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DEFINITIONS.

I.

EVERY right-angled parallelogram is called a rectangle, and is said to be contained by any two of the straight lines which form one of its right angles.

II.

In every parallelogram, any of the parallelograms about a diagonal, together with the two complements, is called a gnomon.

Thus the parallelogram H G together with the
complements AF, FC, is a gnomon, which is H
more briefly expressed by the letters A GK,
or EHC, at the opposite angles of the paral-
lelograms which make up the gnomon.

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PROP. I. (THEOREM.)—If there be two straight lines (A and BC), one of which (BC) is divided into any number of parts at the points (D and E); the rectangle contained by the two straight lines (A and BC), is equal to the rectangle contained by the undivided line (▲), and the several parts of the divided line (BD, DE, and EC).

B

Make

D

EC

From the point B draw BF at right angles (I. 11) to BC.
BG equal (I. 3) to A. Through G draw GH
parallel (I. 31) to BC; and through D, E, and
C, draw DK, EL, and CH parallel to BG.
Then BH, BK, DL and EH, are rectangles (I.
Def. 36).

G

K LH

A

The rectangle BH is equal (II. Ax. 1) to the rectangles BK, DL, and EH. But the rectangle BH is contained by the straight lines A and BC, because GB is equal to A. The rectangle BK is contained by the straight lines A and BD, because GB is equal to A. The rectangle DL is contained by the straight lines A and DE, because DK is equal to BG (I. 34) and BG is equal to A. In like manner, it is proved that the rectangle EH is contained by the straight lines A and EC. Therefore the rectangle contained by the straight lines A and BC, is equal to the several rectangles contained by the straight lines A and BD, A and DE, and A and E C. Wherefore, if there be two straight lines, &c. Q. E.D.

PROP. II. (THEOREM.)—If a straight line (AB) be divided into any two parts (AC and CB), the rectangles contained by the whole (AB) and each of its parts (AC and CB), are together equal to the square of the whole line (AB).

Upon A B describe (I. 46) the square A E. Through C draw CF. parallel (I. 31) to AD or BE.

A

C B

The square AE is equal (II. Ax. 1) to the rectangles AF and CE. But AE is the square of AB. The rectangle AF is contained by BA, AC; because DA is equal to AB. The rectangle CE is contained by AB, BC, because BE is equal to AB. Therefore the rectangles contained by AB, AC and AB, BC are equal (Ax. 1) to the square of AB. Wherefore if a straight line, &c. Q. E.D.

A

TE

PROP. III. (THEOREM.)-If a straight line (AB) be divided into any two parts (AC and CB), the rectangle (AB. BC) contained by the whole and one of the parts is equal to the rectangle (AC. CB) contained by the two parts, together with the square of (BC) the foresaid part. Upon BC describe the square CE (I. 46). Through A draw AF parallel (I. 31) to CD or BE.

The rectangle AE is equal (II. Ax. 1) to the rectangles AD and CE. But AE is the rectangle contained by AB. BC, because BE is equal (Const.) to BC. The rectangle AD is contained by AC, CB, because CD is equal to CB. And DB is the square of BC. Therefore the rectangle AB.BC is equal to the rectangle AC.CB, together with the square of BC. If therefore a straight line be divided, &c. Q. E. D.

Produce ED to F.

A C

D

B

PROP. IV. (THEOREM.)--If a straight line (A B) be divided into any two parts (at C), the square of the whole line (A B) is equal to the squares of the two parts (AC,CB) together with twice the rectangle contained by the parts (AC, CB).

Upon AB describe the square AE (I. 46). Join BD. Through C draw CGF parallel to AD or BE (I. 31), and through G draw HGK parallel to AB or DE.

A

D

C B

F E

K

Because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (I. 29) to the interior and opposite angle ADB. But the H angle ADB is equal to the angle ABD (I. 5), because BA is equal to AD, being sides of a square. Therefore the angle CGB is equal (I. Ax. 1) to the angle CBG, and the side BC (I. 6) to the side CG But CB is equal (I. 34) also to GK, and CG to BK. Therefore the figure CK is equilateral. Again, because CG is parallel to BK, and CB meets them, the angles KBC, GCB are equal (I. 29) to two right angles. But the angle KBC is a right angle (Const.) Therefore GCB is a right angle. And the angles CGK, GKB, opposite to these, are also (I. 34) right angles. Therefore CK is rectangular. But it is also equilateral. Therefore it is a square, and it is described upon the side CB. For the same reason HF is a square, and it is described upon the side HG, which is (I.34) equal to AC. Therefore the figures HF and CK are the squares of AC and CB. Because the complement A Gis 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 AC. 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. From the demonstration, it is manifest, that the parallelograms about the diameter of a square, are likewise squares.

PROP. V. (THEOREM.)—If a straight line (AB) be divided into two equal parts (at C), and also into two unequal parts (at D); the rectangle contained by the uneqal parts (AD, DB), together with the square of the line between the points of section (CD), is equal to the square of half the line (CB).

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

K

Join BE. Though D and through H draw

A.

C D B

+

I

H

IT.

E

G F

The complement CH is equal (I. 43) to the complement HF. To each of these 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 DD. 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 GMG 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.

PROP. VI. (THEOREM.)-If a straight line (AB) be bisected (at C), and produced to any point (D), the rectangle (AD. DB) contained by the whole line thus produced, and the part of it produced, together with the square of (CB) half the line bisected, is equal to the square of the straight line (CD) which is made up of the half and the part produced.

Upon CD describe the square CF (I. 46). Join DE. Through B draw BHG parallel to CE or DF (I. 31); and through H draw KLM parallel to AD or EF. Also, through A draw AK parallel to CL or DM.

K

A C

B D

+

H

M

G F

Because AC is equal to CB, the rectangle AL is equal (I. 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 A M 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. DB, together with the square of CB, is equal to the square of CD. Wherefore a straight line, &c. Q. E. D.

PROP. VII. (THEOREM.)—If a straight line (AB) be divided into any two parts (at C), the squares of the whole line (AB), and of one of the parts (BC) are equal to twice the rectangle (AB, BC) contained by the whole and that part, together with the square of the other part (AC).

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

A.

D

C B

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E

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 E double of AK. But AK and CE are the gnomon AKF and the square CK. Therefore the gnomon AKF and the square CK are together double of AK. But twice the rectangle AB. BC, is double of AK, for BK is equal (II. 4. Cor.) to BC. Therefore the gnomon AKF and the square CK, are together 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, and the squares CK and HF, are equal to twice the rectangle AB. BC, and the square of AC. But the gnomon AKF, and the squares CK and HF, make up the squares AE and CK, which are the squares of A B 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.

PROP. VIII. (THEOREM.)-If a straight line (AB) be divided into any two parts (at C), four times the rectangle (AB, BC) contained by the whole line, and one of the parts, together with the square of the other part (AC), is equal to the square of the straight line (AD), which is made up of the whole and that part (BC).

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

GK

N

P

E

HLF

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 KO 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 AG; therefore the gnomon AOH is four times the rectangle AK; that is, four times the rectangle AR. BC, because BK is equal to BC. To these equals add the square of

EGF

AC, or its equal the square XH (I. 34). Therefore four times the rectangle AB. BC and the square of AC, are together equal to the gnomon AOH and the square XH; that is, to the square AF. But the square AF is the square of AD, or of AB and BC together. Therefore four times the rectangle AB. BC and the square of AC, are together equal to the square of the straight line made up of AB and BC together. Q. E. D.

PROP. IX. (THEOREM.)-If a straight line (AB) be divided into two equal parts (at C), and also into two unequal parts (at D), the squares of the two unequal parts (AD and DB) are together double of the square of half the line, and of the square of the line between the points of section AC, CD).

From the point C draw (I. 11) CE at right CE equal (I. 3) to AC, and join EA, EB. parallel (1.31) to CE, meeting EB in F. parallel to BA, and join AF.

angles to AB. Make Through D draw DF Through F draw FG

A.

E

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 (I. 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 AEB is a right angle. Because the angle GEF is half a right angle, and EFG 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 EG (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 (I. 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 BFD, 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 AC 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 AC. Again, because EG is equal to GF, the square of EG is equal to the square of GF. Therefore the squares of EG and GF are double of the square of GF. But the square of EF is equal (I. 47) to the squares of EG and GF. Therefore the square of EF is double of the square of GF. But GF is equal (I. 34) to CD. Therefore the square of EF is double of the square of CD. But it has been proved that the square of AE is double of the square of AC. Therefore the squares of AE and EF ale double of the squares of AC and CD. But the square of AF is equal (I. 47) to the squares of AE and EF. Therefore the square of AF is double of the squares of AC and CD. But the squares of AD and DF are equal (I. 47) to the square of AF. Therefore the squares of AD and DF are double of the squares of AC and CD. But DF is equal to DB. Therefore the squares of AD and DB are double of the squares of AC and CD. If, therefore, a straight line be divided, &c. Q. E. D.

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