† 2 Ax.

gether with the square of CB, is equal to the gnomon CMG, and the figure LG: but the gnomon CMG and LG make up the whole figure CEFD, which is the square of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c.

Q. E. D.

PROPOSITION VII.

THEOR.-If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part.

Let the straight line AB be divided into any two parts in the point C: the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.

*

A.

H

C

B

G

K

D

F E

Upon AB, describe the square ADEB, and construct the figure as in the preceding propositions. And because AG is equal to GE, add to each of them CK; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK: but AK, CE are the gnomon AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, ist double of AK: but twice the rectangle AB, BC is double of AK, for BK is equal* to BC; therefore the gnomon AKF, together with the square CK, is equal † to twice the rectangle AB, BC: to each of these equals add HF, which is equal to the square of AC; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle, AB, BC, and the square of AC: but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB and BC; therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC. Wherefore, if a straight line, &c.

PROPOSITION VIII.

Q. E. D.

THEOR.-If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.

*

*

† 46. 1.

34. 1.

1 Ax.

36. 1.

Let the straight line AB be divided into any two parts in the point C: four times the rectangle AB, BC, together with the square of AC, shall be equal to the square of the straight line made up of AB and BC together. Produce + AB to D, so that BD be equal † to CB; and upon + 2 Post. AD, describe † the square AEFD; and construct two figures 3.1. such as in the preceding. Because CB is equal to BD, and that CB is equal to GK, and BD to KN, therefore GK is equal to KN: for the same reason, PR is equal to RO: and † because CB is equal to BD, and GK to KN, the rectangle CK is equal to BN, and GR to RN: but CK is equal to RN, because they are the complements of the parallelogram CO; therefore also BN is equal to GR: therefore the four rect- †1 Ax. angles BN, CK, GR, RN are equal to one another, and so are quadruple of one of them CK. Again, because CB is equal to BD, and that BD is equal to BK, that is t, to CG, and CB equal † to GK, that is †, to GP, therefore CG is equal † to GP: and because CG is equal to GP, and PR to RO, the rectangle AG is equal + to MP, and PL to RF: but MP is equal to PL, because they are

*

* 43. 1.

* Cor. 4. 2. & 30 Def.

A

C BD

GK

M

† 34. 1. N† 34. 1.

P R

X

↑ Cor. 4. 2.

& 30 Def. + 1 Ax.

E

43. 1.

the complements of the parallelogram ML; wherefore AG is equal † also to RF: therefore the four rect- +1 Ax. angles AG, MP, PL, RF are equal to one another, and so are quadruple of one of them AG. And it was demonstrated that the four CK, BN, GR, and RN are quadruple of CK: therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK: and because AK is the rectangle contained by AB, BC, for BK, ‡ is equal to BC, therefore four times the rectangle AB, BC, is quadruple of AK: but the gnomon AOH was demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC, is equal to the gnomon AOH: to each of these add XH, which is equal to the square of AC; therefore four times the rectangle AB, BC, together with the square of AC, is equal to the gnomon AOH and the † 2 Ax. square XH: but the gnomon AOH and XH make up the figure AEFD, which is the square of AD; therefore four times the rectangle AB, BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c. Q. E. D.

Because BN is a + square, therefore BK is equal to BD: but BD ist Cor. 4. 2. equal to BC; therefore BK is equal † to BC.

† 30 Def. + Constr. † 1 Ax.

* 11. 1. † 3. 1. * 31. 1.

* 5. 1.

* 32. 1.

* 29. 1.

6. 1.

* 29. 1.

* 6. 1.

. 47. 1.

† 47. 1.

* 34. 1.

PROPOSITION IX.

THEOR.-If a straight line be divided into two equal, and also into two unequal parts, the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

Let the straight line AB be divided into two equal parts at the point C, and into two unequal parts at the point D: the squares of AD, DB, shall together be double of the squares of

AC, CD.

*

From the point C, draw✶ CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; through D, draw * DF parallel to CE; and through F, draw FG parallel to BA; and join AF. Then, because AC is equal to CE, the angle EAC is equal to the angle AEC: and because the angle ACE is a right angle, the two others AEC, EAC together make one * right angle; and they are equal to one another; therefore each of them is half of a right angle. For the same reason, each of the angles CEB, EBC is half a right angle, and therefore the whole AEB is a right angle. And because the angle GEF is half a right angle, and EGF a right angle, for it is equal to the interior and opposite angle ECB, the remaining angle EFG is half a right angle; therefore the angle GEF is equal to

E

F

A

CD B

the angle EFG, and the side EG equal to the side GF.
Again, because the angle at B is half a right angle, and FDB
a right angle, for it is equal to the interior and opposite
angle ECB, the remaining angle BFD is half a right angle;
therefore the angle at B is equal to the angle BFD, and the
side DF to the side DB. And because AC is equal to CE,
the square of AC is equal to the square of CE; therefore the
squares of AC, CE are double of the square of AC: but the
square
of AE is equal to the squares of AC, CE, because ACE
is a right angle; therefore the square of AE is double of the
square of AC: again, because EG is equal to GF, the square
of EG is equal to the square of GF; therefore the squares of
EG, GF are double of the square of GF: but the square of EF
is equal to the squares of EG, GF; therefore the square of
EF is double of the square of GF; and GF is equal
* to CD;
therefore the square of EF is double of the square of CD: but
the square of AE is likewise double of the square of AC;

therefore the squares of AE, EF are double of the squares of AC, CD: but the square of AF is equal to the squares of AE, EF, because AEF is a right angle; therefore the square of AF is double of the squares of AC, CD: but the squares of AD, DF are equal to the square of AF, because the angle ADF is a right angle; therefore the squares of AD, DF are double of the squares of AC, CD: and DF is equal to DB; therefore the squares of AD, DB are double of the squares of AC, CD. If, therefore, a straight line, &c. Q. E. D.

47. 1.

PROPOSITION X.

THEOR.-If a straight line be bisected, and produced to any point, the square of the whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square of the line made up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the point D: the squares of AD, DB shall be double of the squares of AC, CD.

11. 1. † 3.1.

31. 1.

From the point C draw CE at right angles to AB, and make it equal to AC or CB, and join AE, EB; through E, draw* EF parallel to AB; and through D, draw DF parallel to CE. And because the straight line EF meets the parallels EC, FD, the angles CEF, EFD are equal to two right angles; 29. 1. and therefore the angles BEF, EFD are less than two right angles but straight lines, which with another straight line make the interior angles upon the same side less than two right angles, will meet if produced far enough; therefore 12 Ax. EB, FD will meet, if produced, towards B, D: let them meet in G, and join AG. Then, because AC is equal to CE, the angle CEA is equal to the angle EAC; and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a right angle. For the same reason, each of the angles CEB, EBC is half a right angle; therefore AEB is a right angle. And because EBC is half

5. 1.

32. 1.

B

A

D

G

a right angle, DBG is also* half a right angle, for they are 15. 1. vertically opposite: but BDG is a right angle, because it is equal to the alternate angle DCE; therefore the remaining

• 29. 1.

angle DGB is half a right angle, and is therefore equal to
the angle DBG; wherefore also the side BD is equal to the
side DG. Again, because EGF is half a right angle, and that the
angle at F is a right angle, because it is equal to the op-
posite angle ECD, the remaining angle FEG is half a right
angle, and therefore equal to the angle EGF; wherefore also
the side GF is equal to the side FE. And because EC is
equal to CA, the square of EC is equal to the square of CA ;
therefore the squares of EC, CA are double of the square of
CA; but the square of EA is equal to the squares of EC,
CA; therefore the square of EA is double of the square of
AC: again, because GF is equal to FE, the square of GF is
equal to the square of FE; and therefore the squares of GF, FE
are double of the square of EF: but the square of EG is
equal to the squares of GF, FE; therefore the square of EG
is double of the square of EF; and EF is equal † to CD;
wherefore the square of EG is double of the square of CD:
but it was demonstrated, that the square of EA is double of
the
square of AC; therefore the squares of AE, EG are double
of the squares of AC, CD: but the square of AG is equal * to
the squares of AE, EG; therefore the square of AG is double
of the squares of AC, CD: but the squares of AD, GD are
equal to the square of AG; therefore the squares of AD, DG
are double of the squares of AC, CD: but DG is equal to DB;
therefore the squares of AD, DB are double of the squares of
AC, CD. Wherefore, if a straight line, &c. Q. E. D.

*

*

PRO POSI TION XI.

PROBLEM. To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.

Let AB be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.

Upon AB, describe the square ABDC; bisect* AC in E, and join BE; produce CA to F, and make* EF equal to EB; and upon AF describe *the square FGHA: AB shall be divided in H, so that the rectangle AB, BH is equal to the square of AH.