Meteorology By Aristotle

Let A be a hemisphere resting on the circle of the horizon, let its
centre be K and let H be another point appearing on the horizon. Then,
if the lines that fall in a cone from K have HK as their axis, and,
K and M being joined, the lines KM are reflected from the hemisphere
to H over the greater angle, the lines from K will fall on the circumference
of a circle. If the reflection takes place when the luminous body
is rising or setting the segment of the circle above the earth which
is cut off by the horizon will be a semi-circle; if the luminous body
is above the horizon it will always be less than a semicircle, and
it will be smallest when the luminous body culminates. First let the
luminous body be appearing on the horizon at the point H, and let
KM be reflected to H, and let the plane in which A is, determined
by the triangle HKM, be produced. Then the section of the sphere will
be a great circle. Let it be A (for it makes no difference which of
the planes passing through the line HK and determined by the triangle
KMH is produced). Now the lines drawn from H and K to a point on the
semicircle A are in a certain ratio to one another, and no lines drawn
from the same points to another point on that semicircle can have
the same ratio. For since both the points H and K and the line KH
are given, the line MH will be given too; consequently the ratio of
the line MH to the line MK will be given too. So M will touch a given
circumference. Let this be NM. Then the intersection of the circumferences
is given, and the same ratio cannot hold between lines in the same
plane drawn from the same points to any other circumference but MN.

Draw a line DB outside of the figure and divide it so that D:B=MH:MK.
But MH is greater than MK since the reflection of the cone is over
the greater angle (for it subtends the greater angle of the triangle
KMH). Therefore D is greater than B. Then add to B a line Z such that
B+Z:D=D:B. Then make another line having the same ratio to B as KH
has to Z, and join MI.