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Let ABC be a right angled triangle, the fquare of the fide Book I• BC fubtending the right angle is equal to the fquares of the fides BA, AC, containing the right angle. Upon BC defcribe the fquare BDEC; upon BA, AC, the fquares DG, AK; a 46. through A draw AL parallel to BD, or EC; join AD, FC, BK, AE.

b 31.

Then, because BAC, BAG, are each right angles, GAC is a right line. For the fame reafon BAH is a right line; like- © I.. wife the angles DBC, ABF, are right angles; add ABC to both, then the whole angle FBC is equal to ADD) ; and AB, d Ax. 2. BD, are equal to FB, BC; and the angle FBC to ABD; therefore the triangles ABD, FBC, are equal ; but the paral- e 4. e; lelogram BL is double the triangle ABDF; and BG is double f 41, FBC, or ABD; therefore the parallelograms GB, BL, are equal. For the fame reafon LC is equal to CH; but BL, 8 Ax. 6. LC, are equal to the fquare of BC; therefore the fquares of BA, AC, are equal to the fquare of BC. Wherefore, &c.

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PROP. XLVIII. THE OR.

a

F the fquare defcribed upon one of the fides of a triangle be equal to the fquares of the other two fides, the angle contained by these two fides is a right angle.

Let the fquare of the fide BC of the triangle ABC be equal to the fquares of the fides BA, AC, the angle BAC is a right angle.

a

For, let AD be drawn from the point A, at right angles 2, to a 11. AC, and equal to AB; jein DC. Then, because the angle DAC is a right one, the fquare of DC is equal to the fquares of DA, AC. But DA is equal to AB, and AC is common; b 47. therefore the fquares of DA, AC, are equal to the fquares of BA, AC; but the fquare of BC is equal to the fquares of BA, AC, or of DA, AC: Therefore the fquare of BC is equal to c Conft. the fquare of DC; therefore BC is equal to DC; but BA is a Ax, 1, equal to AD, and AC common; therefore BA, AC, are equal to DA, AC, and the bales BC, DC, equal; therefore the angle BAC is equal to the angle DAC“. But DAC is a right e 8. angle; therefore BAC is a right angle. Wherefore, &c.

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E

DEFINITION S.

I.

BOOK II. VERY right angled parallelogram is faid to be containLed by two right lines containing the right angle.

II.

In every parallelogram, either of the two parallelograms that are about the diameter, together with the complement, is called a gnomon.

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PRO P. I. THE OR.

IF
F there be two right lines, and one of them divided into any
number of parts, the rectangle contained by the whole, and di-
vided line, is equal to all the rectangles contained by the whole
line, and the feveral parts of the divided line.

Let A, BC, be the two right lines, one of which, viz. BC, is divided into any number of parts, as D, E; the rectangle contained by A, BC, is equal to the rectangles contained by A, BD; A, DE; A, EC.

For, from the point B draw BF, at right angles, to BC"; make BG equal to A; through G draw GH parallel to BC; and through the points D, E, C, draw DK, EL, CH, each parallel to BGʻ.

The rectangle BK is that contained by BD, BG; for BG is equal to A; the rectangle DL is contained by A, DE; and

EH

EH by A, EC; for DK, EL, CH, are each equal to BG, that Book II, is, equal to A; but the rectangle BH is equal to the rectangles BK, DL, EH; and BH is contained by A, BC; therefore the e 3+. 1. rectangle by A, and BC, is equal to the rectangles by A, BD; A, DE, and A, EC. Wherefore, &c.

IF a

PRO P. II. THE OR.

a right line be any how cut, the rectangles contained by the whole line, and each of the fegments, are equal to the fquare of the whole line.

a

b 31. I.

Let the right line AB be any how cut in C, the rectangles contained by AB, BC, and AB, AC, together, are equal to the fquare of AB. Upon AB defcribe the fquare ADEB ; thro' a 46. 1. C draw CF parallel to AD, or BE; then, because AD is equal to AB, the rectangle under AD, AC, is equal to the rectangle under AB, AC; and the rectangle under EB, BC, is equal to the rectangle under AB, BC; but the rectangle under AD, AC, that is, the rectangle AF, together with the rectangle under EB, BC, that is, CE, are equal to the fquare of AB; that is, the fquare AE. Wherefore, &c.

PR O'P. III. THE O R.

Ianrignone of the parts, is equal to the rectangle under the two parts, together with the fquare of the first mentioned part.

Fa right line be any how cut, the rectangle under the whole

Let the right line AB be any how cut in C, the rectangle under AB, BC, is equal to the rectangle under AC, CB, together with the fquare of BC. Upon BC defcribe the fquare BCDE; produce ED to F, and draw AF parallel to CD or BE; the rectangle under AB, BC, that is, AE, is equal to the rectangles AC, CD; that is, the rectangle under AC, CB, and the fquare of CB. Wherefore, &c.

PRO P. IV. THE OR.

Fa right line be any how cut, the fquare of the whole line is equal to the fquares of the two parts, with twice the rectangle under these parts.

Let

a 46. I.

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