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Produce LK and ON to meet FG in P and Q ;

and let OQ meet KH in R.

Then, (E. 36. 1. E. 34. 1. and constr.) the

i. e. the

FHQH + □ PR+FK

FH=□NM+PR+

FK.

PH and

But the PR is the difference of the QH or (E. 36. 1.) of KI and NM; and the FK is the difference of the FH and PH, or of FH and KI: Whence it is manifest that the proposition is true, when three rectilineal figures are taken : And it may, in the same manner, be proved to be true, when more than three are taken.

PROP. LXXII.

95. PROBLEM. To find a rectangle, which shall have one of its sides equal to a given finite straight line, and which shall be equal to the excess of the greater of two given rectilineal figures above the less.

To the given finite straight line, and on the same side of it, apply (E. 45. 1. cor.) two rectangles, the one equal to the greater and the other to the less, of the given rectilineal figures: And it is manifest that the rectangle which is the difference of the two rectangles so described, will have one of its sides equal to the given straight line, and will be equal to the excess of the greater of the two given figures above the less.

PROP. LXXIII.

96. THEOREM. If two right-angled triangles have two sides of the one equal to two sides of the other, each to each, the triangles shall be equal, and similar to each other.

If the two sides about the right-angle of the one ▲, be equal to the two sides about the right-angle of the other, each to each, it follows, (from E. 4. 1.) that the are equal and similar.

But, let now, the hypotenuses of the right 4, in the two, be equal, and also let one other side, of the one A, be equal to another side of the other; .. (E. 47. 1.) the squares of the two remaining sides of the one, will be equal to the squares, taken together, of the two remaining sides of the other A; from these equals take away the equal squares of the two other sides, which, by the hypothesis, are equal, and there remains the square of the third side, of the one, equal to the square of the third side, of the other A; .. the third side of the one is equal to the third side of the other; .. (E. 4. 1.) the two are equiangular, and are, also, equal to one another.

PROP. LXXIV.

97. PROBLEM. To find a square which shall be equal to any number of given squares.

First, let there be three given square, and let their sides be equal to the three straight lines A, B and C.

[blocks in formation]

Take any straight line DX, indefinite towards X; from D draw (E. 11. 1.) DY ¦ to DX, and produce DY indefinitely towards Y: From DX cut off (E. 3. 1.) DE A, and from DY cut off DFB; and join E, F: Again, from DY cut off DH=EF, and from DX cut off DG = C, and join G, H: The squares described (E. 46. 1.) upon GH shall be equal to the three given squares to the sides of which A, B and C are respectively equal.

For (E. 47. 1. and constr.) EF2=ED2+DF2; i. e. (constr.) DH*= Aˆ+B2;

•. (constr.) DH2+DG2=A*+ B2+ C2

i. e. (E. 47. 1.) GH*= A2+B+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.

Let AC and CB placed in the same straight

F D

A

C B

E

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 1 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,.. (E. 47. 1.) CD'=CB1+BD'

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

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 A is rightangled, .. (E. 47. 1.) the square of the hypotenuse

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