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BOOK II.

On Equality of Area of Rectangular Figures.

Since all rectangles which have two adjoining sides equal to two given straight lines are identically equal, a rectangle is determined when two adjoining sides are given; it is spoken of as the rectangle contained by those two lines, as rect. AB. BC, which is read, "the rectangle contained by AB, BC."

When the adjoining sides of a rectangle are equal it is a square (I. 46); thus rect. AB. AB is AB”, which is read, the square on AB.

PROP. 1. THEOR. The rectangle contained by two given lines is equal to the sum of the rectangles contained by one of the lines and the several parts into which the other is divided. Let L and AB be the two L

с D
given lines, of which
A

B
AB is divided into any
number of parts, as at
C and D.

G H To prove that rect. L.AB

= sum of rects. L. AC, L. CD, and L. DB. From A draw AE 1 AC (1. 11), and make AE =

L. Through E draw a line || AB, also through C, D, and

B, draw lines || AE (1. 31).
Then AH AF + CG + DH.
But AH is rect. EA. AB rect. L. AB,
and AF is rect. EA. AC rect. L. AC,

and CG is rect. L.CD, : FC = EA (I. 34)=L (cons.). Similarly, DH is rect. L.DB. :: rect. L. AB = L. AC + L.CD + L.DB.

Q.E.D.

PROP. 2. THEOR.

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

с A

B into any two parts at C. To prove rects. AB. AC + L

AB. CB = A B2. Let another line L be taken AB. By Prop. 1, rect. L. AB = L. AC + L. CB. :. AB. AB (i.e., ABP) = AB. AC + AB. CB. Q.E.D.

PROP. 3. THEOR. If a straight line be divided into any two parts, the rectangle contained by the whole line and one of the parts will be equal to the square on that part together with the rectangle contained by the two parts. Let AB be divided into any

А.

B two parts at C. To prove rect. AB.BC = BCL

+ AC.CB. Let another line L be taken BC. By Prop. 1, rect. L. AB = L. AC + L.CB.

:rect. AB. BC = AC.CB + CB”. Q.E.D. Note. Propositions 2 and 3 are particular cases of Prop. i, and may be ranked as corollaries to that proposition.

Ex. 1. Prove II. 2 and 3 by means of figures as in II. 1.

parts at C.

PROP. 4. THEOR. If a straight line be divided into any two parts, the square on the whole line will be equal to the squares on the two parts together with twice the rectangle contained by the two parts. Let AB be divided into any two

A

B To prove ABP = AC° + CB? + 2 rect. AC.CB.

H

K
Describe the square on AB

(1. 46). Through C draw
CF || AD (I. 31).

D
Cut off CG CB; and :: CF

= BE (1. 34) = AB, : GF = AC (Ax. 3). Through G draw HGK | AB (1. 31). Then all the figures are rectangular parallelograms by

construction : and since CG = CB, CK is a square, the sq. on CB. Similarly, HF is HG' = AC (1. 34). Also AG is rect. AC.CG = AC. CB, and GE is rect. FG. GK AC.CB, :: HF + CK + AG + GE = ACP + CBP + 2 AC.CB. But HF + CK + AG + GE make up AE, which is the sq. on AB,

:: AB? = ACP + CB? + 2 AC. CB. Q.E.D. COR.—The square on a line is four times the square on half the line.

Note.-Prop. 4 may be expressed thus:-" The square on the sum of two lines (as AC and CB) is equal to the sum of the squares on the lines together with twice the rectangle contained by them;" then Prop. 4 corresponds to the algebraic proposition (a + b)2 = ax + b2 + 2 ab.

2. Prove Prop. 4 by means of Propositions 2 and 3.

3. Show that the straight line joining BD in the above figure passes through G.

PROP. 7. THEOR.

If a straight line be divided into any two parts, the squares on the whole line and on one of the parts will be equal to twice the rectangle contained by the whole and that part, together with the square on the other part. Let AB be divided into any two parts at C.

A To prove ABP + CBP = 2 rect.

AB. BC + AC?. :: AB

AC + CBP + 2 rect. AC.CB (II. 4). To each add CB?, :. AB + CB2 = AC + 2 CB? + 2 AC. CB. Now CB? + AC. CB AB. BC (II. 3). :: AB? + CB AC + 2 AB. BC.

Q.E.D.

Note.-This proposition may be expressed thus:-AC2 < ABP + BC2 by 2 AB. BC. As AC is the difference of AB and BC, the enunciation might then be given: " The square on the difference of two lines is less than the sum of the squares on the two lines by twice the rectangle contained by them;" then Prop. 7 corresponds to the algebraic proposition (a - b)2 = a2 + 32 – 2 ab.

4. Prove Prop. 7 by the help of a figure as in II. 4, but independently of that proposition.

5. The square on the sum of two lines (as AB + BC) is equal to the square on their difference (AC), together with four times the rectangle contained by the lines (4 AB. BC). (Eu. II. 8.)

6. If a líne be divided internally into unequal segments, the distance of the point of section from the middle point of the line is half the difference of the segments ; if the point of section be external, its distance from the middle point of the line is half the sum of the segments.

7. Given the sum and difference of two straight lines, to find them.

8. Prove geometrically that the square on a whole line is nine times the square on a third part of it.

PROP. A. THEOR.

Euclid II., 5 and 6. The difference of the squares on two given lines is equal to the rectangle contained by the sum and difference of the lines.

F

Let AB and BC be

B

А.
the two given lines,
of which BC is

M M the greater, and let

K
H

L
them be placed in
the same straight
line. Cut off BD

E

G AB. Then AC is the sum of the two lines, and DC is their

difference. To prove the difference of BC and AB2 rect.

AC.CD. Describe the square on BC (I. 46). Draw DG || BE (1. 31). Cut off DL DC. Through L draw KM | AC, and through A draw AK

|| BE (I. 31). As in II. 4, HG is HL - BD2 = AB, and BF is BC?. :: the difference of BC and AB? is the difference of

the sum of BM and MG. AB and LM

DL (11. 4) :: MG

AH, and :: BM + MG AM. :: the difference of BC? and ABP = AM rect.

rect. AC.CD. Q.E.D. Note The algebraic proposition corresponding to the above is a? – 62=

BF and HG But :: GL

BH,

AC.CM

(a + b) (a - b).

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