PART IV.

ON SQUARES AND RECTANGLES IN CONNECTION WITH THE SEGMENTS OF A STRAIGHT LINE.

A rectangle whose adjacent sides are AB, AD is denoted by the rect. AB, AD; this is equivalent to the product AB. AD. Similarly a square drawn on the side AB is denoted by the sq. on AB, or AB2.

2. If a point X is taken in a A straight line AB, or in AB produced, then X is said to divide AB into the two segments AX, XB; the segments being in either case the distances of the dividing point X from the extremities of the given line AB.

X B

Fig. 1.

B

X

Fig. 2.

In Fig. 1, AB is said to be divided internally at X.
In Fig. 2, AB

divided externally at X.

Obs. In internal division the given line AB is the sum of the segments AX, XB.

In external division the given line AB is the difference of the segments AX, XB.

THEOREM 50. [Euclid II. 1.]

If of two straight lines, one is divided into any number of parts, the rectangle contained by the two lines is equal to the sum of the rectangles contained by the undivided line and the several parts of the divided line.

Let AB and K be the two given st. lines, and let AB be divided into any number of parts AX, XY, YB, which contain respectively a, b, and c units of length; so that AB contains a+b+c units.

Let the line K contain k units of length.

It is required to prove that

the rect. AB, K=rect. AX, K+rect. XY, K+rect. YB, K, namely that

(a+b+c) k= ak + bk +

ck.

Construction. Draw AD perp. to AB and equal to K.
Through D draw DC par1 to AB.

Through X, Y, B draw XE, YF, BC par1 to AD.

Proof. The fig. AC = the fig. AE + the fig. XF + the fig. YC; and of these, by construction,

fig. AC=rect. AB, K; and contains (a+b+c)k units of area, (fig. AE=rect. AX, K; and contains ak units of area,

or,

the rect. AB, K = rect. AX, K+ rect. XY, K+rect. YB, K;

(a+b+c) k=

ak + bk

+ ck.

Q.E.D.

* COROLLARIES. [Euclid II. 2 and 3.]

Two special cases of this Theorem deserve attention.

(i) When AB is divided only at one point X, and when the undivided line AD is equal to AB.

Then the sq. on AB = the rect. AB, AX+the rect. AB, XB.
That is,

The square on the given line is equal to the sum of the rectangles contained by the whole line and each of the segments.

Or thus:

AB2 = AB. AB

= AB(AX + XB)

= AB. AX + AB. XB.

(ii) When AB is divided at one point X, and when the undivided line AD is equal to one segment AX.

E

с

Then the rect. AB, AX = the sq. on AX the rect. AX, XB.
That is,

The rectangle contained by the whole line and one segment is equal to the square on that segment with the rectangle contained by the two segments.

Or thus:

AB. AX = (AX + XB) AX
=AX2+AX. XB.

THEOREM 51. [Euclid II. 4.]

If a straight line is divided internally at any point, the square on the given line is equal to the sum of the squares on the two segments together with twice the rectangle contained by the segments.

A+2 + XB.AX+XBAX

Let AB be the given st. line divided internally at X; and let the segments AX, XB contain a and b units of length respectively. Then AB is the sum of the segments AX, XB, and therefore contains a+b units.

It is required to prove that

namely that

AB2= AX2 + XB2 + 2AX. XB;

(a+b)2 = a2 + b2 + 2ab.

Construction. On AB describe a square ABCD. From AD cut off AE equal to AX, or a. Then EDXB = b. Through E and X draw EH, XF par1 respectively to AB, AD and meeting

at G.

Proof. Then the fig. AC = the figs. AG, GC+the figs. EF, XH.
And of these, by construction,

fig. AC is the sq. on AB, and contains (a+b)2 units of area;
(fig. AG=sq. on AX, and contains a2 units of area;

fig. GC=sq. on XB,

b2

;

fig. EF = rect. EG, ED

=rect. AX, XB

THEOREM 52. [Euclid II. 7.]

If a straight line is divided externally at any point, the square on the given line is equal to the sum of the squares on the two segments diminished by twice the rectangle contained by the segments.

Let AB be the given st. line divided externally at X; and let the segments AX, XB contain a and b units of length re spectively.

Then AB is the difference of the segments AX, XB, and therefore contains a b units.

It is required to prove that

namely that

AB2 = AX2 + XB2 - 2AX. XB; (a - b)2= a2 + b2

2ab.

Construction. On AX describe a square AXGE. From AE cut off AD equal to AB, or a - b. Then ED = XB = b. Through D and B draw DF, BH par1 respectively to AX, AE, meeting

at C.

Proof. Then the fig. AC = the figs. AG, CG – the figs. EF, XH. And of these, by construction,

fig. AC is the sq. on AB, and contains (a - b)2 units of area;

(fig. AG=sq. on AX, and contains a2 units of area;