THE COLLECTION. BOOK VII 
413 
Draw EG perpendicular to BF. 
Then the triangles ECU, EGF are similar and (since 
BG = EG) equal in all respects: therefore EF = BE. 
Now BF 2 = BE 2 + EF 2 , 
or BG. BF+ BF. FC = BE. BE+BE. EE+EF 2 . 
But, the angles ECF, EEF being right, E, 0, F, E are 
concyclic, and BG. BF — BE. BE. 
, Therefore, by subtraction, 
BF .FC = BE .EE+EF 2 
= BE. EE + BE 2 
= BE. EE + EE 2 + BE 2 
= EB.BE+EE 2 
= FB.BG + EE 2 . 
Taking away the common part, BG . GF, we have 
CF 2 = BG 2 + EE 2 . 
Now suppose that we have to draw BEE through B in 
such a way that EE = k. Since BG, EE are both given, we 
have only to determine a length x such that x 2 = BG 2 + k 2 , 
produce BG to F so that GF = x, draw a semicircle on BF as 
diameter, produce AD to meet the semicircle in E, and join 
BE. BE is thus the straight line required. 
Prop. 73 (pp. 784-6) proves that, if D be the middle point 
of BG, the base of an isosceles triangle A BG, then BG is the 
shortest of all the straight lines through D terminated by 
the straight lines AB, AG, and the nearer to BG is shorter than 
the more remote. 
There follows a considerable collection of lemmas mostly 
showing the equality of certain intercepts made on straight 
lines through one extremity of the diameter of one of two 
semicircles having their diameters in a straight line, either 
one including or partly including the other, or wholly ex 
ternal to one another, on the same or opposite sides of the 
diameter.