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angle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D.

PROP. III. THEOR.

* 46. 1.

A C

B

If a straight line be divided into any two parts, the rect

angle 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.

Let the straight line AB be divided into any two parts in the point C: the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square of BC.

Upon BC describe * the square CDEB, and produce ED to F, and through A draw* AF parallel to CD or BE. Then the rectangle AE is equal to the rectangles Al), CE: but AE is the rectangle contained by AB,

BC, for it is contained by AB, BE, of which BE is + 30 Def. equal+ to BC; and AD is contained by AC, CB, for

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, &c. Q. E. D.

* 31. 1.

F

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 shall be equal to the squares of AC, CB, together with twice the rectangle contained by AC, CB.

Upon AB describe * the square ADEB, and join BD, and through C draw* CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And 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 ADB is equal* to the angle ABD, because BA is equal to AD t, being sides of a

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+1 Ax.

* 6. 1.

G

K

34. 1. +1 Ax.

F E

square; wherefore the angle CGB is equal to the angle CBG; and therefore

A C B the side BC is equal* to the side CG: but CB is equal * also to GK, and CG II to BK; wherefore the figure CGKB ist equilateral: it is likewise rectangular; D for, since CG is parallel to BK, and CB meets them, therefore the angles KBC, GCB are equal+ to two right angles: but KBC is at right angle; + 29. 1. wherefore GCB is at right angle: and therefore also 1 30 Def. the angles * CGK, GKB, opposite to these, are right . 34. 1. angles; and therefore CGKB is rectangular: but it is and 1 Ax. also equilateral, as was demonstrated; wherefore it is at square, and it is upon the side CB: for the same + 30 Def. reason HF also is a square, and it is upon the side HG, which is equal + to AC: therefore HF, CK are the + 34. 1. squares of AC, CB: and because the complement AG is equal to the complement GE, and that AG is the • 43. 1. rectangle contained by AC, CB, for GC is equal + to + 30 Def. CB; therefore GE is also equal + to the rectangle AC, † 1 Ax. 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, CB, and to 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 equalt to the squares of AC, + 1 Ax. CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c. Q. E. D.

Cor. From the demonstration, it is manifest, that parallelograms about the diameter of a square are like

wise squares.

PROP. V. THEOR.
If a straight line be divided into two equal paris 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 the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D: the rectangle AD, DB, together with the square of CD, shall be equal to the square of CB.

Upon CB describe* the square CEFB, join BE, and * 46. 1. through D draw * DHG parallel to CE or BF; and * 31. 1. through H draw KLM parallel to CB or EF; and also

+ 2 Ax.

A

L

H н

K

through A draw AK parallel to CL or BM. And be* 43. 1.

cause the complement CH is equal * to the complement HF, to each of these add DM; therefore the whole CM

is equalt to the whole DF: but CM is equal* to AL, * 36. 1. + Hyp.

because AC is equalt to CB; therefore also AL is equalt +1 Ax.

to DF: to each of these add CH, 2 Ax. and the whole AH is equal+ to DF

D B and CH : but AH is the rectangle

M contained by AD, DB, for DH is * Cor. 4. 2. equal * to DB; and DF together

E G F & 30 Def. with CH is the

gnomon

CMG; +1 Ax.

therefore the gnomon CMG is equal to the rectangle * Cor. 4. 2. AD, DB: to each of these add LG, which is equal * to and 34. 1. the square of CD; therefore the gnomon CMG, toge+ 2 Ax. ther with LG is equalt to the rectangle AD, DB toge

ther 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.

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

Let the straight line AB be bisected in C, and produced to the point D: the rectangle AD, DB, together with the square of

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

Upon CD describe * the square CEFD, join DE, and through B draw * BHG parallel to CE or DF, and through H draw KLM parallel to AD or EF, and also through A draw AK pa

rallel to CL or DM. And because + Hyp. AC is equal + to CB, the rectangle

H
AL' is equal * to CH; but CH is

equal * to HF; therefore also AL +1 Ax. is equal + to HF: to each of these

add CM; therefore the whole AM

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# 46. 1.

* 31. 1.

A

с

B D

L

K

M M

* 36. 1.

3. 14.

E

G F

is equal + to the gnomon CMG: but ĀM is the rect- † 2 Ax. angle contained by AD, DB, for DM is equal * to DB: * Cor. 4. 2. therefore the gnomon CMG is equal + to the rectangle & 30 Def. AD, DB: add to each of these LG, which is equal+ to + Cor. 4.2. the square of CB; therefore the rectangle AD, DB, to- and 34. 1. gether with the square of CB, is equal to the gnomon + 2 Ax. 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. Wherefore, if a straight line, &c.

Q. E. D.

PROP. VII. THEOR.

A

с

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.

Let the straight line AB be divided into any two parts in the point C: the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square

of AC. Upon AB describe * the square ADEB, and con- * 46. 1. struct the figure as in the preceding propositions. And because AG is equal* to GE, add to each of them CK; * 43. 1. therefore the whole AK is equal to the whole CE; therefore AK, CE, are double of AK: but AK, CE, are the goomon AKF, to

-В gether with the square CK; therefore the gnomon AKF, together with the H н

K square CK, is + double of AK: but twice

+ 1 Ax. the rectangle AB, BC is double of AK, D' for BK is equal * to BC; therefore the gnomon AKF, together with the square CK, is equal + to twice the rectangle AB, BC: to each of these equals add HF, which is equalt to the square of AC: there- Cor. 4. 2. fore the gnomon AKF, together with the squares CK, & 34. 1. HF, is equal to twice the rectangle, AB, BC, and the † 2 Ax. square of AC: but the gnomon AKF, together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares of AB 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.

E

* Cor. 4. 2. & 30 Def.

71 Ax.

Q. E. D.

PROP. VIII. THEOR.

* 34. 1.

* 43. 1.

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

Let the straight line AB be divided into any two parts in the point C: four times the rectangle AB, BC, together with the square of AC, shall be equal to the square of the straight line made up of AB and BC

together. + 2 Post. Produce + AB to D, so that BD be equal + to CB, + 3. 1. and upon AD describe t the square AEFD; and con† 46.1.

struct two figures such as in the preceding. Because

CB is equal to BD, and that CB is equal * to GK, + 1 Ax.

and BD to 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 36.1. equal * to BN, and GR to RN: but CK is equal * to

RN, because they are the complements of the parallel+ 1 Ax. ogram CO; therefore also BN is equal to GR: there

fore the four rectangles BN, CK, GR, RN, are equal to one another, and so are quadruple of one of them CK.

Again, because CB is equal to BD, and * Cor. 4. 2. that BD is equal * to BK, that is t, to А & 30 Def. CG, and CB equal + to GK, that is to + 34. 1.

to GP; therefore CG is equal + to GP: + Cor. 4.2. and because CG is equal to GP, and & 30 Def. PR to RO, the rectangle AG is equal + † 56. 1.

to MP, and PL to RF: but MP is
equal * to PL, because they are the complements of
the parallelogram ML; wherefore AG is equal + also
to RF: therefore the four rectangles AG, MP, PL,
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 contain the gno-
mon AOH, are quadruple of AK: and because AK is
the rectangle contained by AB, BC, for BK is equal to
BC; therefore four times the rectangle AB, BC is
quadruple of AK: but the gnomon AOH was demon-
strated to be quadruple of AK; therefore four times

CBD

GKAN

+ 34. 1.

+1 Ax.

HLF

i* 13. 1.

+1 Ax.

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