AG is equal (I. 43) to the complement GE; and AG is the rectangle contained by AC. CB, for GC is equal to CB. Therefore GE is also equal to the rectangle AC.CB. Wherefore AG and GE are equal to twice the rectangle A C.CB. Also HF and CK are the squares of AC and CB. Therefore the four figures HF, CK, AG and GE, are equal to the squares of AC and CB, with twice the rectangle AC.CB. But the figures HF, CK, AG and GE make up the whole figure AE, which is the square of AB. Therefore the square of AB is equal to the squares of AC and CB, and twice the rectangle AC. CB. Wherefore, if a straight line be divided, &c. Q. E. D.

COR. 1.-From the demonstration, it is manifest, that the parallelograms about the diameter of a square, are likewise squares.

Corollary 2.—The square of any straight line is equal to four times the square of half the straight line.

The demonstration of the preceding proposition might have been shortened by the application of some corollaries to the propositions in Book I. It will be a useful exercise for the student to discover this abridgment himself.

The following demonstration is founded on the preceding propositions of this book. Because AB is divided into any two parts at C, the square of AB is equal (II. 2) to the two rectangles AB.AC, and AB.BC. But because AB is divided into any two parts at C, the rectangle AB. AC is equal (II. 3) to the rectangle AC.CB, and the square of AC; also the rectangle AB.BC is equal to the rectangle AC.CB, and the square of CB. Adding these equals, the rectangles AB.AC and AB.BC, are together equal to twice the rectangle AC.CB, and the squares of AC and CB. But it has been proved that the square of AB is equal to the same two rectangles. Therefore the square of AB is equal (I. Ax. 1) to the squares of AC and CB, and twice the rectangle AC.CB.

If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section (or division), is equal to the square of half the line.

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 rectangle AD,DB, together with the square of CD, is equal to the square of CB.

Upon CB describe (I. 46) the square CF. Join BE. Through D draw DHG parallel to CE K or BF (I. 31); and through H draw KLM parallel to CB or EF. Also through A draw AK parallel to CL or BM.

A

I

C D B

E. G R

M

Because the complement CH is equal (I. 43) to the complement HF. To each of them, add DM. Therefore the whole CM is equal to the whole DF. But CM is equal to AL (I. 36) because AC is equal to CB. Therefore AL is equal to DF. To each of these equals, add CH. Therefore the whole AH is equal to DF and CH; that is, to the gnomon CMG. But AH is the rectangle AD.DB, for DH is equal to DB. Therefore the gnomon CMG is equal to the rectangle AD.DB. To each of these equals add LG, which is equal (II. 4. Cor.) to the square of CD. Therefore the gnomon CMG, together with LG,

is equal to the rectangle AD.BD, together with the square of CD. But the gnomon CMG and LG make up the whole figure CF, which is the square of CB. Therefore the rectangle AD.DB, together with the square of CD, is equal to the square of CB. Wherefore, if a straight line, &c. Q. E. D.

Another Demonstration: Because CB is divided into any two parts at D, the rectangle CD.DB, and the square of DB are together equal (II. 3) to the rectangle CB.BD, or to the rectangle A C.DB; for, AC is equal (Const.) to CB. To each of these equals, add the rectangle CD.DB. Therefore twice the rectangle CD.DB, and the square of DB are together equal to the rectangles AC.DB, and CD.DB, which are equal (II. 1) to the rectangle AD. DB. Again, to each of these equals add the square of CD. Therefore the squares of CD and DB, and twice the rectangle CD.DB are together equal to the rectangle A D.D B, and the square of CD. But the squares of CD and DB and twice the rectangle CD.DB, are together equal (II. 3) to the square of CB. Therefore the rectangle A D.D B, and the square of CD, are together equal to the square of CB.

Corollary 1.-From this proposition it is manifest, that the difference of the squares of two unequal lines AC and CD, is equal to the rectangle contained by their sum A D, and their difference D B.

Corollary 2.-The rectangle contained by the segments of any straight line, is a maximum when the point of section is the middle point.

Corollary 3.-The sum of the squares of the two parts into which a straight line is divided is a minimum, when it is bisected.

Corollary 4.-The square of one of the legs of a right-angled triangle is equal to the rectangle contained by the sum and the difference of the hypotenuse and the other leg.

If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced.

Let the straight line AB be bisected at C, and produced to the point D. The rectangle AD.DB, together with the square of CB, is equal to the square of CD.

Upon CD describe the square CF (I. 46). K

Join D E. Through B draw BHG parallel to
CE or DF (L. 31); and through H draw
KLM parallel to AD or EF. Also, through
A draw AK parallel to CL or DM.

A C

B D

H

M

G K

Because AC is equal to CB, the rectangle AL is equal (1. 36) to the rectangle CH. But CH is equal (I. 43) to HF. Therefore AL is equal to HF. To each of these equals add CM. Therefore the whole AM is equal to the gnomon CMG. But AM is the rectangle AD.DB, because DM is equal (II. 4. Cor.) to DB. Therefore the gnomon CMG is equal to the rectangle AD.DB. To each of these equals add LG, which is equal to the square of CB. Therefore the rectangle AD.DB, together with the square of CB, is equal to the gnomon CMG, and LG. But the gnomon CMG and LG make up the square CF, which is the square of CD. Therefore the rectangle AD.D B, together with the square of CB, is equal to the square of CD. Wherefore a straight line, &c. Q. E. D.

N A

C

B D

Another Demonstration: Produce CA to N, making CN equal to C D. To these equals, add the equals CA and CB respectively. Therefore NB, is equal to AD. But the rectangle N B.BD and the square of CB is equal (II. 5) to the square of CD. And it has been proved that AD is equal to NB. A D.D B, and the square of CB is equal to the square of CD.

PROP. VII. THEOREM.

Therefore the rectangle

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 at the point C. The squares of AB and BC are equal to twice the rectangle AB.BC, together with the square of AC.

Upon AB describe the square AE (I. 46) and join BD. Through C, draw CF parallel to AD or BE, H cutting BD in G (I. 31), and through G draw HGK parallel to AB or DE.

A

D

C B

F E

K

Because AG is equal to GE (I. 43); to each of them add CK. Therefore the whole AK is equal to the whole CE; and AK and CE together, are double of AK. But AK and CE are the gnomon AKF and the square CK. Therefore the gnomon AK F and the square CK are together double of A K. But twice the rectangle A B.BC, is double of AK, for BK is equal (II. 4 Cor.) to B C. Therefore the gnomon AK F and the square CK, are together equal to twice the rectangle AB.B C. To each of these equals, add_HF, which is equal to the square of A C. Therefore the gnomon AKF, and the squares CK and HF, are equal to twice the rectangle AB.B C, and and the square of A C. But the gnomon AKF, and the squares CK and HF, make up the squares AE 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. Q. E. D.

Otherwise.-Because A B is divided into any two parts at C, the square of AB is equal (II. 4) to the squares of AC and CB, and twice the rectangle A C.C B. To these equals, add the square of C B. Therefore the squares of AB and C B are together equal to the square of A C, twice the square of CB, and twice the rectangle A C.C B. But the rectangle A B. B C is equal (II. 3) to the square of CB, and the rectangle A C.C B. Therefore twice the rectangle AB.B C is equal (I. Ax. 6) to twice the square of CB, and twice the rectangle AC.CB. To these equals, add the square of A C. Therefore the square of A C and twice the rectangle A B.B C are together equal (I. Ax. 2) to the square of A C, twice the square of CB and twice the rectangle AC.CB. But it was proved that the squares of AB and CB are equal to the square of AC, twice the square of CB, and twice the rectangle AC.CB. Therefore the squares of AB and CB are together equal (I. Ax. 1) to twice the rectangle A B.B C and the square of A C.

Corollary 1.-The square of the difference of two straight lines is equal to the difference between the sum of their squares and twice the rectangle contained by them.

Corollary 2.-The square of the same of two straight lines is equal to the square of their difference, together with four times the rectangle contained by them.

PROP. VIII. THEOREM.

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.

Let the straight line AB be divided into any two parts at the point C. Four times the rectangle AB.B C, together with the square of A C, are equal to the square of the straight line made up of A B and BC together.

Produce AB to D, making (I. 3) BD equal to BC. Upon AD describe (I. 46) the square AF. Join its opposite points DE. Through the points B and C, draw the straight lines (I. 31) BKL and CPH parallel to AE or DF. Through the points K and P, where they meet the diagonal, draw MGN and XPO parallel to AD or EF.

M

X

A CBD

E

GK

N

P R

HL F

Because CB is equal to BD (Const.) and also to GK (I. 34), the squares (II. 4 Cor.) GR and BN are equal. Because the sides of these squares are all equal, and all their adjacent angles are right angles (I. 13). Therefore the complements CK and KO are (I. 43) equal squares, and the four squares (I. 36) CK, BN, GR and `K O ́are all equal. Because CG, and GP, are sides of equal squares, the rectangles AG and MP are I. 36) equal. For the same reason, the rectangles PL and RF are equal. But the rectangle MP is (I. 43) equal to the rectangle PL. Therefore the four rectangles AG, MP, PL, and RF are all equal. Because the four squares CK, BN, GR, and KO are four times CK; and the four rectangles AG, MP, PL, and RF are four times A G. Therefore the gnomon AOH is four times the rectangle AK; that is, four times the rectangle AB. BC, because BK is equal to B C. To these equals, add the square of A C, or its equal the square XH (I. 34). Therefore four times the rectangle AB. BC and the square of A C, are together equal to the gnomon A OH and the square XH; that is, to the square AF. But the square AF is the square of A D, or of AB and BC together. Therefore four times the rectangle AB. BC and the square of A C, are together equal to the square of the straight line made up of A B and B C together. Q. E. D.

Dr. Thomson, in his edition, objects to this demonstration on the ground that it tacitly assumes the truth of the first proposition of Book V. This objection may be removed by the following demonstration :-Because the triangle ADE is equal (I. 34) to the triangle FDE, and the triangle XPE equal to the triangle HPE. Therefore the remaining figure A DPX is equal (I. Ax. 3) to the remaining figure FDPH; and the gnomon AOH is double (II. Ax. 2) the figure ADPX. Because CB is equal (Const.) to BD, the square GR is equal (I. 34) to the square BN, and the triangle BDK to the triangle K PR (I. Ax. 7). To these equals, add the figure A BKPX. Therefore the figure A DP X is equal to the rectangle AR. But the rectangle AR is double the rectangle AK, because BK is equal to KR. Therefore the figure ADPX is double the rectangle AK. But it has been proved that the gnomon AOH is double the figure ADP X. Therefore the gnomon AOH is four times the rectangle A K; that is, four times the rectangle A B. B C; because BK is equal to BC. The rest of this demonstration is the same as that of the preceding; viz.," To these equals add, &c"

Otherwise.-Produce AB to D making BD equal to BC. Because the square of CD is equal (II. 4 Cor. 2) to four times the square of CB; and twice the rectangle A C. CD is equal (II. Ax. 2) to four times the rectangle A C. C B. Therefore, adding these equals, the square of CD and twice the rectangle A C. CD are together equal to four times the square of CB, and four times the rectangle A C. C B. But the rectangle A B. BC is equal (II. 3) to the square of CB and the rectangle A C. CB. Therefore four times the rectangle A B. B C is equal to four times the square of C B and four times the rectangle A C. CB. But it has been proved that the square of CD, and twice the rectangle A C. CD are together equal to the same magnitudes. Therefore the square of CD and twice the rectangle A C. CD are together equal to four times the rectangle A B. BC. To these equals add the square of AC. Therefore the squares of AC and CD, and twice the rectangle AC. CD, are together equal to four times the rectangle AB. B C and the square of A C. But the squares of AC and CD, and twice the rectangle AC. CD are equal (II. 4) to the square of A D. Therefore four times the rectangle AB. BC and the square of AC, are together (I. Ax. 1) equal to the square of A D, that is, of the straight line made up of AB and B C together.

PROP. IX. THEOREM.

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 Cand into two unequal parts at the point D. The squares of AD and and DB together, are double the squares of AC and CD together.

From the point C draw (I. 11) CE at right angles to AB. Make CE equal (I. 3) to AC, and join EA, EB. Through D draw DF parallel (1. 31) to CE, meeting EB in F. Through F draw FG parallel to BA, and join AF.

A

G

E

C D B

Because AC is equal to CE, the angle EAC is equal (I. 5) to the angle AEC. Because ACE is a right angle, the two other angles AEC, EAC of the triangle AEC are together equal (1.32) to a right angle. But they are equal to one another. Therefore each of them is half a right angle. For the same reason, each of the angles CEB, EBC is half a right angle. Therefore the whole angle A E B is a right angle. Because the angle GEF is half a right angle, and E GF is a right angle, being equal (I. 29) to the interior and opposite angle ECB. Therefore the remaining angle EFG is half a right angle. Wherefore the angle GEF is equal to the angle EFG, and the side E G (I. 6) to the side GF. Again, because the angle at B is half a right angle, and FDB is a right angle, being equal (1. 29) to the interior and opposite angle ECB. Therefore the remaining angle BFD is half a right angle. Wherefore the angle at B is equal to the angle B FD, and the side DF (I. 6) to the side DB. Because AC is equal to CE, the square of AC is equal to the square of CE. Therefore the squares of AC and CE are double of the square of A C. But the square of AE is equal (I. 47) to the squares of AC and CE. Therefore the square of AE is double of the square of A C. Again, because E G is equal to GF, the square of EG is equal to the square of GF. of E G and G F are double of the square of GF. is equal (I. 47) to the squares of EG and GF.

Therefore the squares But the square of EF Therefore the square of