448 A Demonstration of 
Diracram IX. 
Since EC? = DC* — DE’; .-. AC* 
= AK*?+ 2AE. EC + EC*= AE’? + 
AE .EC + AE .BC+ AE. EB+ 
EC?— AE. AC + AE. BC+ DE’ + 
EC?= DC: + AE.(AC+BC). Also 
AC” — AE? + 2AE x EC’ + EC7= AE 
.AC’+ AE. EB — AE. CB EC? 
AE x (AC’— C’B) + DE*+ EC* =DC* 
46 + AE x (AC’— C’B). Which is Propo- 
sition the 46th. 
DraGrRaM X. 
Having through the given rectangle 
ABCD drawn AE meeting DC produced 
in E, and cutting BC in F, let fall BL 
perpendicular thereto. Then by similar 
Pror. LV. In every right-angled triangle, as the differ- 
ence between the hypothenuse and one side is to the differ- 
ence between the same side and its adjacent segment, so‘is 
the same side to the same segment. : 
Pror. LVI. If HC bea tangent toa circle meeting the 
diameter DB produced in H, and from the point of contact 
Ca perpendicular CK to that diameter be drawn, and like- 
wise a line from H cutting the circle in Fand G, and the 
perpendicular CK in I, and F be the nearest point to H; 
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