440 A Demonstration of 
42 
the angles at 9’ and T’ being right, the 
points g’, 'T’, b, n are ina circle, and the 
Znbg’ = n'T’g’, consequently nbg’ = nag’ 
and ng” bisects both ab and the chord of 
the circle. Which is the a ae 
Proposition. 
Let fall TC david dicdtin to T’n then 
by similar triangles TG =iTT’: Tn: : 
TC : TT, therefore Tn. T’C— £TT”. 
_ Which is the forty-third Proposition. 
If any diameter AB be drawn to this 
circle, and TA, T’B be drawn intersect- 
ing in O, through which drawing OF 
intersecting AB in P and producing BT, 
AT’ till they intersect in H; then because 
BA. is a diameter, the angles at T and 
T’ are right, and the points T, O, T’, H 
inacircle; .. the 2THT’ = TOB =the 
complement of TBO; but because F'T' is 
i 
isosceles triangle a perpendicular be drawn to the opposite 
side; then Isay that the rectangle contained under that 
side and it’s segment intercepted by the perpendicular and 
the base is equal to half the square of the base. 
Prop. XLIV. If ina line AB two points C and D be 
taken; then I say that 
AB-+ AD x BC + BC* = 2ABC + BCD. 
And moreover that ACV. DEES 
. 
AB+AD X CD-+- CD* = 2ADC + BCD. 
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Oe fan ee Pl 
