and EF; because the Angle AEF is a Right Angle, and confequently the Square of A F is double to the Squares of AC and CD. But the Squares of A D, and D F, are equal to the Square of A F: For the Angle at D is a Right Angle. Therefore the Squares of A D, and D F, together, fhall be double to the Squares of AC, and CD, together. But DF is equal to D B. Therefore the Squares of A D, and D B, together, will be double to the Squares of A C, and CD, together. Wherefore, if a Right Liné be any how cut into two equal, and two unequal Parts ; then the Squares of the unequal Parts together, are double to the Squares of the half Line, and the Squares 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. LET the Right Line A B be bifected in C, and any

ftrait Line B D added directly thereto. I fay, the Squares of A D, and D B, together, are double to the Squares of A C, and CD, together.

For,

1

#11. I. + 31 I.

29. I.

For, draw *CE from the Point C at Right Angles to A B, which make equal to A C, or CB; and draw A E, EB; likewife through E let E F be + drawn pat rallel to A D, and through D, DF+ parallel to CE.

Then, because the Right Line E F falls upon the Parallels EC, FD, the Angles CEF, EFD, are t equal to two Right Angles. Therefore the Angles FEB, EFD, are together lefs than two Right Angles. But Right Lines making, with a third Line, Angles together less than two Right Angles, being in* Ax. 12. finitely produced, will meet*. Wherefore E B, FD, produced, will meet towards B D. Now let them be produced, and meet each other in the Point G, and let A G be drawn.

And then, because A C is equal to CE, the Angle AEC will be equal to the Angle EAC+: But the Angle at C is a Right Angle. Therefore the Angle EAC, or A E C, is half a Right one. By the fame Way of Reafoning, the Angle CEB, or EBC, is half a Right one. Therefore A E B is a Right Angle. And fince EBC is half a Right Angle, DBG will alfo be half a Right Angle, fince it is vertical to E B C. But BDG is a Right Angle alfo; for it is * équal to the alternate Angle DCE. Therefore the remaining Angle DGB is half a Right Angle, and fo equal to D B G. Wherefore the Side B D is + equal to the Side D G. Again, because EGF is half a Right Angle, and the Angle at F is a Right Angle, for it is equal to the oppofite Angle at C; the remaining Angle FEG will be alfo half a Right one, and is equal to the Angle EGF; and fo the Side GF is + equal to the Side E F. And fince CE is equal to CA, and the Square of EC equal to the Square of CA; therefore the Squares of E C, C A, together, are double to the Square of CA. But the Square of EA is equal to the Squares of E C and CA. Wherefore the Square of E A is double to the Square of A C. Again, because G F is equal to FE, the Square of G F alfo is equal to the Square of F E. Wherefore the Squares of G F and F E are double to the Square of F E. But the Square of E G is equal to the Squares of GF, FE. Therefore the Square of E G is double to the Square of EF: But EF is equal to CD. Wherefore the Square of E G fhall be

double