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these As (say the AA BC), and having an < equal to the given Z, F. (Prop. XLIII.)

Let G HKL be the parallelogram so constructed, in which the < G is equal to the given < F.

Construct a parallelogram HMNK, having the line H K for one of its sides, having an < equal to the given _ F, and equal in area to the A ACD; and let it be so placed that the < equal to the given Z, F, may be adjacent to the KH G in the parallelogram GHKL, which is not equal to the _ F.* (Prop. XLIV.)

Construct a parallelogram MOPN, having the line MN for one of its sides, having an < equal to the given ZF, and equal in area to the A ADE; and let it be so placed that the < equal to the given Z, F, may be adjacent to the _ NMH in the parallelogram HMNK, which is not equal to the given _ F.

The figure LG O P will be the parallelogram required.

To show this, it is necessary to prove : 1. That it is a parallelogram. 2. That it is equal in area to the figure ABCDE.

First, then, to prove that it is a parallelogram. For this purpose the first thing that we have to show is, that the lines L K, K N, and N P form one continued straight line, and also that the lines GH, HM, and M 0, form one continued straight line.

This is shown as follows:

Since G L and H K are parallel (being, by construction, opposite sides of a parallelogram), and G H crosses them, the interior <8 LGH and G H K are together equal to two right Zs. (Prop. XXIX.)

* This may always be accomplished by one or other of the various constructions by which the problem proposed in Prop. XLIV. may be solved.

But the < KHM is by construction equal to the ZF, and is therefore also equal to the 2 LGH, which was likewise constructed equal to the < F.

Therefore the sum of the <s KHM and KHG is equal to the sum of the <s LGH and KHG.

But the sum of the Zs LGH and KH G is equal to two right Zs.

Therefore the sum of the s KHM and KH G is also equal to two right s.

We have therefore two right lines, GH and MI, meeting a third right line, KH, at the same point H, and on opposite sides of the line KH, and the two adjacent s so formed are together equal to two right <s.

It follows (according to the 14th proposition), that the lines G H and H M form one continued right line.

In a similar manner it may be shown that the s NM O and NMH are together equal to the sum of the <s KHM and NMH, and are therefore together equal to two right <s, and consequently that the lines H M and MO form one continued right line.

Next, to show that the lines LK, KN, and NP form one continued straight line.

Since the opposite <s of parallelograms are equal, it follows that the <s LKH, K N M, and N PO are equal respectively to the <s L GH, KHM, and NM O, each of which was made equal to the angle F.

Consequently the <s LKH, KNM, and N P O are each equal to the < F; and therefore are equal to each other.

Now, because PO and N M are parallel and NP meets them, the interior Zs NPO and PNM are together equal to two right Zs.

But the < KNM is equal to the < NPO.

Therefore the sum of the s KNM and PNM is equal to the sum of the <s NPO and PNM.

But the sum of the <s NP() and PNM is equal to two right Zs.

Therefore the sum of the <s KNM and PNM is also equal to two right Zs.

We have therefore two right lines K N and PN, meeting a third right line at the same point N, and on opposite sides of the line N M, and making the two adjacent s together equal to two right Zs.

It follows (according to the 14th proposition), that the lines K N and P N form one continued right line.

In the same way it may be shown that the lines L K and K N form one continued right line.

And since L K is parallel to GH, the whole lines L P and G O are parallel.

Moreover, since K H is parallel to L G, and N M is parallel to K H, it follows that N M is parallel to LG.

And since N M is parallel to L G, and P O is parallel to NM, it follows that PO is parallel to L G.

The figure LP OG is therefore a parallelogram that is, a quadrilateral figure whose opposite sides are parallel).

It remains to show that it is equal in area to the figure ABCDE.

This is evident. For since the three parallelograms, of which the whole parallelogram GO PL is composed, were made equal respectively to the three As into which the figure A BCD E was divided, it follows that the whole parallelogram L O is equal to the whole figure ABCDE.

PROPOSITION XLVI. To describe a square upon a given straight line.

For the construction necessary, we must be able 1. From the greater of two given straight lines to

cnt off a part equal to the less. (Prop. III.)

2. From a given point in a given straight line to

draw a perpendicular to that line. (Prop. XI.) 3. Through a given point to draw a straight line

which shall be parallel to a given straight line. To prove that the construction effects what is re

quired we must know,1. That things which are equal to the same are

equal to each other. (Ax. I.) 2. That the opposite sides and Z8 of a parallelo

gram are equal. (Prop. XXXIV.)

E

c

D

A

B

Let A B be the given straight line. It is required to construct upon it a four-sided figure which shall have all its sides equal, and all its <s right Zs.

From the point A draw the line A E at right Zs to A B, and cut off from it a part A C, equal to A B.

Through the point C draw a straight

line C D parallel to A B. Through B draw a straight line parallel to A C, and meeting the line C D in the point D.

A B D C is the figure required. For it is obvious from the mode in which it was described, that it is a parallelogram. Consequently, its opposite sides and <s are equal. That is to say, the side C D is equal to the side A B, and the side D B to the side A C; the LCDB to the CA B, and the < ACD to the < A B D.

Hence, since A B and A C are equal to each other, it follows that all the four sides are equal to each other.

Again, since the LCD B is equal to the < C AB, and C A B is a right Z, the CD B is also a right Z.

And since C D and A B are parallel, and C A inter

sects them, the two interior Zs D C A and CAB are together equal to two right Zs.

But C A B is a right 2. Therefore A C D is also a right Z.

Therefore, also, the Z A B D is a right Z, since it is equal to the < A CD.

It has thus been shown that A B D C is a four-sided figure, which has all its sides equal, and all its Zs right Zs.

PROPOSITION XLVII. If squares be described upon the three sides of a right-angled triangle, the square which is described upon the hypotenuse (or side subtending the right angle) is equal to the sum of the squares described on the sides which contain the right angle.

For the construction employed in this proposition

we must be able, 1. To join two given points by a straight line.

(Post. 1.) 2. From a given point to draw a straight line

which shall be parallel to a given straight line.

(Prop. XXXI.) 3. On a given straight line to describe a square.

(Prop. XLVI.) To prove the proposition by the aid of the con

struction we must know, 1. That if equals be added to equals, the sums

are equal. (Ax. II.) 2. That if two straight lines meet a third at the

same point, but on opposite sides of the right line, and make the two adjacent s together equal to two right <s, those two right lines forni one continued right line. (Prop. XIV.)

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