438 A Demonstration of 
33 consequently y, D’, h are in a right line. 
Which is the thirty-third Proposition. 
Also because °D’h—?Fh=,D’W’ — »FH 
therefore Fh, FE’ make equal angles with 
dy and EE’ is perpendicular to the diameter 
DH, and parallel to TT’; draw FD’ cut- 
ting T'T’ in G’, and the circle in N, then 
FD’. FN = FT? = FG? + GT? =FG* 
4 GG’ +TG@. GT =FG" +DG’. 
GN. Therefore T'T’ is the locus of all 
points G’ dividing lines intercepted be- 
tween F and the periphery so that FD’. 
FN — GF’ + DG’. GN. From E and 
h through G’ draw lines meeting the circle 
again in S and R, and describe as in the 
2d Diagram, the circle FTnT’ cutting 
FN inl. Then because FG’. G1=TG’. 
G'T’, and this = SG’. G’E = hG@’ . GR, 
therefore the points F, 1,8, E are in a 
other, and another angle B in the first equal to the sum of 
the angles D and E in the second; then shall the sides AC’ 
BC, DE, EF be proportional. 
Prop. XLI. The square of the line bisecting the ver- 
tical angle of any triangle is a mean proportional between 
the differences of the squares of each side including that 
angle, and the square of the adjacent segment of the base 
made thereby. 
Prop. XLII. If from the same point two tangents be 
drawn to a circle, and a line be drawn joining the points 
