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BC is equal to the sum of the rectangles contained by the undivided line (A) and the several parts (BD, DE, and E C) of the divided line (B C).

The beginner must carefully observe that the gist of the proof does not consist in showing that the rectangle BH is equal to the sum of the rectangles BK, DL, and EH, because that is self-evident; but in showing that the rectangles BH, BK, DL, and EH answer to the description of them given in the enunciation.

PROPOSITION II.

If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the rectangles contained by the whole and each of the parts.

For the construction employed in this proposition we must be able:

1. On a given straight line to describe a square. (I. 46.)

2. Through a given point to draw a straight line parallel to a given straight line. (I. 31.)

Let AB be a right line divided into two parts at the point C. We have to show that the on AB is equal to the sum of rect. A B, A C and rect. A B, BC. On A B describe the □ A E. Through EC draw CF parallel to AD, and meeting D E in the point F.

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A

C

B

It is self-evident that the AE is equal to the sum of the rectangles A F and CE.

But the AE is the on AB.
The rectangle AF is the rectangle contained by

A B and A C, for it is contained by AD and AC, and AD and A B are equal, being sides of the same.

The rect. CE is the rectangle contained by A B and BC, for it is contained by B E and B C, and BE is equal to A B.

Consequently the on A B is equal to the sum of the rectangles contained by the whole line (A B), and each of the parts (A C and CB) into which it is divided.

Here again the beginner must observe that the main point in the above proof is, to show that the A E, and the rectangles A F and CE answer to the description of them given in the enunciation. We need no proof to show that the AE is equal to the sum of the rects. A F and CE.

If in Prop. I. the two lines A and B C were of the same length, and BC were divided into two parts, it is obvious that the second proposition might be considered as only a particular case of the first; for the rect. under A and B C would then be the same as the☐ on BC, and the two rects. under A and the parts of the divided line would be the same as the rectangles under the whole line BC and each of its parts.

PROPOSITION III.

If a right line be divided into any two parts, the rectangle contained by the whole line and one of the parts is equal to the square on that part together with the rectangle contained by the two parts.

The propositions required for the construction em

ployed in this proposition are the same as in the first and second propositions.

Let A B be a right line divided into two parts at C. We have to prove that the rect. A B, A C is equal to the on AC together with the rect. A C, CB.

B

E

ADF C.

On A C describe the
Produce D F, and through the point
B draw B E, parallel to A D, and
meeting DF produced in the point

E.

It is self-evident that the rect. A E is equal to the sum of the AF and the rect. CE.

But A E is the rectangle contained by A B and AC, for it is contained by AB and AD, and AD and A C are equal, being sides of the same ☐.

AF is the on A C.

CE is the rectangle contained by A C and C B, for it is contained by F C and C B, and F C and A C are equal, being sides of the same ☐.

Consequently the rectangle under the whole line (AB) and one part (A C) is equal to the square on that part, together with the rectangle contained by the two parts (A C and CB.)

Here again the gist of the proof consists in showing that the rectangle A E, and its parts A F and CE answer to the description given of them in the enunciation.

[The following proposition is made use of in several succeeding propositions.]

PROPOSITION A.

The parallelograms about the diagonal* of a square, are themselves squares.

For the proof of this proposition we must know the following,

1. Magnitudes that are equal to the same are equal to each other.

* That is parallelograms whose diagonals are parts of the diagonal of the square. See Book I., Prop. 43.

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2. The angles at the base of an isosceles triangle
are equal. (I. 5.)

3. If two angles of a triangle are equal, the sides
opposite those angles are also equal. (I. 6.)
4. If a right line intersect two parallel right lines,
the external angle is equal to the internal
opposite angle on the same side of the inter-
secting line, and the two internal angles on the
same side are together equal to two right
angles. (I. 29.)

5. The opposite sides and angles of a parallelo-
gram are equal. (I. 34.)

Let ABCD be

a□, and AC one of its diagonals. Through a point K G in AC the lines HG and EF are drawn parallel respectively to A B and AD. HE and F G are parallelograms about the diagonal of the square. We have to prove that they are themselves squares.

A

E B

First take the parallelogram HE. Because DC and HG are parallel and AC intersects them, therefore the HKA is equal to the DCA. But the lines AD and DC are equal because they are sides of the same ☐; therefore the

DOA.

Consequently the

DAC is equal to the

HAK (i.e., DA C) and the ZHKA are equal to each other, for they are both equal to the DCA.

It follows, therefore (according to I. 6), that the line HA is equal to the line HK.

But HE is a parallelogram; therefore its opposite sides are equal. Consequently HA and K E are equal, and HK and A E are equal. Therefore HE is equilateral, that is, has all its sides equal.

Moreover the 8 HAE and HKE are opposite s

of a parallelogram, and are therefore equal, and HAE is a right, therefore HKE is a right . Also since A B and H G are parallel and A H intersects them, the sHA E and A HK are together equal to two rights; and since HAE is a right, AHK must be a right, and the KEA, which is opposite and equal to the ▲ AHK, must also be a right ▲.

We have thus shown that the parallelogram HE has all its sides equal, and all its angles right angles; therefore it is a square.

In a similar way it may be shown that the parallelogram F G is a square.

The preceding proposition is usually incorporated in the proof of the Fourth Proposition.

PROPOSITION IV.

If a right line be divided into any two parts, “ the square on the whole line is equal to the sum of the squares on the parts together with twice the rectangle under the parts.

For the construction employed in this proposition we must be able,—

1. To join two points by a right line. (Post. I.) 2. On a given straight line to describe a square. (I. 46.)

3. Through a given point to draw a line parallel
to a given straight line. (I. 31.)

For the proof of the proposition we must know,--
1. That the parallelograms about the diagonal of
a square are themselves squares. (Prop. A.)
2. That the opposite sides of a parallelogram are
equal. (I. 34.)

Let A B be a straight line divided into two parts at C.

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