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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
GH BC; and DK, EL, CH || BG. Prop. 30.
Since, rect. BH =
rect. BH =:
[rect. BK + rect. DL +
rect. BK BG. BD = A. BD,
rect. DL DK. DE = A. DE,
rect. EH = EL.ECA. EC;
A. BC=A. BD+A.DE+A. EC. Wherefore, if there be two str. lines, &c.
THEOR. 3. 2 Eu.
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 + BC2.
draw AF || CD or BE AFCD=BE = BC. rect. AD+rect. CE, (rect. AE = AB. BE = AB. BC,
and Def. Since, rect. AE
AC. CD = AC. BC,
AB. BC= AC. BC + BC2.
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.a=ab+a2.
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 AB2 = AC + CB2 + 2.AC. CB.
CGF AD or BE,
HGK || AB or DE;
BD meets ||s AD, CF,
ext. BGC = int. oppo. ADB; Prop. 28.
BA = AD,
▲ ABD = / ADB ;
CBG = ▲ BGC,
BC = CG,
In the same way it may be proved, that
fig HF AC2.
fig. AG= fig. GE;
fig. AGAC.CG = AC.CB.
fig. AG + fig. GE = 2 AC.CB;
(the 4 figs. HF, CK,
AG, GE, i. e. the
whole fig. AE, or
JAC2 + CB2 + 2. AC. CB.
Wherefore, if a str. line, &c.
ms about the dia. of a sq. are
Note. If the parts of the divided line be a and b, the proposition algebraically expressed is (a+b)2=a2+2ab+b2.
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 + CD2 =
fig. AL= fig. DF,
add to each the fig. CH,
.. whole fig. AH = fig. DF+fig. CH,
but fig. AH=
AD. DB, for DH Cor.
and fig. DF + fig. CH = gnomon CMG,
add to each the fig. LG, which is CD3. gnomon CMG+fig. LG = AD . DB+CD2, i. e. whole fig. CEFB or CB2=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 be + (ab)2=a2; or be + (c− a)2 = a2.
PROP. XLV. THEOR. 6. 2 Eu.
If a straight line be bisected, and produced to