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equal As DCB and DAB, the remainders will be equal, that is to say, the parallelogram FCHE is equal to the parallelogram KAGE.

PROPOSITION XLIV. To construct a parallelogram which shall have a given straight line for one of its sides,* which shall have an angle equal to a given rectilineal angle, and which shall be equal in area to a given triangle.

For the construction in this proposition we must

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

(Post. I.) 2. To produce a straight line to any length.

(Post. II.) 3. Through a given point to draw a straight line

parallel to a given straight line. (Prop. XXXI.) 4. To construct a parallelogram equal in area to

a given A, and having an equal to å given

rectilineal Z. (Prop. XLIII.) 5. We are also supposed to be able to move a

given figure from one position to any other

that may be required. (See § 9, p. 8.) To prove that the construction effects what is

required, we must know, 1. That if two parallel straight lines are inter

sected by a third, the two interior <8 are

together equal to two right _8. (Prop. XXIX.) 2. That if two straight lines are cut by a third,

and the two interior _8 so formed are together less than two right Zs, those two straight lines will meet if produced. (Ax. XIII.)

* This first condition is usually expressed thus :-To a given straight line to apply a parallelogram which, &c.

3. That if two straight lines cut one another, the

vertically opposite Zs are equal. (Prop. XV.) 4. That the complements of the parallelograms

about the diagonal of a parallelogram are equal. (Prop. XLIII.)

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Let PT be the given straight line, A B C the given A, and D the given Z. We have to construct a parallelogram which shall have P T for one of its sides, which shall have an < equal to the Z D, and which shall be equal in area to the A ABC.

Construct a parallelogram, AF D E, equal to the ^ A B C, and having an equal to the < D.

Let this parallelogram be removed to the position APDE, so as to have one of its sides, AP, in the same straight line with TP, and the APD (equal to the given _ D) at the point P.

Produce the side E D, and through the point T draw a straight line, TL, parallel to A E, and meeting ED produced in the point L.

Join the points L and P by the straight line L P.

Produce the lines E A and LP. These lines, when produced, will meet; for since LT is parallel to EA, the <s AEL and ELT are together equal to two right Zs. (Prop. XXIX.)

Consequently, the Zs AEL and ELP are less than two right 28, and therefore the lines EA and LP will meet when produced. (Ax. XIII.)

Let them meet in the point M. Through the point M draw the line M 0, parallel to A Tor E L.

Prodụce the line D P to meet M O in the point N, and the line L T to meet the line M O in the point 0.

By this construction we get a parallelogram, ELOM, subdivided into four other parallelograms, two of which are about the diagonal L M, and the other two are the complements of these.

PTON is the parallelogram required. For, 1. It has P T for one of its sides. 2. The vertically opposite <S, APD and N PT, formed by the intersection of the straight lines D N and AT, are equal. (Prop. XV. But the LAPD is equal to the < D. Therefore the < NPT is equal to the < D.

3. The parallelogram PTON is equal to the a ABC.

For, since PTON and APDE are the complements of the parallelograms about the diagonal L M in the parallelogram ELOM, they are equal. (Prop. XLIII.)

But the parallelogram APDE was made equal to the A ABC.

Therefore the parallelogram PTON is also equal to the A ABC.

In the construction employed in this proposition it is perfectly indifferent which side of the parallelogram AFDE be placed so as to be in the same straight line with the given line PT; nor does it matter at which end of the line PT it be placed, if the following directions be attended to:

Place one of the sides of the parallelogram, AFDE, so as to be in the same straight line with PT. Produce the opposite side of the parallelogram in the direction of the line PT. Through that extremity of the line PT at which the parallelogram AFDE was not placed, draw a line parallel to either of those sides of the parallelogram which are not in the same straight line as PT, or parallel to it, so to meet that side of the parallelogram which was produced. Draw thai diagonal

as

of the parallelogram thus formed, which when produced will pass through that extremity of the line PT, at which the parallelogram AFD E was placed. Produce this diagonal to meet a produced side of the first parallelogram. The mode of completing the figure will then be obvious. The following are some of the various constructions that will result. N

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M

M

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PROPOSITION XLV. To describe a parallelogram equal to å given rectilineal figure, and having an angle equal to a given rectilineal angle.

For the construction in this proposition we must

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

(Post. I.) 2. To describe a parallelogram equal to a given

A, and having an < equal to a given recti

lineal Z. (Prop. XLIII.) 3. To construct a parallelogram which shall have

a given line for one of its sides, which shall have an equal to a given rectilineal Z, and which shall be equal in area to a given A. (Prop. XLIV.)

To prove that the construction accomplishes what

is required, we must know,1. That things that are equal to the same are

equal to each other. (Ax. I.) 2. That if equals be added to equals the sums are

equal. (Ax. II.) 3. That if two straight lines meet a third at the

same point, but on opposite sides of it, and the two adjacent Z8 so formed are together equal to two right Zs, those two straight lines lie in

one and the same straight line. (Prop. XIV.) 4. That if two parallel straight lines be inter

sected by a third straight line, the two interior Zs are together equal to two right 8. (Prop.

XXIX.) 5. That if two straight lines are both parallel to

the same straight line, they are parallel to each other. (Prop. XXX.)

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Let ABCDE be the given rectilineal figure, and F the given <. We have to construct a parallelogram equal in area to the figure A B C D E, and having an < equal to the < F.

Divide the rectilineal figure ABCDE into As by joining the points A and C by the straight line AC, and the points A and D by the straight line A D.*

Construct a parallelogram equal in area to one of

* It does not matter how the figure ABCDE is divided into triangles, nor into how many it is divided.

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