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.