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added Line, together with the Square of half the Line, is equal to the Square of the Line compounded of half the Line, and the added Line taken as one Line; which was to be demonstrated.

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PROPOSITION VII.

THEOREM.
If a Right Line be any bow cui, the Square of

the whole Line, together with the Square of one
of the Segments, is equal to double the Rectangle
contained under the whole Line, and the said
Segment, together with the Square, made of the

other Segment. L

ET the Right Line A B be any how cut in the

Point C. I say the Squares of AB, BC, together, are equal to double the Rectangle contained under AB, BC, together with the Square, made of AC.

For let the Square of AB be * described, viz.

ADE B, and construct s the Figure. + 43. 1. Then because the Rectangle AG, is f equal to

the Rectangle GE. IF CF, which is common, be added to both, the whole Rectangle AF shall be equal to the whole Rectangle CE, and so the Rectangles AF, CE, are double to the Rectangle AF; but AF, CE, make up the Gnomon KLM, and the Square CF. Therefore the Gnomon KLM, together with the Square CF, shall be double to the

Rectangle A F. But double the Rectangle under AB, 1 Cor. 4. BC, is double the Rectangle AF, for BF is I equal

to BC. Therefore the Gnomon KLM, and the Square CF, are equal to twice the Rectangle contained under AB, BC. And if HF, which is common, being the Square of AC, be added to both; then the Gnomon KLM, and the Squares CF, HF, are equal to double the Rectangle contained

§ A Figure is said to be constructed, when Lines, drawn in a Paral. lelogram parallel to the Sides thereof, cut tbe Diameter in one Point, and make two Parallelograms about the Diameter, and two Complements. So likewise a double Figure is said to be constructed, when two Right Lines parallel to the Sides, make four Parallelograms about tbe Diameter, and Four Complements

under

under AB, BC, together with the Square of A C. But the Gnomon KLM, together with the Squares CF, HF, are equal to ADE B, and CF, viz. the Squares of AB, BC. Therefore the Squares of AB, BC, are together equal to double the Rectangle con tained under AB, BC, together with the Square of AC. Therefore, if à Right Line be any how cut, the Square of the whole Line, together with the Square of one of the Segments, is equal to double the Rectangle contained under the whole Line, and the faid Segment, together with the Square, made of the other Segment; which was to be demonstrated.

PROPOSITION VIII.

THEOREM.
If a Right Line be any how cut into two Parts,

four times the Rectangle, contained under the
whole Line, and one of the Parts, together with
the Square of the other Part, is equal to the
Square of the Line, compounded of the whole
Line, and the first Part taken as one Line.
ET the Right Line AB be cut any how in C.

I say four times the Rectangle contained under AB, BC, together with the Square of AC, is equal to the Square of AB, and BC taken as one Line.

For let the Right Line A B be produced to D, fo that BD be equal to BC, describe the Square AE FD, on AD, and construct the double Figure.

Now fince CB is * equal to BD, and also to * Hypo +GK, and BD is equal to KN: GK shall be + 34. I. likewife equal to KN; by the fame Reasoning, PR is equal to RO. And fince CB is equal to BD, and GK to KN, the Rectangle CK will I be I 36. 1. equal to the Rectangle BN, and the Rectangle GR to the Rectangle RN. But CK is * equal to RN;* 42. To for they are the Complements of the Parallelogram CO. Therefore BN is equal to GR, and the four Squares B N, KC, GR, RN, are equal to each other; and so they are together Quadruple CK. Again, because CB is equal to BD, and BD to BK, that is, equal to CG; and the faid CB is equal

also

E 2

also to GK, that is, to GP; therefore CG shall
be equal to GP. But PR is equal to RO; there-
fore the Rectangle AG shall be equal to the Rect-
angle MP, and the Rectangle PL equal to RF.
But MP is equal to PL; for they are the Comple-
ments of the Parallelogram ML. Wherefore AG
is equal also to RF. Therefore the four Parallelo-
grams AG, MP, PL, RF, are equal to each other,
and accordingly they are together Quadruple of AG.
But it has been proved that the four Squares CK,
BN, GR, RN, are Quadruple of CK. There-
fore the four Rectangles,

and the four Squares, mak-
ing up the Gnomon STY, are together Quadruple
of AK; and because A K is a Rectangle contained
under AB, and B C, for B K is equal to BC; four
times the Rectangle under AB, BC, will be Qua-
druple of A K. But the Gnomon STY has been
proved to be Quadruple of AK. And fo four times
the Rectangle contained under AB, BC, is equal to

the Gnomon STY. And if XH, being equal to + Cor. 4. † the Square of AC, which is common, be added

to both: Then four times the Rectangle contained
under AB, BC, together with the Square of A C, is
equal to the Gnomon STY, and the Square XH,
But the Gnomon STY and HX, make AEFD,
the whole Square of AD. Therefore four times the
Rectangle contained under A B, BC, together with
the Square of AC, is equal to the Square of AD,
that is, of A B and B C taken as one Line. Where-
fore, if a Right Line be any how cut into two Parts,
four times the Rectangle contained under the whole
Line, and one of the Parts, together with the Square
of the other Pari, is equal to the Square of the Line,
compounded of the whole Line, and the first Part
taken as one Line; which was to be demonstrated.

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PROPOSITION IX.

THEORE M.
If a Right Line be any how cut into two equal, and

two unequal Parts; then the Squares of the un-
equal Parts together, are double to the Square
of the balf Line, and the Square of the inter-
mediate Part.

ET any Right Line A B be cut unequally in D,

and equally in C. I say the Squares of AD, DB, together, are double to the Squares of AC and CD together.

For let * CE be drawn from the Point C at Right * 11. 1. Angles to AB, which make equal to AC, or CB, and join EA, EB. Also thro' D let + DF be drawn t 31. 1. parallel to CE, and F G thro' F parallel to AB, and draw AF.

Now because AC is equal to CE, the Angle EAC will be equal to the Angle AEC; and since the # 5. I. Angle at C is a Right one, the other Angles AEC, EA C, together, shall * make one Right Angle, and * 3. Car. are equal to each other : And so AEC, EAC, are each 32. I. equal to half a Right Angle. For the same Reason are also CEB, EBC, each of them half Right Angles. Therefore the whole Angle AEB is a Right Angle. And since the Angle GEF is half a Right cne, and EGF is a Right Angle; for it is t equal to the † 29. I. inward and opposite Angle ECB, the other Angle EFG will be also equal to half a Right one. Therefore the Angle GEF is equal to the Angle EF G. And so the Side EG is I equal to the Side GF. Again, I 6. I. because the Angle at B is half a Right one, and FDB is a Right one, because equal to the inward and oppofite Angle ECB, the other Angle BFD will be half a Right Angle. Therefore the Angle at B is equal to the Angle BFD; and so the Side DF is equal to the Side DB. And because AC is equal to CE, the Square of AC will be equal to the Square of CE. Therefore the Squares of AC, CE, together, are double to the Square of AC; but the Square of EA is 4 equal to the Squares of AC, CE, together, since 1 47. Is E 3

ACE

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+ 47. I,

ACE is a Right Angle. Therefore the Square of
E A is double to the Square of AC. Again, because
EG is equal to GF, and the Square of E G is equal
to the Square of GF: Therefore the Squares of EG,
GF, together, are double to the Square of GF. But
the Square of EF is t equal to the Squares of EG, GF,

Therefore the Square of EF is double the Square of
GF: But GF is equal to CD; and so the Square of
EF double to the Square of CD. But the Square
of AE is likewise double to the Square of A C.
Wherefore the Squares of A E, and EF, are double
to the Squares of AC and CD. But the Square of
AF is tequal to the Squares of AE and EF; be-
cause the Angle AEF is a Right Angle, and conse-
quently the Square of AF is double to the Squares
of AC, and CD. But the Squarés. of AD, DF,
are equal to the Square of AF: For the Angle at D
is a Right Angle. Therefore the Squares of AD,
and DF, together, shall be double to the Squares of
AC and CD together. But DF is equal to D B.
Therefore the Squares of AD, and DB, together,
will be double to the Squares of AC and CD, to-
gether. Wherefore, if a Right Line be any bow cut
into two equal, and two unequal Parts,' then the
Squares of the unequal Parts together, are double to
the Square of the half Line, and the Square of the in-
termediate Part; which was to be demonstrated.

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; THEOREM.
If a Right Line be cut into two equal Parts, and to

it be direEfly added another; the Square made on
[tbe Line compounded of ] the whole Line, and
the added one, together with the Square of the
added Line, fall be double to the Square of the
half Line, and the Square of [that Line which is
compounded of ] the balf, and the added Line.
ET the Right Line ĄB bę bifected in C, and

e any straight Line BD added directly thereto.
say, the Squares of AD, DB, together, are double
to the Squares of AC, CD, together,

For

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