A B and A D together are equal to twice the squares of B E and EA

together (II. Prop. a).

(II. 4, Cor. 2).

Therefore the squares of A B, BC, CD, and D A are together equal to the square of BD together with twice the squares of EA and

E C.

Again, because EAC is a triangle, of which the base AC is bisected in E (Hyp.), therefore the squares of EA and EC are together equal to twice the squares of cr and EF together (II. Prop. A).

And the doubles of these equals are equal; therefore the doubles of the squares of EA and E C are together equal to four times the squares of C F and EF together.

Therefore the squares of A B, BC, CD, and D A are together equal to the equare of BD together with four times the squares of CF

and E F.

But four times the square of CF is equal to the square of a c (IL. 4, Cor. 2).

Therefore the squares of AB, BC, CD, and DA are together equal to the squares of BD and A c together with four times the square of E F. Q.E.D.

Prop. D. Problem. To divide a given straight line into two parts, so that the rectangle contained by its segments shall be equal to a given square, not greater than the square of half the given straight line.

Produce н G to к making нK equal to G B (I. 3).

With centre and distance x describe the circle M KL, cutting A B in M. The straight line AB is divided in м, so that the rectangle A M, M B is equal to the square CDEF.

Join HM. Then because A B is divided into two equal parts in G, and two unequal parts in x, therefore the square of GB is equal to the square of GM together with the rectangle a M, M B.

But GB is equal to I, by construction, and HK is equal to HM (I. Def. 15).

Therefore (I. Ax. 1) G B is equal to Hм, and the square of G B to the square of H м.

Therefore the square of HM is equal to the square of GM together with the rectangle a M, MB.

But the square of HM is equal to the squares of H G and GM together (I. 47).

Therefore (I. Ax. 1) the squares of HG and GM are together equal to the square of a together with the rectangle AM, M B.

From each of these equals take the square of GM, then the rectangle A M, MB is equal to the square of H G (I. Ax. 3). But the square of H G is equal to the square CDEF (Cons.).

Therefore the rectangle ▲ M, M B is equal to the square CDEF (I. Ax. 1). Q.E.F.

Prop. E. Problem. To produce a given straight line, so that

Let A B be the given straight line, and C D E F the given square. the whole line thus produced, and the part of it produced may be It is required to produce AB so that the rectangle contained by equal to the given square C D E F.

From the point B in the straight line AB draw вH at right angles to AB (I. 11) and equal to CD (I. 3) one of the sides of the given square CDE F.

Bisect AB in the point G and join G H.

With centre G and distance G H describe the circle H M N. Produce A B to meet the circumference in L. Then the rect

angle A L, LB is equal to the square CDEF.

Because в H is equal to CD (Cons.), therefore the square of BH is equal to the square of CD, that is the square C DEF. Again, because G is the centre of the circle H M N, therefore G H is equal to GL (I. Def. 15) and therefore the square of GH is equal to the square of G L.

Again, because HBG is a right angled triangle having the right angle H BG (Cons.), therefore the square of G H is equal to the squares of G B and B H together (I. 47).

But the square of G H has been above shown to be equal to the square of G L. Therefore (I. Ax. 1) the square of G L is equal to the squares of G B and в H together.

But the square of GL is equal to the square of GB together with the rectangle A L, L B (II. 6).

Therefore (I. Ax. 1) the squares of GB and RH are together equal to the square of G B together with the rectangle A L, LB. From each of these equals take away the common square of & B, then (I. Ax. 3) the rectangle AL, LB is equal to the square of

BH.

But the square of BH is equal to the square CDE F(Cons.). Therefore the rectangle A L, LB is equal to the square C D E F (L. Ax. 1). Q.E.F..

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SIMPLICITAS (Wemyss, Fifeshire) has solved the first thirty-two of the Second Centenary of Algebraical Problems; D. HORNBY (Driffield) the first fifty-five, with the exception of Nos. 7, 12, 35, 37, 39, 45, and 54; GEORGE and Q Q. Fenchurch-street) the first thirty-two, except Nos. 16, 23, 26, WILD (Dalton-on-Tees) the whole of the second portion from No. 73 to 70; 28, 30, and 31, besides Nos. 40, 41, and 51 of the second portion. D. HORNBY (Driffield): The following is George Wild's solution of Problem 49:

54
Therefore = 6 days
8

is the time they would both require to finish it.

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