Certainly the rectangle AE is equal to the rec- A tangles AF and CE: and the rectangle AE is the fquare of AB; and the rectangle AF is a rectangle contained by BA, AC; for (by def. 1. 2.) it is contained by DA, AC; and AD is equal to AB (being fides of a square): and the rectangle CE D is contained by AB, BC; for EB is equal to AB:

wherefore the rectangle contained by BA, AC together with that contained by AB, BC is equal to the square of AB.

If therefore a straight line be cut as it may happen, the rectan gles contained by the whole and each of the segments are equal to the fquare of the whole line.

PROP. III.

If a straight line be cut as it may happen, the rectangle contained by the whole and one of the fegments is equal to the rectangle contained by the two segments; and to the fquare of the forementioned fegment.

For let the straight line AB be cut as it may happen in the point C: fay that the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB; together with the square of BC.

For let CDEB the fquare of BC be described (by 46. 1.); and let ED be produced to F: and through the point A let AF be drawn (by 31. 1.) parallel to either of the straight lines CD, BE. Certainly the rectangle AE is equal to the rectangles AD, CE; and the rectangle AE is a rectangle contained by AB, BC; for (by def. 1. B. 2.) it is contained by AB, BE; but BE is equal to BC (being fides of a

A

C

B

fquare): And the rectangle AD is contained

by AC, CB; for DC is equal to CB ; but F D the rectangle DB is the fquare of CB: there

E

fore the rectangle contained by AB, CB is equal to the rectangle contained by AC, CB; together with the fquare of BC.

If therefore a straight line be cut as it may happen, the rectangle, contained by the whole and one of the fegments, is equal to the rectangle

VOL. I.

G

Book II. rectangle contained by the two fegments; and to the square of the forementioned fegment. Which was to be demonftrated.

If a straight line be cut as it may happen, the square of the whole line is equal to the fquares of the two fegments, and to the rectangle, contained by the two fegments, taken twice.

For let the straight line AB be cut as it may happen in the point C: I fay that the fquare of AB is equal to the fquares of AC, CB and the rectangle contained by AC, CB taken twice.

For let the fquare ADEB be defcribed upon AB (by 46. 1.); and let BD be joined, and through the point C let CGF be drawn parallel to either of the lines AD, BE; and through the point G let HK be drawn parallel to either of the lines AB, DE.

A

D

C B

G

K

FE

And because CF is parallel to AD and BD hath fallen upon them; the outward angle BGC is equal to the inward and oppofite ADB but the angle ADB (by 5.1.) H is equal to ABD, because the fide BA is equal to AD; therefore (by com. not. 1.) the angle CGB is equal to GBC; so that alfo (by 6. 1.) the fide BC is equal to the fide CG but (by 34. 1.) CG is equal to BK and CB to GK; wherefore alfo GK is equal to KB: wherefore CGKB is equilateral; I fay also that it is rectangular For because CG is parallel to BK and CB hath fallen upon them; therefore (by 29. 1.) the angles KBC, BCG are equal to two right angles; but KBC is a right angle; wherefore alfo GCB is a right angle; fo that alfo (by 34. 1.) the oppofite angles CGK, GKB are right angles: wherefore the figure CGKB is rectangular; and it has been demonftrated to be equilateral; therefore (by def. 30. 1.) it is a fquare and it is defcribed upon BC: Certainly for the same reason also the figure HF is a square; and it is defcribed upon HG that is upon AC: therefore the figures HF, CK are the fquares of AC, CB. And because AG is equal to GE (by 43. 1.); and AG is the rectangle contained by AC, CB; for

GC is equal to CB; and GE is therefore equal to the rectangle con- Book II. tained by AC, CB: wherefore AG, GE are equal to the rectangle contained by AC, CB taken twice; but also HF, CK are the squares of AC, CB; therefore the four figures HF, CK, AG, GE are equal to the fquares of AC, CB and the rectangle contained by AC, CB taken twice: but HF, CK, AG, GE are the whole figure ADEB, which is the fquare of AB; wherefore the fquare of AB is equal to the fquares of AC, CB and the rectangle contained by AC, CB taken twice.

Wherefore if a straight line be cut as it may happen, the square of the whole line is equal to the fquares of the two fegments, and to the rectangle contained by the two segments taken twice.

Another Demonstration.

I fay that the fquare of AB is equal to the fquares of AC, CB and to the rectangle contained by AC, CB taken twice.

For, making use of the fame figure, because BA is equal to AD, (by 5. 1.) the angle ABD is equal to the angle ADB; and fince (by 32. 1.) the three angles of every triangle are equal to two right angles; therefore the three angles of the triangle ABD viz. ABD, ADB, BAD are equal to two right angles; but BAD is a right angle (being an angle of a fquare); therefore the remaining angles ABD, ADB are equal to one right angle; and they are equal: each, therefore, of the angles ABD, ADB is the half of a right angle: but BCG is a right angle (by prop. 29. 1.) for it is equal to the inward and oppofite angle at A; therefore the remainder CGB is the half of a right angle: wherefore the angle CGB is equal to CBG; fo that alfo (by 6. 1.) the fide BC is equal to the fide CG; but (by 34. 1.) CB is equal to KG, and CG to BK: wherefore the figure CK is equilateral; but it has a right angle, the angle CBK wherefore CK is a square; and it is defcribed upon CB. Certainly for the fame reafon alfo HF is a fquare; and is equal to the fquare of AC: wherefore CK, HF are squares, and are equal to the squares of AC, CB: and because AG is equal to GE (by 43. 1.) and AG is the rectangle contained by AC, CB; for CG is equal to CB; and EG is equal to the rectangle contained by AC, CB; therefore the figures AG, GE are equal to the rectangle con

G 2

tained

Book II. tained by AC, CB; taken twice: but CK, HF are equal to the Squares of AC, CB; therefore the figures CK, HF, AG, GE are equal to the fquares of AC, CB and the rectangle contained by AC, CB taken twice: But the Squares CK, HF and the rectangles AG, GE are the whole Square AE, which is the fquare of AB.

Wherefore the fquare of AB is equal to the squares of AC, CB; and to the rectangle contained by AC, CB taken twice. Which was to be demonstrated.

Cor. Certainly from thefe demonftrations it is manifeft that in fquare spaces the parallelograms about the diameter are fquares.

If a straight line be cut into equal and unequal fegments; the rectangle contained by the unequal fegments of the whole line, together with the fquare of the line between the points of section is equal to the fquare of half the line..

For let any straight line AB be cut into equal fegments at the point C; and into unequal fegments at the point D; I say that the rectangle contained by AD, DB together with the square of CD is equal to the fquare of CB.

For let the fquare CEFB be defcribed upon BC; and let BE be joined; and let DHG be drawn, through the point D, parallel to either of the lines CE, BF; and again through the point H let KLM be drawn parallel to either of the lines CB, EF; and again through the point A let AK be drawn parallel to either of the lines CL, BM.

And fince the complement CH is A
equal to the complement HF; let DM
which is common be added; therefore K
the whole CM is equal to the whole
DF: but CM is equal to AL (by prop.
36.1.) because AC is equal to CB:

C

D B

H

M

L

E

G F

therefore alfo (by com. not. 1.) AL is
equal to DF; let CH which is common be added; therefore the
whole AH is equal to DF and DL: but AH is the rectangle con-
tained by AD, DB for DH is equal to DB (by cor. to 4. 2.) but

FD,

FD, DL or FDL is called a Gnomon (by def. 2. 2.); therefore Book II. the Gnomon FDL is equal to the rectangle contained by AD, DB: Let LG which is common be added; which (by Cor. to 4. 2.) is equal to the fquare of CD: wherefore the Gnomon FDL and the Square LG is equal to the rectangle contained by AD, DB and the fquare of CD: but the Gnomon FDL and the fquare LG is the whole square CEFB, which is the fquare of CB: Wherefore the rectangle contained by AD, DB together with the square of CD is equal to the fquare of CB.

Wherefore if a ftraight line be cut into equal and unequal fegments; the rectangle contained by the unequal fegments of the whole line, together with the square of the line between the points of section, is equal to the fquare of half the line. Which was to be demonftrated.

If a straight line be cut in halves, and any straight line be added to it in a straight line; the rectangle contained by the whole with the part produced, and the part produced, together with the square of half the line; is equal to the fquare defcribed upon the compounded line, that is of half the line and the part produced, as upon one line.

For let any ftraight line AB be cut in halves at the point C, and let any straight line BD be added to it in a straight line: I fay that the rectangle contained by AD, DB together with the fquare of BC is equal to the fquare of CD.

For let the fquare CEFD be described (by 46. 1.) upon CD; and let DE be joined; and through the point B let (by 31. 1. BHG be drawn parallel to either of the lines CE, DF; and through the point H, let KLM be drawn parallel to either of the lines AD, EF; and farther through the point A let AK be drawn parallel to either of the lines CL, DM.

Therefore because AC is equal to CB, A the rectangle AL (by prop.36.1.) is equal to the rectangle CH; but CH is also equal (by 43.1.) to HF; wherefore AL is K equal to HF; let CM which is common be added, therefore the whole AM is equal to CM, MG that is (by def. 2.2.)

C

B

D

H

M

L

E

G F