these As (say the ▲ ABC), and having an equal to the given, F. (Prop. XLIII.)

Let G H KL be the parallelogram so constructed, in which the G is equal to the given

F.

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

Construct a parallelogram M OPN, having the line MN for one of its sides, having an equal to the given F, and equal in area to the ▲ ADE; and let it be so placed that the may be adjacent to the

equal to the given Z, F, NMH in the parallelogram HMNK, which is not equal to the given F.

The figure L GOP 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, KN, and N P form one continued straight line, and also that the lines GH, H M, and M O, 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 interiors LGH and GHK are together equal to two rights. (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 which was likewise constructed equal to the

LGH,
F.

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

But the sum of the s LGH and KH G is equal to two rights.

Therefore the sum of the s KHM and KHG is also equal to two rights.

We have therefore two right lines, G H and MH, meeting a third right line, KH, at the same point H, and on opposite sides of the line KH, and the two adjacents so formed are together equal to two right Zs.

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

In a similar manner it may be shown that the s NMO and NM H are together equal to the sum of the ZS KHM and NMH, and are therefore together equal to two rights, 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 opposites of parallelograms are equal, it follows that the /s LKH, K N M, and N PO are equal respectively to the s LG H, KHM, and NM O, each of which was made equal to the angle F.

Consequently the s LKH, KN M, and NPO are each equal to the F; and therefore are equal to each other.

Now, because PO and N M are parallel and N P meets them, the interiors NPO and PNM are together equal to two rights.

But the KN M is equal to the

NPO.

Therefore the sum of the s KNM and PNM is

equal to the sum of the s NPO and P NM.

But the sum of the s NPO and P NM is equal to

two rights.

Therefore the sum of the s KNM and P N M is also equal to two rights.

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

It follows (according to the 14th proposition), that the lines KN and PN 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 LK is parallel to GH, the whole lines LP and GO are parallel.

Moreover, since KH is parallel to L G, and N M is parallel to KH, it follows that N M is parallel to LG. And since N M is parallel to L G, and PO is parallel to N M, it follows that PO is parallel to L G.

The figure L POG 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 G O PL is composed, were made equal respectively to the three As into which the figure ABCDE was divided, it follows that the whole parallelogram LO 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 cut off a part equal to the less. (Prop. III.)

E

A

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 required 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 s of a parallelogram are equal. (Prop. XXXIV.)

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 AE at rights to A B, and cut off from it a part A C, equal to A B. Through the point C draw a straight line CD 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 CD is equal to the side A B, and the side D B to the side A C; the CDB 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 CD B is equal to the ▲ C A B, and C A B is a right, the CD B is also a right . And since C D and A B are parallel, and C A inter

sects them, the two interiors D C A and CA B are together equal to two rights.

But C A B is a right . Therefore A CD is also

a right .

Therefore, also, the A B D is a right, since it is equal to the AC D.

It has thus been shown that A B D C is a four-sided figure, which has all its sides equal, and all its s 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 construction 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 adjacents together equal to two rights, those two right lines form one continued right line. (Prop. XIV.)