Imágenes de páginas
PDF
EPUB

KBC is a right angle. [I. Def. 30.] Therefore GCB is a right angle. [Ax. 3.] And therefore also the angles CGK, GKB opposite to these are right angles. [I. 34. and Ax. 1.] Therefore CGKB is rectangular ; and it has been proved to be A equilateral; therefore it is a square, and it is on the side CB.

K For the same reason HF is also a square, and it is on the side HG, which is equal to AC. [I. 34.] Therefore HF, CK are the squares on AC, CB. And because the complement D

F AG is equal to the complement GE, [I. 43.] and that AG is the rectangle contained by AC, CB, for CG is equal to CB; therefore GE is also equal to the rectangle AC, CB. [A&. 1] Therefore AG, GE are equal to twice the rectangle AC, CB. And HF, CK are the squares on AC, CB. Therefore the four figures HF, CK, AG, GE are equal to the squares on AC, CB, together with twice the rectangle AC, CB. But HF, CK, AG, GE make up the whole figure ADEB, which is the square on AB. Therefore the square on AB is equal to the squares on AC, CB, together with twice the rectangle AC, CB.

Therefore, if a straight line, &c. Q.E.D.

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

If AC, CB be considered as two lines, then AB is their sum and the proposition may be thus enunciated.

The square on the sum of two lines is equal to the squares on the two lines, together with twice the rectangle contained by the lines.

The corresponding proposition in arithmetic is similarly expressed by substituting numbers' for lines, and 'product' for rectangle. Symbolically :

If m=a+b; m2=a2+62 + 2ab.

The result is established by multiplication. Prop. 4 may be thus deduced from the previous propositions.

The square on AB=AB. BC+AB. AC. (prop. 2.]

and AB. BC=AC: CB#sq: on BC: }(prop. 3.] Therefore, AB. AC + AB. BC=2 rect. AC. CB+sq. on AC + sq. on BC. i.e. The sq. on AB=sq. on AC +sq. on BC +2 rect. AC. CB. Q.E.N.

If this be compared with the fig. AF and FB will be seen to be the rectangles AB. AC and AB. BC respectively; and the former is made up of AG (=AC. CB) and the sq. on AC and the latter of GE (=rect. AC. CB) and the sq. on BC. Ex. 1. Prove that HF is a square. 2. The square on a line is four times the square on half the line.

PROPOSITION 5. THEOREM. 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 on the line between the points of section, is equal to the square on half the line. 1.

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 on CD, shall be equal to the square on CB.

On CB describe A. 2.

the square CEFB; join BE

M through D draw DHG parallel to CE or BF; through H draw KLM parallel to CB or EF; and through A draw AK parallel to CL or BM. [1. 31.] 3,

Then the complement CH is equal to the com

plement HF; [I. 43.] to each of these add DM; therefore the whole CM is equal to the whole DF. [Ax. 2.) But CM is equal to AL, [I. 36.] because AC is equal to CB. [Hyp.] Therefore also AL is equal to DF. [Ax. 1.] To each of these add CH; there

[ocr errors]

H

fore the whole AH is equal to DF and CH. [Ax. 2.] But AH is the rectangle contained by AD, DB, for DH is equal to DB; [II. 4, Cor.] and DF together with CH is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB. To each of these add LG. which is equal to the square on CD. (II. 4, Cor., and I. 34.] Therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square on CD. [Ax. 2.] But the gnomon CMG and LG make up the whole figure 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.

Therefore, if a straight line, &c. Q.E.D.

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

If AC contains m units and CD n units, then AD=m+n, and BD=m-n.

But (m+n) (m n)=m2 – ma by multiplication,
Hence the corresponding proposition in arithmetic is,

The product of the sum and difference of two numbers is equal to the difference of their squares.

i. Let AC. CD be considered as two independent lines, then AD is their sum and DB is their difference (for DB=CB - CD and CB=CA); and in the prop. it is proved that the rectangle AD. DB is less than the sq. on AC by the sq. on CD. i.e.

The difference of the squares on AC and CD is equal to the rectangle contained by (AC+CD) and (AC-CD).

2. Therefore if a line be divided equally and unequally the greater of the unequal segments is the sum of half the line, and the line between the points of section and the lesser segment is the difference of the same two lines.

3. Again, since the rectangle contained by the unequal segments, together with the sq. on CD, always makes the same sum, viz., the square on BC, wherever the point D may be placed, it follows that as the point D approaches the middle of the line and CD grows less, the rect. AD. DB must increase. Therefore when D is in the middle and CD vanishes, the rectangle AD. DB will have its greatest value. Ex. Having given the sum and difference of two lines it is required

to determine the lines.

PROPOSITION 6. THEOREM.

B

[ocr errors]
[ocr errors]
[ocr errors]

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.

Let the straight line AB be bisected at the point 1.

C, and produced to the point D; the rectangle AD, DB, together with the square on CB, shall be equal to the square on CD.

On CD describe 2. the

square CEFD; join DE; through B draw BHG parallel to CE or DF; K through H draw KLM parallel to AD or EF; and through A draw, AK parallel to CL or DM. I. 31.]

E

с F Then because AC is equal to CB, [Hyp.] the

rectangle AL is equal to the rectangle CH; [I. 36.] but CH is equal to HF; [I. 43.] therefore also AL is equal to HF. [Ax. 1.] To each of these add CM; therefore the whole AM is equal to the gnomon CMG. [Ax. 2.] But AM is the rectangle contained by AD, DB, for DM is equal to DB. [II. 4, Cor.] Therefore the rectangle AD, DB is equal to the gnomon GMG. [Ax. 1.] To each of these add LG, which is equal to the square on CB. (II. 4, Cor., and I. 34.] Therefore the rectangle AD, DB, together with the square on 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 on CD. Therefore the rectangle AD, DB, together

[ocr errors]

with the square on CB, is equal to the square on CD.

Therefore, if a straight line, &c.

Q.E.D.

Prop. 6. Let AC. CD be considered as two independent lines, then AD is their sum and DB is their difference, and the proposition proves that the rect. AD. DB is less than the sq. on CD by the sq. on AC., i.e.

The difference of the sqs. on AC and CD is equal to the rectangle contained by (AC+CD) and (CD-CA).

This is the same result as was obtained by the last proposition, and the two may be combined in one statement, thus,

The rectangle contained by the sum and difference of two lines is equal to the difference of the squares on the lines.

PROPOSITION 7. THEOREM.

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.

Let the straight line AB be divided into any 1.

two parts at the point C: the squares on AB, BC shall be equal to twice the rectangle AB, BC, together with the square on AC.

On AB describe the A 2.

B square ADEB, and construct the figure as in the preceding propositions.

Then AG is equal to 3. QE; (I. 43.

) to cach of these add CK; therefore the whole AK is equal to the whole CE; therefore AK, CE are double of AK. But AK, CE are the gnomon AKF, to

E gether with the square CK; therefore the gnomon AKF, together with the square CK, is double of AK. But twice the rectangle AB, BC is double of AK, for

[ocr errors]

F

« AnteriorContinuar »