H
E
E
L
E
M
E
N
T
S
the parallelogram BE equal to the triangle ABC, [I.44]
and to CE
the parallelogram CM equal to D and in the angle FCE which is equal to the angle CBL. [I.45]
Therefore
BC is in a straight line with CF, and LE with EM.
Now
let GH be taken a mean proportional to BC, CF [VI.13],
and
on GH let KGH be described similar and similarly situated to ABC.
