# Proof: By Euclid

• For let $BE$, $CE$, and $GE$ have been joined.
• And since for every triangle (any) two sides are greater than the remaining (side) [Prop. 1.20], $EB$ and $EF$ is thus greater than $BF$.
• And $AE$ (is) equal to $BE$ [thus, $BE$ and $EF$ is equal to $AF$].
• Thus, $AF$ (is) greater than $BF$.
• Again, since $BE$ is equal to $CE$, and $FE$ (is) common, the two (straight lines) $BE$, $EF$ are equal to the two (straight lines) $CE$, $EF$ (respectively).
• But, angle $BEF$ (is) also greater than angle $CEF$.1 Thus, the base $BF$ is greater than the base $CF$.
• Thus, the base $BF$ is greater than the base $CF$ [Prop. 1.24].
• So, for the same (reasons), $CF$ is also greater than $FG$.
• Again, since $GF$ and $FE$ are greater than $EG$ [Prop. 1.20], and $EG$ (is) equal to $ED$, $GF$ and $FE$ are thus greater than $ED$.
• Let $EF$ have been taken from both.
• Thus, the remainder $GF$ is greater than the remainder $FD$.
• Thus, $FA$ (is) the greatest (straight line), $FD$ the least, and $FB$ (is) greater than $FC$, and $FC$ than $FG$. I also say that from point $F$ only two equal (straight lines) will radiate towards (the circumference of) circle $ABCD$, (one) on each (side) of the least (straight line) $FD$.
• For let the (angle) $FEH$, equal to angle $GEF$, have been constructed on the straight line $EF$, at the point $E$ on it [Prop. 1.23], and let $FH$ have been joined.
• Therefore, since $GE$ is equal to $EH$, and $EF$ (is) common, the two (straight lines) $GE$, $EF$ are equal to the two (straight lines) $HE$, $EF$ (respectively).
• And angle $GEF$ (is) equal to angle $HEF$.
• Thus, the base $FG$ is equal to the base $FH$ [Prop. 1.4].
• So I say that another (straight line) equal to $FG$ will not radiate towards (the circumference of) the circle from point $F$.
• For, if possible, let $FK$ (so) radiate.
• And since $FK$ is equal to $FG$, but $FH$ [is equal] to $FG$, $FK$ is thus also equal to $FH$, the nearer to the (straight line) through the center equal to the further away.
• The very thing (is) impossible.
• Thus, another (straight line) equal to $GF$ will not radiate from the point $F$ towards (the circumference of) the circle.
• Thus, (there is) only one (such straight line).
• Thus, if some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight lines radiate from the point towards the (circumference of the) circle, then the greatest (straight line) will be that on which the center (lies), and the least the remainder (of the same diameter).
• And for the others, a (straight line) nearer to the (straight line) through the center is always greater than a (straight line) further away.
• And only two equal (straight lines) will radiate from the same 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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