H
E
E
L
E
M
E
N
T
S
let the angle BGH be constructed equal to the angle at DFE,
and
the angle GBH equal to the angle FDE. [I.23]
Therefore
the remaining angle E is equal to the remaining angle at H; [I.32]
therefore
the triangle FDE is equiangular with the triangle GBH;
therefore, proportionally,
as FD is to GB,
so is FE to GH,
and
ED to HB. [VI.4]
