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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.

Then I is the pole of the circle on which the lines from K fall. For

the ratio of D to IM is the same as that of Z to KH and of B to KI. If

not, let D be in the same ratio to a line indifferently lesser or

greater than IM, and let this line be IP. Then HK and KI and IP will

have the same ratios to one another as Z, B, and D. But the ratios

between Z, B, and D were such that Z+B:D=D: B. Therefore

IH:IP=IP:IK. Now, if the points K, H be joined with the point P by the

lines HP, KP, these lines will be to one another as IH is to IP, for

the sides of the triangles HIP, KPI about the angle I are

homologous. Therefore, HP too will be to KP as HI is to IP. But this

is also the ratio of MH to MK, for the ratio both of HI to IP and of

MH to MK is the same as that of D to B. Therefore, from the points

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