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INDICATIVE MOOD.

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Future.

Cadrò or caderò, I shall or will
fall

cadrai or caderdi, thou wilt fall
cadrà or caderà, he will fall

cadrémo or caderémo, we will

fall

cadréte or caderéte, you will fall
cadranno or caderanno, they

will fall

Conditional Present.

From the point A draw A E perpendicular to BC (I. 12). Then, because the angle A EB is a right angle (Cons.), therefore the square of A B is equal to the squares of B E and EA together (I._47).

For the same reason the square of A c is equal to the squares of AE and EC together.

Therefore, adding equals to equals (I. Ax. 2), the squares of AB and AC together are equal to the squares of BE and EC together with twice the square of EA.

But because BC is divided into two equal parts in D and two unequal parts in E, therefore the squares of BE and EC are together equal to twice the square of BD together with twice the square of D E (II. 9).

Therefore the squares of AB and AC together are equal to Cadréi, or caderei, caderia, ca-twice the squares of B D, D E, and E A together. dria, I should or would fall cadrésti or caderésti, thou wouldst fall

cadrébbe or caderebbe, caderia,
cadría, he would fall

cadde, cadéo, cadette, or cade, cadrémmo or caderémmo, we

he fell

cadémmo, we fell

cadéste, you fell

caddero, cadéro, cader, cadettero,

cadérono, they fell

would fall

cadréste or caderéste, you would
fall

cadrébbero or caderebbero, cade-
riano, they would fall

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But the squares of D E and EA are together equal to the square of AD (I. 47), and the doubles of these equals are equal. Therefore twice the squares of D E and EA are equal to twice the square of A D. Therefore the squares of AB and A c together are equal to twice the squares of B D and DA together. Q.E.D.

Prop. B. Theorem. The squares of the two diagonals of a parallelogram are together equal to the squares of its four sides.

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Let ABCD be any parallelogram of which the diagonals are A C and BD cutting each other in E. Then the squares of a C and BD together are equal to the squares of AB, BC, CD, and DA together. By Exercise 2 to Proposition XXXIV. of the first book, the diagonals bisect each other. And because A B D is a triangle of which the side BD is bisected in E, therefore, by the last proposition, the squares of AB and AD together are equal to twice the squares of B E and A E together.

For the same reason the squares of BC and CD are together equal to twice the squares of BE and E C together. Therefore, adding equals to equals, the squares of AB, BC, CD, and DA together are equal to four times the square of B E together with twice the squares of AE and E C (I. Ax. I. 2).

But AE is equal to E C, because the diagonals bisect each other, and therefore the square of AE is equal to the square of EC. Therefore the squares of A B, BC, CD, and DA together are equal to four times the squares of B E and AE together.

But the square of any straight line is equal to four times the square of half the line (II. 4, Cor. 2). Therefore the square of BD is equal to four times the square of B E, and the square of a c is equal to four times the square of A E.

Therefore the square of AB, BC, CD, and D A are together equal to the squares of AC and BD together. Q.E.D.

Prop. c. Theorem. The squares of the four sides of a tra pezium are together equal to the squares of its two diagonals and four times the square of the straight line which joins the points of the bisection of the diagonals.

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Again, because EA C 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 CF and EF together (II. Prop. A).

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

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

and E F.

But four times the square of or is equal to the square of A C (II. 4, Cor. 2).

Therefore the squares of AB, BC, CD, and DA are together equal to the squares of B D 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.

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Produce H G to K making и x equal to GB (I. 3).

With centre H and distance HK describe the circle M KL, cutting A B in M. The straight line AB is divided in M, so that the rectangle A M, M B is equal to the square C D E F.

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

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

Therefore (1. Ax. 1) GB is equal to HM, and the square of G B to the square of H M. Therefore the square of HM is equal to the square of GM together with the rectangle ▲ M, MB.

But the square of HM is equal to the squares of 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, MB.

E

II

A

G

B

F

M

N

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

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

Bisect A B in the point & and join G H.

With centre G and distance GH describe the circle H MN. Produce A B to meet the circumference in L. Then the rectangle A L, L B 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 CDEF.

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 GL.

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

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

Therefore (I. Ax. 1) the squares of &B and BH are together equal to the square of a 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 A L, LB is equal to the square of

BH.

But the square of BH is equal to the square CD E F (Cons.). Therefore the rectangle AL, LB is equal to the square C D E F (I. Ax. 1). Q.E.F.*

ANSWERS TO CORRESPONDENTS.

JOHN BURROUGH: The POPULAR EDUCATOR has been published in the United States. We cannot promise to act upon your suggestion. GEORGE YOUNG will find our Lessons in Italian answer his purpose. UN FRANÇAIS: It is doubtful whether we shall be able to find room for any lessons in Spanish, but we should be glad if it were possible.

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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. 23 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:

5

9

:

9 Days Days

:: 10: 18

9

From each of these equals take the square of GM, then the rectangle A M, M B is equal to the square of H G (I. Ax. 3). But the number of days in which A can do the whole. Therefore he would the square of H G is equal to the square CDEF (Cons.). Therefore the rectangle A M, M B is equal to the square CDEF require 8 days to do the remaining (I. Ax, 1). Q.E.F.

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

of the work alone.

The above five exercises were solved by J. H. EASTWOOD (Middleton).

408

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