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angle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line.

Let A and BC be two str. lines, and let BC be divided into any parts in D and E; then A.BC=A. BD+A. DE+A. EC

[blocks in formation]

Since, rect. BH = {

Draw BF I BC,

Prop. 10. make BG = A;

Prop. 3. draw GH || BC; and DK, EL, CH || BG. Prop. 30. .. A = = BG = DK = EL.

Prop. 33. Frect. BK + rect. DL +

rect. EH,
(rect. BH = BG .BC = A.BC,
And

rect. BK = BG .BD= A.BD,
rect. DL = DK. DE = - A.DE,
rect. EH = EL. EC = A.EC;

A. BC = A.BD+A. DE+A.EC.
Wherefore, if there be two str. lines, &c.

3. 2 Eu.

PROP. XLII. THEOR.
If a straight line be divided into any two

parts, the rectangle contained by the whole
and one of the parts, is equal to the rect-

angle contained by the two parts, together with the square of the aforesaid part.

Let the str. line AB be divided into two parts in the pt. C; then AB. BC = AC. BC + BC%.

[blocks in formation]

30.

Prop. 38.

On BC descr. sq. CE.

prod. ED to F. Prop. 30.

draw AF || CD or BE Prop. 33

... AF

- CD BE=BC. and Def. Since, rect. AE = rect. AD + rect. CE,

(rect. AE = AB. BE = AB. BC,
And {rect. AD= AC.CD= AC .BC,

(rect. CE = BC;
... AB. BC = AC. BC + BC%.
Therefore, if a str. line be, &c.

Note.—If the parts of the divided line be a and b; the proposition algebraically expressed, is a+b.arab ta'.

PROP. XLIII. THEOR.

4. 2 Eu.

If a straight line be divided into any two

parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.

Let str. line AB be divided into any two parts in C, then ABP = AC + CB2 + 2.AC. CB.

[blocks in formation]

{

Ax. 1.

Ax. l.

Upon AB descr. sq. ABED.

Prop. 38. join BD

CGF || AD or BE,
draw
.: BD meets || AD, CF,
:. ext. _ BGC = int. oppo. _ ADB; Prop. 28.

BA= AD,
LABD=LADB;

Prop. 5.
i. e.
Z CBG

_BGC,
..
BC = CG,

Prop. 6. but BC = GK,

Prop. 33. ..

CG = GK;
BC=CG = GK = BK,

fig. CK is equilat. Again :: CB meets Ils CG, BK,

.. LKBC + ZGCB = 2 rt. Zs; Prop. 28. but

LKBC
..

< GCB = rt. L
opp. Łs CGK, GKB are rt. _s.

Prop. 33. . . fig. CK is rectang.

fig. CK = CB. In the same way it may be proved, that

fig HF = AC-. Also fig. AG = fig. GE;

Prop. 37. i. e. fig. AG= AC.CG -- AC.CB.

or

rt. L

Ax.l.

fig. GE= AC.CB
:. fig. AG + fig. GE: = 2 AC.CB;
(the 4 figs. HF, CK,

AG, GE, i. e. the SAC + CB +
whole fig. AE, or

2. AC. CB.
AB?.
Wherefore, if a str. line, &c.

Cor. The about the dia. of a sq. are likewise squares.

Note. If the parts of the divided line be a and b, the proposition algebraically expressed is (a+b)'=a + 2ab +62.

yms

PROP. XLIV. THEOR.

5. 2 Eu.

If a straight line be divided into two equal

parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the whole line.

Let the str. line AB be divided into two equal parts at the pt. C, and into two unequal parts at the point D; then AD. DB + CD = CB2. .

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fig. CM ={fig. AL, for AC=

CB;

On CB descr. sq. CEFB,

Prop. 38. join BE; ŞAK, DHG || CE or BF

Prop 30. draw

KLHM || AB.
Since, comp. CH= comp. HF Prop.37.

add to each the fig. DM.
.. fig. CM = fig. DF,
but,

Prop. 35. .. fig. AL = fig. DF,

add to each the fig. CH, :: whole fig. AH = fig. DF+fig. CH, but fig. AH = JAD. DB, for DH Çor.

Prop. 43.

= DB;
and fig. DF + fig. CH = gnomon CMG,

:: gnomon CMG = AD. DB,
add to each the fig. LG, which is CD'.

Prop. 43. :: gnomon CMG+fig. LG = AD.DB+CD2, i. e. whole fig. CEFB or CB’=AD.DB+CD2.

Wherefore, if a str. line, &c.

Cor.—Since CB2 – CD=AD.DB; the diff. of the Squares of two unequal lines is equal to the rectangle of their sum and diff.

Note.--If the unequal parts of the divided line be b and c, and the half line be a; the proposition algebraically expressed is bc + (a - b)=a”; or bc + (c-a) = a.

Cor.

6. 2 Eu.

PROP. XLV. THEOR.
If a straight line be bisected, and produced to

E

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