A history of Greek mathematics - Wilbourhall.org

THE COLLECTION. BOOK VII 413
Draw EG perpendicular to BF.
Then the triangles BCH, EGF are similar and 'since
BC = EG) equal in all respects ; therefore EF = BH.
Now BF* = BE 2 + EF 2 ,
BF+ BF . FC = BH . BE+ BE . EH+EF
or BC .
2 .
But, the angles HCF, HEF being right, H, C, F, E are
BF = BH . BE.
concyclic, and BC .
Therefore, by subtraction,
BF.FC= BE. EH+EF 2
= BE.EH+BH 2
= BH.HE+EH 2 + BH 2
= EB.BH+EH 2
= FB.BC+EH 2 .
Taking away the common part, BC .
CF 2 = BC 2 + EH 2 .
GF, we have
Now suppose that we have to draw BHE through B in
such a way that HE = k. Since BC, EH are both given, we
have only to determine a length x such that x 2 = BC 2 + Jc 2 ,
produce BC to F so that CF = 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 BC, the base of an isosceles triangle ABC, then BC is the
shortest of all the straight lines through D terminated by
the straight lines A B, AC, and the nearer to BC 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 external
to one another, on the same or opposite sides of the
diameter.