shall be of a given altitude, and equiangular with, and also equal to, a given parallelogram.

Let ABCD be a given, and E a given

straight line: It is required to describe a which shall be equal to the ABCD, and also equiangular with it; and which shall have its altitude equal to the given line E.

From the point C draw (E. 11. 1.) CF 1 to BC, and make CF=E; through F draw (E. 31. 1.) HG parallel to BC; produce BA and CD to meet HG, in H and G; join H, C, and let HC cut AD in I; through I draw (E. 31. 1.) KIL parallel to HB or GC: The KLCG, which (constr. and E. 29. and 34. 1.) is equiangular with the ABCD, and has its altitude equal to E, is also equal to the ABCD.

For, since BI and IG are compliments about the diameter HC of the HBCG, they are (E. 43. 1.) equal to one another; to each of these equals add the LD; and it is plain that the KLCG=ABCD.

93. COR. Hence, a rectangle may very readily be found, which shall be equal to a given square,

and shall have one of its sides equal to a given straight line.

PROP. LXXI.

94. THEOREM. If there be any number of rectilineal figures, of which the first is greater than the second, the second than the third, and so on, the first of them shall be equal to the last together with the aggregate of all the differences of the figures.

First let there be three such given rectilineal figures. Make (E. 45. 1.) the

FH equal to the

greatest of the given figures, having its / FGH of any given magnitude; produce GH to X; from HX cut off (E. 3. 1.) HI=GH; find (S. 57. 1. cor.) a ▲ equal to the next greatest of the given figures, and apply (E. 44. 1.) to HI a equal to that A, having its 2 IHK = HGF: Again, from IX cut off IM=GH or HI, and, in like manner, to IM apply a□ IO, equal to the least of the given figures, and having its MIN= 2 HGF.

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 ☐ FH QH +PR+FK

i.e. the

FH ONM+PR+

But the PR is the difference of the

FK.

PH and

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 A, 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, 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.

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.) EF=ED2+DF'; i. e. (constr.) DH'= A'+B';