Book II. right line C B falls upon them, the angles KB C, GCB are [by 29. 1.] equal to two right angles. But [by 30. def. 1.] the angle KB C is a right angle, and fo the angle GCB is a right angle too: wherefore [by 34. 1.] the opposite angles CGK, GKP will be right angles. Therefore C G K B is a rectangle. But it has been proved to be equilateral: wherefore CG KB is a fquare, defcribed upon в C. By the fame reason also the parallelogram HF is a fquare, viz. the fquare of HG, that is, of AC: therefore HF, C K are the fquares of AC, C B. And because the rectangle AG [by 43. 1.] is equal to the rectangle G E, and A G is contained under the right lines A C, CB; for G C is equal to CB; therefore GE is equal to the rectangle contained under AC, CB: Wherefore the rectangles AG, GE are equal to twice the rectangle under the right lines AC, C B. But HF, CK are the fquares of A C, C B. Therefore the four parallelograms HF, CK, A G, GE are equal to the squares of AC, CB, together with twice the rectangle under the right lines A C, B C. But HF, CK, AG, GE make up the whole A D E B, viz. the fquare of A B, Wherefore the fquare of AB is equal to both the fquares of A C, B C together, with twice the rectangle contained under AC and

св.

Therefore if a right line be divided into any two parts, the fquare of the whole line is equal to both the fquares of the parts, together with twice the rectangle contained under the parts. Which was to be demonftrated.

Otherwise:

I fay, the fquare of AB is equal to both the fquares of A C, CB, together with twice the rectangle under A c, C B. For becaufe, in the fame figure, BA is equal to A D2 the angle A B D [by 5. 1.] will be equal to the angle A D B ; and fince [by 32. 1.] the three angles of every triangle, all together, are equal to two right angles. But the angle BAD is a right angle. Therefore the remaining angles A BD, ADB are both together equal to one right angle. But they are equal to one another: Wherefore the angles ABD, A DB are each one half a right angle. But BCG is a right angle, for it is equal to the inward opposite angle at A; and fo the remaining angle CGB is one half a right angle: accordingly, the angle CGE is equal to the angle

BG,

CB G, and the fide B C [by 6. 1.] equal to the fide C G. But CB [by 34. I.] is equal to KG, and CG to BK: wherefore CK is equilateral. But it has one right angle CB K; therefore it is a square, viz. that of C B. For the fame reason HF is a fquare too, being equal to the square of AC: Therefore the fquares C K, HF are equal to the squares of A C, CB. Again, because [by 43. 1.] the rectangle AG is equal to EG, but AG is the rectangle contained under the right lines A C, CB, for C G is equal to CB; the rectangle E G is equal to that under the right lines AC, CB: Wherefore A G, G E are equal to twice the rectangle under the right lines A C, C B; but alfo C K, HF, are equal to the fquares of A C, CB: therefore C K, HF, AG, GE are equal to the fquares of AC, CB, together with twice the rectangle under A C, CB. But CK, HF and A G, GE, make up the whole square of A B.

Therefore the fquare of A B is equal to both the squares of AC, CB, together with twice the rectangle contained under A C, C B. Which was to be demonftrated.

Corol. From hence it is manifeft, that the parallelograms in fquares that are about the diameter, are squares themselves.

If a right line be divided into two equal, and into two unequal parts, the rectangle contained under the unequal parts of the whole, together with the Square of the right line between the two points of divifion, will be equal to the square of half the

line c.

For let any right line A B be divided at the point c into equal parts, and into unequal ones at the point D: I fay, the rectangle contained under the right lines A D, D B, together with the fquare of c D, is equal to the fquare of the half line C B.

For [by 46. 1.] defcribe the fquare CEFB of BC; join BE; through D draw [by 31. 1.] DHG parallel to CE or B F; through H draw K L M parallel to C B or E F ; and through a draw A K parallel to CL or B M.

Now

Now because [by 43. 1.] the complement c H is equal to the complement HF, add D M, which is common to

A

C

D

B

NH

K

L

E

G

both then is the whole CM equal to the whole DF. But C M [by 36. 1.] M is equal to AL, fince AC is equal to C B. And fo AL is also equal to. D F. Add CH, which is common to

both. And then the whole AH will be equal to D F, DL. But A H is the rectangle under A D, DB, for DH is equal to DB; but FD, DL is the gnomon Nxo: Therefore the gnomon NX O is equal to the rectangle contained under the right lines A D, D B. Add L G, which [by Cor. 4. 2.] is equal to the fquare of CD, to both, and then the gnomon NXO and LG are equal to the rectangle under the right lines AD, DB, together with the fquare of C D. But the gnomon NX O and LG, make up the whole fquare C E F B, viz. that of CB: Therefore the rectangle AD, DB, toge-. ther with the fquare of c D, is equal to the fquare of C B.

If therefore a right line be divided into two equal, and into two unequal parts, the rectangle contained under the unequal parts, together with the fquare of the right line between the two points of divifion, will be equal to the fquare of half the line. Which was to be demonstrated.

This theorem may be demonftrated otherwise thus: Because the fquare of CB is [by 4. 2.] equal to the fquares of CD, DB, C D B together with twice the rectangle under D B, CD, and the rectangle under DB, CB [by 3. 2.] equal to the rectangle under D B, C D, together with the fquare of DB; the fquare of cв will be equal to the remaining iquare of c D, together with the remaining rectangle under D B, CD, and the rectangle under D B, C B, Or under D B, A C. But the rectangle under D B, and the whole a D, is equal to the rectangles under D B, A C, and under DB, C D [by 1. 2.]. Therefore the fquare of CB will be equal to the fquare of c D, together with the rectangle under D B, AD; that is, the rectangle under A D, D B, together with the fquare of CD, will be equal to the fquare of c B. Which was to be demonftrated.

PROP.

If a right line be divided into two equal parts, and. any other right line be joined to it in the fame direction; the rectangle contained under the line, compofed of the whole line and the joined line, and the joined line, together with the jquare of the kalf line, is equal to the fquare of the line compofid of the half and the joined line, taken as one line.

For let any right line A B be bifected in the point c, and any other right line BD be joined to it in the fame direction: I fay, the rectangle contained under AD, DB together with the fquare of c B, is equal to the fquare

of C D.

I.

A

For [by 46. 1.] defcribe the fquare CEFD of CD: join DE; thro' the point B [by 31. 1.] draw BHG parallel to CE or DF; through the point H draw K L M parallel to A D or EF; and through a draw AK parallel to CL or D M.

Now because A c is equal to CB, the rectangle A L will be [by 36. 1.] equal to the rectangle C H. But [by 43. I.] CH

K

C

B D

H

M

L

N

E

G

F

is equal to HF; and therefore will AL be equal to HF. Add CM, which is common to both; then the whole A M is equal to the gnomon N Xo. But AM is the rectangle contained under the right lines A D, DB; for DM [by cor. 4. 2.] is equal to DB: therefore the gnomon N x o is equal to the rectangle contained under the right lines A D, D B. Again, add LG, which is common to both, being equal to the fquare of c B; then the rectangle under A D, D B, together with the fquare of BC, is equal to the gnomon NXO and to LG. But the gnomon Nxo and LG make up the whole fquare CEFD, that is, the fquare of CD: wherefore the rectangle under A D, DB, together with the fquare of BC, is equal to the fquare of c D.

If therefore a right line be divided into two equal parts, and any right line be joined to it in the fame direction; the rectangle contained under the whole line and the joined G

line,

line, and the joined line, together with the fquare of the half line, is equal to the fquare of the right line compofed of the half and the joined line, taken as one line. Which was to be demonstrated.

d This theorem may be otherwife demonftrated thus: Let the right line AB be bifected in the point c, and let any right line B D be added to it, having the fame direction: I fay, the rectangle under the whole line A D, and the added line в D, together with the fquare of the half line C B, will be equal to the fquare of the right line c D. For let the right line A È be added to AB on the contrary fide E A equal to BD; then the whole line ED will be cut equally in

+

C

B J

2.]

c, and unequally in B; fince EA, AC are equal to D B, BC. The right line E B alfo will be equal to the right line a D, for E A is equal to DB and A B is common to both. Therefore [by 5. 2. the rectangle under the unequal parts E B, BD, together with the fquare of the intermediate fegment CB; that is, the rectangle under A D, BD, together with the fquare of the half line CB, will be equal to the fquare of the half line CD; that is, to the fquare of the right line compounded of half the given line A B, together with the added line в D, taken as one line. Which was to be demonstrated.

PROP. VII. THEOR.

If a right line be divided into any two parts, the Square of the whole line, together with that of one of the parts, will be equal to twice the rectangle contained under the whole line and the faid part, together with the Square of the remaining parte.

For let any right line A B be divided by the point c into any two parts: I fay, both the fquares of A B, BC are equal to twice the rectangle contained under the right lines A B, BC, together with the fquare of a c.

For [by 46. 1.] defcribe the fquare A DEB of A B, and conftruct for make] the figure.

Now because [by 43. 1.] the rectangle AG is equal to the rectangle GE, add CF, which is common, to both; then the whole AF is equal to the whole CE: therefore