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the rectangle AD DB, together with the square of CD, shall be equal to the square of CB.

*46. 4. #31. 1.

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Upon CB describe the square CEFB, draw BE, through D draw* DHG parallel to CE or BF; through H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM. The complement CH is equal 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, because AC is equal to CB; therefore also AL is equal to DF:§ to each of 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 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 of these add LG, which is equal to the square *Cor. of CD; therefore the gnomon CMG, together with LG, is equal to the rectangle AD BD, together 12 Ax. 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 BC. Wherefore, if a straight line, &c. Q. E. D.

*Cor. 4. 2. and 26. Def.

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Cor. From this proposition it is manifest, that the differ

The student will find it useful to remember that the construction in this and in each of the three propositions next following, always commences with describing the greatest square mentioned in the enunciation, and then completing in it the construction of the 4th prop. The remainder of the construction in each particular case will at once be suggested from the proposed points of division, since each of these is to be at the angle of a rectangle.

§ Take each of these equals from the entire figure, then the remainder CF will be equal to the remainder AH, LG, that is, to the rectangle AD DB (for DB is equal to DH) and the square of CD, since LH is equal to CD. In a similar way the demonstration of the next proposition may be a little abridged.

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

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Upon CD describe the square CEFD, join DE, *46. 1. 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 parallel to CL or DM. Then because AC is equal to CB, the rectangle AL is equal to CH; but CH is equal to HF; therefore also AL is equalt to HF to each of these add CM; therefore the whole AM is equal to the gnomon CMG; but AM is the rectangle con*Cor. 4. tained by AD, DB, for DM is equal to DB: therefore the gnomon CMG is equal to the rectangle AD·DB: add to each of these LG, which is equal to the square of CB, therefore the rectangle AD BD, 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. Wherefore, if a straight line, &c.

2. and 26 Def. +1 Ax.

+Cor. 4. 2. and 34. 1.

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PROP. VII. THEOR.

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.

#46. 1.

#43. 1.

Upon AB describe the square ADEB, and construct the figure as in the preceding propositions. Because AG is equal to GE, if CK be added to each the whole AK will be equal to the whole CE; therefore AK, CE are double of AK: but AK, CE are the gnomon AKF, together with the square CK; therefore the gnomon AKF, together with the square CK, ist double of AK: but twice the rectangle AB BC is D *Cor. 4. double of AK, for BK is equal*

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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 +Cor. 4. equal to the square of AC: therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB BC, and the 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. Q. E. D.

+2 Ax.

PROP. VIII. THEOR.

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. 13. 1. 146. 1.

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*36. 1.

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Prolong† AB to D, so that BD be equal† to CB, and upon AD describet the square AEFD; and construct two figures such as in the preceding. Because CB is equal to BD, and that CB is equal* to GK, and BD to KN; therefore GK is equalt to KN: for a like reason, PR is equal to RO: and because CB is equal to BD, and GK to KN, the rectangle CK is equal to BN, and GR to RN: but CK is equal to RN, because they are the complements of the parallelogram CO: therefore also BN is equal to GR: therefore 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. that BD is equal to BK, that A is,t to CG, and CB equal to GK, that ist, to GP; therefore M CG is equal to GP, and because X +Cor. 4, CG is equal to GP, and PR to RO, the rectangle AG is equalt to MP, and PL to RF: but MP is equal to PL, because they are the complements of the parallelogram ML; wherefore AG is equalt 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

2. and 26

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was demonstrated that the four CK, BN, GR, and RN are quadruple of CK: therefore the eight rectangles which compose the gnomon 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 demonstrated to be quadruple of AK; therefore four times the rectangle

+1 Ax. *Cor. 4. 2. and 34. 1.

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AB BC is equal to the gnomon AOH: to each of these add XH, which is equal to the square of AC; therefore four times the rectangle AB⚫BC, together with the square of AC, is equal to the gnomon AOH and the square XH: but the gnomon AOH and XH make up the figure AEFD, which is the square of AD: therefore four times the rectangle AB BC, together with the square of AC, is equal to the square of AD, that is, of AB and BC added together in one straight line. Wherefore, if a straight line, &c.

Q. E. D.

PROP. IX. THEOR.

If a straight line be divided into two equal, and also into two unequal parts; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section.

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 squares of AD, DB shall together be double of the squares of AC, CD.

*11. 1.

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From the point C draw* CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; through D draw* DF parallel to CE, and through draw FG parallel to BA; and join AF. Then, because AC is equal to CE, the angle EAC is equal to the angle AEC: and because the angle ACE is a right angle, the two others AEC, EAC together make one* right angle; and they are equal to one another; therefore each of them

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