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.. (constr.) DHˆ+DG2=A2+ B'+ C2
i. e. (E. 47. 1.) GH*= A2+B2+C2.

And in the same manner, it is evident, a square may be found, which shall be equal to the aggregate of any number of given squares.

PROP. LXXV.

98. PROBLEM. Two unequal squares being given, to find a third square, which shall be equal to the excess of the greater of them above the less.

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line, be the sides of the two given squares, of which the square of AC is the greater: From the centre C, at the distance CA, describe the circle ADE, meeting AB, produced, in E; from B draw (E. 11. 1.) BD to AB, and let BD meet the circumference in D: The square of BD is equal to the excess of the square of AC above the square of BC.

For join D, C: And since (constr.) the B is a right 2, .. (E. 47. 1.) CD2= CB2+BD'

i. e. (E. def. 15. 1.) AC2=CB'+BD'

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Whence it is manifest, that the square of BD is equal to the excess of the square of AC above the square of CB.

PROP. LXXVI.

99. THEOREM. If the side of a square be equal to the diameter of another square, the former square shall be the double of the latter.

For (E. def. 30. 1. and E. 47. 1.) the square of the diameter of a square is equal to the squares of its two sides; i. e. to the double of the square itself:.. the square of any straight line which is equal to the diameter of a square, is the double of that square.

PROP. LXXVII.

100. THEOREM. In any right-angled triangle, the square which is described on the side subtending the right angle, as a diameter, is equal to the squares described upon the other two sides, as diameters.

For, (S. 76. 1.) the squares described on the hypotenuse, and on the two sides of a ▲ as diameters, are, respectively, the halves of the squares of those lines: But since (hyp.) the ▲ is rightangled,.. (E. 47. 1.) the square of the hypotenuse

is equal to the squares of the two sides; .. the square described on the hypotenuse as a diameter, is equal to the squares described on the other two sides as diameters.

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SUPPLEMENT

TO THE

ELEMENTS OF EUCLID.

BOOK II.

PROP. I.

1. THEOREM. If two given straight lines be divided, each into any number of parts, the rectangle contained by the two straight lines, is equal to the rectangles contained by the several parts of the one and the several parts of the other.

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any parts in the points E, F, and let the given straight line CD be divided first into two parts in the point G The rectangle contained by AB and CD is equal to rectangles contained by AE and CG, by EF and CG, by FB and CG, by AE and GD, by EF and GD, and by FB and GD, taken together.

From the point A draw (E. 11. 1.) AX 1 to AB; from AX cut off (E. 3. 1.) AI CG, and from IX cut off IH GD, so that AH = CD; through I and H draw (E. 31. 1.) IN and HK parallel to AB, and through B, F, E, draw BK, FM, EL, parallel to AH: Then (E. 1.2.) the rectangle AN is equal to the rectangles contained by AE and CG, by EF and CG, and by FB and CG; also the rectangle IK is equal to the rectangles contained by HL and GD, by LM and GD, and by MK and GD; but (E. 34. 1.) HL AE; LM = EF; and MK = FB; ... the rectangle IK is equal to the rectangles contained by AE and GD, by EF and GD, and by FB and GD; but the two rectangles AN and IK make up the rectangle AK, which is contained by AB and AH or CD; .. the rectangle contained by AB and CD is equal to the rectangles contained by AE and CG, by EF and CG, by FB and CG, by AE and GD, by EF and GD, and by FB and GD, taken together,

And, in the same manner, the proposition may be proved to be true, when the given straight line CD is divided into more than two parts.

2. COR. If the parts EF, FB, &c., into which

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