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EU CLI D.
I. Every right-angled parallelogram is said to be Boor II. contained by any two of the straight lines which contain one of the right angles.
II. In every parallelogram, any of the parallelograms about a diameter, together with
D the two complements, is called a Gnomon. Thus the parallelogram HG, ' together with the complements AF, FC, is the
F 'gnomon, which is more 'briefly expressed by the • letters AGK, or EHC,
C which are at the opposite angles of the parallelograms which make the gnomon.'
PROP. I. THEOR. 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 eqnal to the rectangles contained by the undivided line, and the several parts of the divided line.
a 11. 1.
Book II. Let A and BC be two straight lines; and let BC be di
o vided into any parts in the points D, E; the rectangle contained by the straight lioes A, B
D Е С
From the point B drawa G!
H н \ 3. 1. and make BG equal to A; <31. 1. and througl G drawe GH pa-F
A A rallel to BC; and through D, E, C, drawe DK, EL, CH parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A; and BK is contained by A, BD, for it is contained by GB, BD, of
which GB is equal to A; and DL is contained by A, DE, * 34. 1. because DK, that isd BG is equal to A; and in like man
ner 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 also by A, EC. Wherefore, if there be two straight lines, &c. Q. E.1).
PROP. II. THEOR.
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 the straight line AB be divided into any two parts in the point C; the A СВ rectangle contained by AB, BC, together with the rectangle* AB, AC, shall be equal to the square of AB.
Upon AB described the square b 31.1.
ADEB, and through C drawb CF,
E the rectangle contained by BA, AC; for it is contained by DA, AC of which AD is equal to AB; and CE is contain
* N. B. To avoid repeating the word contained too frequently, the rect: angle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.
* 46. 1.
ed by AB, BC, for BE is equal to AB; therefore the rect- Book II. angle contained by AB, AC, together with the rectangle AB, BC is equal to the square of AB. If therefore a straight line, &c. Q.E.D.
PROP. III. THEOR. -
Let the straight line AB be divided into two parts in the
B 10.1. CDEB, and produce ED to F, and through A draw! AF parallel to CD or BE; then the rectangle AE is equal to the rectan. gles AD, CE; and AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and F
E AD is contained by AC, CB, for CD is equal to BC; 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, &c. Q. E. D.
b 31. 1.
PROP. IV. THEOR. 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 parts.
Let the straight line AB be divided into any two parts in C; the square of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB.
Upon AB describe the square ADEB, and join BD, and - 46. 1. through C drawb CGF parallel to AD or BE, and through o 31. 1. G draw HK parallel to AB or DE: And because CF is parallel to AD, and BD falls upon them, the exterior angle
• 29. 1.
Book II. BGC is equal to the interior and opposite angle ADB ;
but ADB is equald to the angle ABD, because BA is equal
to AD, being, sides of a square;
K * 34. 1. CG: But CB is equalf also to GK,
and CG to BK; wherefore the
F E therefore equal to two right angles; and KBC is a right angle; wherefore GCB is a right angle: and therefore also the angles CGK, GKB opposite to these, are right angles, and CGKB is rectangular; but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB: For the same reason HF also is a square, and it
the side HG, which is equal to AC: Therefore HF, CK are the squares of AC, CB; and because the comple• 43. 1. ment AG is equals to the complement GE, and that AG
is the rectangle contained by AC, CB, for GC is equal to
Cor. From the demonstration, it is manifest, that the parallelograms about the diameter of a square are likewise squares.
a 36. 1.
If a straight line be divided into two equal parts,
Upon CB describea the square CEFB, join BE, and . 46. 1. through D drawb DHG parallel to CE or BF; and through S1. 1. H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM: And because the complement CH is equal to the complement HF, to each of 43. 1. these add DM; therefore the whole CM is equal to the whole DF; but Misequald
С DB to AL ; because AC is equal to CB; therefore also ALis
L H Η equal to DF. To each of K
M these add CH, and the whole AH is equal to DF and CH: But AH is the rectangle contained by AD, DB, for DH is equale to DB; and DF together with CH Cor. 4. 2. is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB: To each of these add LG, which is equale 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. Wherefore if a straight line, &c. Q. E. D.
From this proposition it is manifest, that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference.