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Let ABC be a right angled triangle, the square of the fide
BOOK I. BC fubtending the right angle is equal to the squares of the fides BA, AC, containing the right angle. Upon BC defcribe the square BDEC a; upon BA, AC, the squares BG, AK; a 16. through A draw AL parallel to BD, or EC"; join AD, FC, b žr. BK, AE.
Then, because BAC, BAG, are each right angles?, GAC is a right line. For the same reason BAH is a right line; like- C 14. wise the angles DBC, ABF, are right anglesa; add ABC to both, then the whole angle FBC is equal to ABD d; and AB, Ax. 2. BD, are equal to FB, BC; and the angle TBC to ABD; therefore the triangles ABD, FBC, are equal"; but the paral- € 4. lelogram BL is double the triangle ABD'; and BG is double f 41. FBCf, or ABD; therefore the parallelograms GB, BL, are equal 8. For the same reason LC is equal to CH; but BL, 8 Ax. 6. LC, are equal to the square of BC; therefore the squares of BA, AC, are equal to the square of BC. Wherefore, &c.
PRO P. XLVIII. THE O R.
F the square described upon one of the sides of a triangle bee
qual to the squares of the other two sides, the angle contained by these two sides is a right angle.
Let the square of the side BC of the triangle ABC be equal to the squares of the sides BA, AC, the angle BAC is a right angle.
For, let AD be drawn from the pcint A, at right anglesa, to a 11. AC, and equal to AB; join DC. Then, because the angle DAC is a right one, the square of DC is equal to the squares of DA, ACB. But DA is equal to AB, and AC is common; b 47. therefore the squares of DA, AC, are equal to the squares of BA, AC; but the square of BC is equal to the squares of BA, AC, or of DA, AC: Therefore the square of BC is equal to c Const, the square of DC4; therefore BC is equal to DC; but BA is d Ax, si 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 Dace. But DAC is a right e 8. angle; therefore BAC is a right angle. Wherefore, &c.
Ε L Ε Μ Ε Ν Τ S
E U C L I D.
1. Book II. VERY right angled parallelogram is said 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.
PRO P. I. THEOR.
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 several 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 Ab; through G draw GH parallel to BC; and through the points D, E, C, draw DK, EL, CH, each Tallel to BGC.
The rectangle BK is that contained by BD, BG; for BG is equal to A; the rectangle DL is contained by A, DE ; and.