446 4 Demonstration of 
41 
under the two greater and two less, minus 
the sum of the rectangles under each of — 
the greater and each of the less, There- 
fore (AC*—AF* ) x(CB*—FB?) — AC* 
. CB’ + AF’. FB*— AC’, FB* —CB? 
. AF*. But CF bisects the angle ACB, 
consequently CF? = AC. CB —AF. FB, 
AC .FB= AF .CB, and AC*. FB* + 
CB? . AF = 2AC* . FB* = 2AC . FB. 
AF. CB, therefore the quantity above = 
AC* . CB* + AF? . FB*— 2AC.CB.. 
AF. FB= the square of AC. CB— 
AF. FB, and consequently = CF* x CF” ; 
therefore AC? — AF* : CF’: : CF* : CB 
— FB*. Which is Proposition 41. 
If upon CF produced be taken E so 
that BE = BF, then the angle BEF — 
BFE = AFC, and the angles at C being 
equal, the two triangles ACF, BCE are 
rectangle under the whole and Ashe lesser segments are con- 
tinual proportionals. 
Prop. Lil. If three lines are continual proportionals : 
the sum of the squares of the mean and the greater ex- 
treme is to the rectangle contained under the same, as the 
sum of the extremes is to the mean, 
Prop. LIU. In every right-angled triangle, as the 
hypothenuse is to the sum of the sides about the right 
eee 
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