AD DB, together with the square on CD. But the gnomon CMG and LG make up the whole figures CEFB, which is the square on CB; therefore the rectangle AD DB, together with the square on CD, is equal to the square on CB.

COR.-From this proposition, it is manifest that the difference of the squares of two unequal lines BC, CD, is equal to the rectangle contained by their sum and difference.

For BC2 AD DB + CD2,

and taking CD2 from both

these equals, there remains BC2 - CD2 = AD· DB = (BC + CD) (BC - CD), for BC+CD = AC + CD AD, and BC-CD = DB.

=

PROPOSITION VI. THEOREM.

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 on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced.

Given the straight line AB, bisected in C, and produced to the point D; to prove that the rectangle AD DB, together with the square on CB, is equal to the square on CD.

join

(Const.) Upon CD describe the square CEFD (I. 46); DE, and through B draw BHG parallel to CE or DF (I. 31), and through H draw KLM parallel to AD or EF, and

also through A draw AK parallel to CL
or DM. (Dem.) And because AC is A
equal to CB, the rectangle AL is
equal to CH (I. 36);

equal to HF (I. 43);

C

B D

but CH is

K

L

H IM

therefore also

To each of

AL is equal to HF.

these add CM;

therefore the whole

AD and DB, for DM is equal to DB (II. 4, Cor.); the gnomon CMG is equal to the rectangle AD⚫ DB; each of these LG, which is equal to the square on CB; fore the rectangle AD DB, together with the square on CB, is equal to the gnomon CMG, together with LG. But the gnomon CMG, together with LG, makes up the whole figure CEFD, which is the square on CD; therefore the rectangle AD DB, together with the square on CB, is equal to the square on CD. Schol. The Corollary to Proposition 5 follows in a similar manner from this one. These two propositions are equivalent to the following: If a line be bisected, and be also cut unequally, either internally or externally, the rectangle under the unequal segments is equal to the difference between the squares on half the line and the mean distance.

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

Given the straight line AB, divided into any_two_parts in the point C; to prove that the squares on AB and BC are

equal to twice the rectangle AB · BC, together with the square on AC.

and

(Dem.)

add to each of them

(Const.) Upon AB describe the square ADEB (I. 46),
construct the figure as in the preceding propositions.
And because AG is equal to GE (I. 43),
CK; the whole AK is therefore equal to
the whole CE; therefore AK and CE are
double of AK. But AK and CE are the

gnomon AKF, together with the square CK; H
therefore the gnomon AKF, together with
But twice

the square CK, is double of AK.

A

с B

K

the rectangle AB BC is also double of AK, for BK is equal to BC (II. 4, Cor.); therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB⚫ BC. of these equals add HF, which is equal to the square on AC;

D

F

E

To each

therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB BC, and the square on AC. Now, the gnomon AKF, together with the squares CK, HF, makes up the whole figure ADEB and CK, which are the squares on AB and BC; therefore the squares on AB and BC are equal to twice the rectangle AB BC, together with the square on AC.

с

Otherwise because the square on AB is equal to the squares on AC and CB, together with twice the rectangle AC CB; if the square on CB be added to both, the squares on AB and CB are equal to the square on AC, together with twice the square of CB, and twice the rectangle AC CB. But the square on CB, together with the rectangle AC CB, is equal to the rectangle AB⚫ BC (II. 3); and therefore twice the square on CB, together with twice the rectangle AC CB, is equal to twice the rectangle AB BC.

A

B

Wherefore the squares on AB and CB are equal to twice the rectangle AB BC, together with the square on AC.

COR. 1.-Hence, the sum of the squares on any two lines is equal to twice the rectangle contained by the lines, together with the square on the difference of the lines.

COR. 2.-The square on the difference of two lines is less than the sum of their squares, by twice their rectangle.

For if AB be one line and BC another,

AC is their differ

ence, and it was proved in the proposition that twice the rectangle AB BC, together with the square on AC, is equal to the squares on AB and BC; take twice the rectangle ABBC from each of these equals, and there remains the square on AC equal to the squares on AB and BC, diminished by twice the rectangle AB⚫ BC.

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 on the other part, is equal to the square on the straight line which is made up of the whole and the first mentioned part.

Given the straight line AB, divided into any two parts in the point C; to prove that four times the rectangle AB BC, together with the square on AC, is equal to the square on the straight line, which is made up of AB and BC together.

(Const.) Produce AB to D, so that BD be equal to CB, and upon AD describe the square AEFD; and construct two

figures such as in the preceding.

(Dem.)

Because CB is equal to BD, and that
CB is equal to GK (I. 34),

and BD to

GK

M

N

P R

X

E

HL F

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 (I. 36), and GR to RN. But CK is equal to RN (I. 43), because they are the complements of the parallelogram CO; therefore, also, BN is equal to GR; and the four rectangles BN, CK, GR, and RN are therefore equal to one another, and so are quadruple of one of them, CK. Again, because CB is equal to BD, and BD equal to BK (II. 4, Cor.), that is, to CG, CG is equal to CB; and CB is 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 (I. 43), because they are the complements of the parallelogram ML; wherefore AG is equal

also to RF; therefore the four rectangles AG, MP, PL, and 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 make up the gnomon AOH, are quadruple of AK. And because AK is the rectangle contained by AB and BC, for BK is equal to BC, 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 on AC (II. 4, Cor.); therefore four times the rectangle AB BC, together with the square on AC, is equal to the gnomon AOH and the square XH.

But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD;. therefore four times the rectangle AB BC, together with the square on AC, is equal to the square on AD; that is, on the line made up of AB and BC added together.

COR. 1.-Hence, because AD is the sum, and AC the difference of the lines AB and BC, four times the rectangle contained by any two lines, together with the square on their difference, is equal to the square on the sum of the lines. COR. 2.-From the demonstration, it is manifest that, since the square on CD is quadruple of the square on CB, the square on any line is quadruple of the square on half that line.

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

Given the straight line AB, divided at the point C into two equal, and at D into two unequal parts; to prove that the squares on AD and DB are together double of the squares on AC and CD.

the

and join EA,

E

(Const.) From the point C draw CE at right angles to AB (I. 11), and make it equal to AC or CB, EB; through D draw DF parallel to CE (I. 31), and through F draw FG parallel to AB; and join AF. (Dem.) Then, because AC is equal to CE, angle EAC is equal to the angle AEC (I. 5); and because the angle ACE is a right angle, the two others, AEC, EAC, together make one right angle (I. 32); and they are equal to one another; half of a right angle. For the same CEB, EBC, is half a right angle;

A

C D B

therefore each of them is reason, each of the angles and therefore the whole

for it is equal to the

AEB is a right angle. And because the angle GEF is half a right angle, and EGF a right angle, interior and opposite angle ECB (I. 29), the remaining angle EFG is half a right angle (I. 32); therefore the angle GEF is equal to the angle EFG, and the side EG equal to the side GF (I. 6). Again, because the angle at B is half a right angle,

BFD,

therefore

Again,

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 (I. 32); therefore the angle at B is equal to the angle and the side DF to the side DB. Now, because AC is equal to CE, the square on AC is equal to the square on CE; therefore the squares on AC and CE are double of the square on AC. But the square on EA is equal to the squares on AC and CE, because ACE is a right angle (I. 47); the square on EA is double of the square on AC. because EG is equal to GF, the square on EG is equal to the square on GF; therefore the squares on EG and GF are double of the square on GF; but the square on EF is equal to the squares on EG and GF; therefore the square on EF is double of the square on GF; and GF is equal to CD (I. 34); fore the square on EF is double of the square on CD. the square on AE is likewise double of the square on AC; therefore the squares on AE and EF are double of the squares on AC and CD. And the square on AF is equal to the squares on AE and EF (I. 47), because AEF is a right angle; therefore the square on AF is double of the squares on AC and CD. But the squares on AD and DF are equal to the square on AF, because the angle ADF is a right angle; therefore, the squares on AD and DF are double of the squares on AC and CD. And DF is equal to DB; therefore the squares on AD and DB are double of the squares on AC and CD.

there

But

Otherwise:

Consider AC as one line, and CD as another,

then AD is

equal to their sum,

and DB is equal to their difference;

and

(A) + (B)

= 4AC CD

(A) — (B)

then by adding the two lines above, the result is

(1) AD2 + DB2 = 2AC2 + 2CD2 which proves the proposition. Subtracting (B) from (A), (2) AD2 - DB2

COR. 1.-From (1). The square on the sum, together with the square on the difference of two lines, is equal to twice the sum of the squares on the lines.

COR. 2.-From (2). The square on the sum of two lines diminished by the square on the difference of the lines, is equal to four times the rectangle contained by the lines.

E