# Proof: By Euclid

• For the straight lines radiating towards the convex (part of the) circumference, $HLKG$, the least is the one between the point and the diameter $AG$, (namely) $DG$, and a (straight line) nearer to the least (straight line) $DG$ is always less than one farther away, (so that) $DK$ (is less) than $DL$, and $DL$ than than $DH$.
• For let the center of the circle have been found [Prop. 3.1], and let it be (at point) $M$ [Prop. 3.1].
• And let $ME$, $MF$, $MC$, $MK$, $ML$, and $MH$ have been joined.
• And since $AM$ is equal to $EM$, let $MD$ have been added to both.
• Thus, $AD$ is equal to $EM$ and $MD$.
• But, $EM$ and $MD$ is greater than $ED$ [Prop. 1.20].
• Thus, $AD$ is also greater than $ED$.
• Again, since $ME$ is equal to $MF$, and $MD$ (is) common, the (straight lines) $EM$, $MD$ are thus equal to $FM$, $MD$.
• And angle $EMD$ is greater than angle $FMD$.1 Thus, the base $ED$ is greater than the base $FD$ [Prop. 1.24].
• So, similarly, we can show that $FD$ is also greater than $CD$.
• Thus, $AD$ (is) the greatest (straight line), and $DE$ (is) greater than $DF$, and $DF$ than $DC$.
• And since $MK$ and $KD$ is greater than $MD$ [Prop. 1.20], and $MG$ (is) equal to $MK$, the remainder $KD$ is thus greater than the remainder $GD$.
• So $GD$ is less than $KD$.
• And since in triangle $MLD$, the two internal straight lines $MK$ and $KD$ were constructed on one of the sides, $MD$, then $MK$ and $KD$ are thus less than $ML$ and $LD$ [Prop. 1.21].
• And $MK$ (is) equal to $ML$.
• Thus, the remainder $DK$ is less than the remainder $DL$.
• So, similarly, we can show that $DL$ is also less than $DH$.
• Thus, $DG$ (is) the least (straight line), and $DK$ (is) less than $DL$, and $DL$ than $DH$. I also say that only two equal (straight lines) will radiate from point $D$ towards (the circumference of) the circle, (one) on each (side) on the least (straight line), $DG$.
• Let the angle $DMB$, equal to angle $KMD$, have been constructed on the straight line $MD$, at the point $M$ on it [Prop. 1.23], and let $DB$ have been joined.
• And since $MK$ is equal to $MB$, and $MD$ (is) common, the two (straight lines) $KM$, $MD$ are equal to the two (straight lines) $BM$, $MD$, respectively.
• And angle $KMD$ (is) equal to angle $BMD$.
• Thus, the base $DK$ is equal to the base $DB$ [Prop. 1.4].
• So I say that another (straight line) equal to $DK$ will not radiate towards the (circumference of the) circle from point $D$.
• For, if possible, let (such a straight line) radiate, and let it be $DN$.
• Therefore, since $DK$ is equal to $DN$, but $DK$ is equal to $DB$, then $DB$ is thus also equal to $DN$, (so that) a (straight line) nearer to the least (straight line) $DG$ [is] equal to one further away.
• The very thing was shown (to be) impossible.
• Thus, not more than two equal (straight lines) will radiate towards (the circumference of) circle $ABC$ from point $D$, (one) on each side of the least (straight line) $DG$.
• Thus, if some point is taken outside a circle, and some straight lines are drawn from the point to the (circumference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straight lines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center.
• For the others, a (straight line) nearer to the (straight line) through the center is always greater than one further away.
• For the straight lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter.
• For the others, a (straight line) nearer to the least (straight line) is always less than one further away.
• And only two equal (straight lines) will radiate from the point towards the (circumference of the) circle, (one) on each (side) of the least (straight line).
• (Which is) the very thing it was required to show.

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