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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 BG, AK; a 46. through A draw AL parallel to BD, or EC1; join AD, FC, b 31. BK, AE.

Then, because BAC, BAG, are each right angles, GAC is aright line. For the fame reafon BAH is a right line; like- c 14. wife the angles DBC, ABF, are right angles'; add ABC to both, then the whole angle FBC is equal to ABD 4; and AB, d Ax. 2. BD, are equal to FB, BC; and the angle TBC to ABD; therefore the triangles ABD, FBC, are equal ; but the parál- e 4. lelogram BL is double the triangle ABD; 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, g 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.

PRO P. XLVIII. THE OR.

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 thefe 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.

For, let AD be drawn from the point A, at right angles, to a 11, AC, and equal to AB; join DC. Then, becaufe 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 47therefore 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 DC4; therefore BC is equal to DC; but BA is a Ax, equal to AD, and AC common; therefore BA, AC, are equal to DA, AC, and the bafes BC, DC, equal; therefore the angle BAC is equal to the angle DAC. But DAC is a right e s. angle; therefore BAC is a right angle. Wherefore, &c.

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DEFINITION S.

1.

VERY right angled parallelogram is faid to be contain ed 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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PROP. I. THE OR.

IF there be two right lines, and one of them divided into any number of parts, the rectangle contained by the whole, and divided 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 BC1; make BG equal to Ab; through G draw GH parallel to BC; and through the points D, E, C, draw DK, EL, CH, each pa-. rallel 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

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