GEOMETRY.-BOOK II.

DEFINITIONS.

I. Every right-angled parallelogram is a rectangle, and is contained by any two of the straight lines which contain one of the right angles.

A E

F

D

II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, 11 is called a gnomon. Thus the parallelogram HG, together with the complements AF, FC, is the gnomon AGK, or EHC.

B G

K

C

I.—If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two straight lines; of which BC is divided into any parts in the points D, E. Then the rectangle contained by A and BC, is equal to that contained by A and BD, together with that contained by A and DE and by A and EC.

B

DEC

KLH

A

From B, draw BF at right angles to BC, and make BG equal to A; through G draw GH parallel to BC, G and through D, E, C, draw DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH. But BH is contained by A and BC, for GB is equal tʊ A: and the rectangle BK is contained by A, BD, for GB is equal to A; also DL is contained by A, DE, because DK, that is, BG, is equal to A; and similarly the rectangle EH is contained by A, EC: therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE, and by A, EC.

II-If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square of the whole line.

Let AB be divided into any two parts in the point C; then the rectangle contained by AB, BC, together with that contained by AB, AC, is equal to the square of AB.

A

C B

FE

Upon AB describe the square ADEB, and through C draw CF parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; and AF is the rectangle contained by BA, D AC; for DA is equal to AB: and CE is contained by AB, BC, for BE is equal to AB: therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. III.-If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

A C

Let AB be divided into any two parts in the point C; then the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. Upon BC describe the square CDEB, and produce ED to F; through A draw AF parallel to CD or BE: then the rectangle AE is equal to the rectangles AD, CE; but AE is the rectangle contained by AB, BC, for BE is equal to BC: and AD is contained by AC, CB, for CD is equal to to CB; and BD is the square of BC: therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC.

F D

IV.--If a straight line be divided into any two parts,

the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the marts.

B

Let AB be divided into any two parts in C; then the square of AB shall be equal to the squares of AC and CB, together with twice the rectangle contained by AC, CB.

A

CB

G

K

D

FE

Upon AB describe ADEB, join BD, and through C draw CGF parallel to AD or BE, and through G draw HGK parallel to AB or DE; then, because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal to the interior and opposite angle ADB; but the angle ADB is equal to the angle ABD, because BA is equal to AD; therefore the angle CGB is equal to CBG, and the side BC to the side CG; but CB is equal also to GK, and CG to BK; wherefore the figure CGKB is equilateral. It is also rectangular, for, since CG is parallel to BK, and CB meets them, therefore the angles KBC, GCB are equal to two right angles; but the angle KBC is a right angle, therefore GCB is a right angle: and therefore also the angles CGK, GKB, opposite to these, are right angles; consequently CGKB is rectangular: wherefore it is a square, and it is upon the side CB. For the same reason HF is a square, and it is upon the side HG, which is equal to AC. Therefore the figures HF, CK, are the squares of AC, CB. And because the complement AG is equal 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, GE are equal to twice the rectangle AC, CB; and HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE, are equal to the squares of AC, and twice the rectangle AC, CB: but HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB; therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB.

CB,

COR. The parallelograms about the diameter of a square, are likewise squares.

V.-If a straight line be divided into two equal, 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, is equal to the square of half the line.

Let AB be divided into two equal parts in the point C, and into two unequal parts in the point D: then the rectangle AD, DB, together with the square of CD, shall be equal to the square of CB.

C DR

L H M

E GF

Upon CB describe the square CEFB, join BE, through D draw DHG parallel to CE or BF; through H draw KLM parallel to CB or EF; also through A draw AK parallel to CL or BM. The complement CH is equal to the complement HF; to each add DM; therefore the whole CM is equal to the whole DF; but because AC is equal to CB, therefore CM is equal to AL, also AL is equal to DF; to each add CH, and therefore the whole AH is equal to DF and CH. But AH is the rectangle contained by AD, DB, for DH is equal to DB; and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB: to each add LG, which is equal to the square of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD: but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB; therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB.

COR. The difference of the squares of two unequa1 lines AC, CD, is equal to the rectangle contained by their sum AD and their difference DB.

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

A

с

DB

L

H M

Let AB be bisected in C, and produced to D; then the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Upon CD describe the square CEFD, join DE, through B draw K BHG parallel to CE or DF, through H draw KLM parallel to AD or EF, and through A draw AK par

E

G F

allel to CL or DM. AC is equal to CB, therefore the rectangle AL is equal to the rectangle CH, but CH is equal to HF; therefore AL is equal to HF; to each add CM; therefore the whole AM is equal to the gnomon CMG; but AM is the rectangle contained by AD, DB, for DM is equal to DB: therefore the gnomon CMG is equal to the rectangle AD, DB: add to each 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 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.

VII.-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.

H

A

Then

CB

G

K

Let AB be divided into any two parts in C. the squares of AB, BC, are equal to twice the rectangle AB, BC, together with the square of AC. Upon AB describe the square ADEB, join BD; through C draw CF parallel to AD or BE cutting BD in G, and through G draw HGK parallel to AB or DE. AG is equal to GE, add to each CK; therefore AK is equal to CE, and AK, CE, are double of AK, but AK, CE are the gnomon AKF and the square CK; therefore the gnomon AKF and the square CK is double cf AK: but twice the rectangle AB, BC, is

D

FE