# Proof: By Euclid

• For let the center of the circle $ABC$ have been found [Prop. 3.1], and let it be (at point) $E$, and let $EA$ and $EB$ have been joined.
• And since $AF$ is equal to $FB$, and $FE$ (is) common, two (sides of triangle $AFE$) [are] equal to two (sides of triangle $BFE$).
• And the base $EA$ (is) equal to the base $EB$.
• Thus, angle $AFE$ is equal to angle $BFE$ [Prop. 1.8].
• And when a straight line stood upon (another) straight line makes adjacent angles (which are) equal to one another, each of the equal angles is a right angle [Def. 1.10] .
• Thus, $AFE$ and $BFE$ are each right angles.
• Thus, the (straight line) $CD$, which is through the center and cuts in half the (straight line) $AB$, which is not through the center, also cuts ($AB$) at right angles.
• And so let $CD$ cut $AB$ at right angles.
• I say that it also cuts ($AB$) in half.
• That is to say, that $AF$ is equal to $FB$.
• For, with the same construction, since $EA$ is equal to $EB$, angle $EAF$ is also equal to $EBF$ [Prop. 1.5].
• And the right angle $AFE$ is also equal to the right angle $BFE$.
• Thus, $EAF$ and $EFB$ are two triangles having two angles equal to two angles, and one side equal to one side - (namely), their common (side) $EF$, subtending one of the equal angles.
• Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
• Thus, $AF$ (is) equal to $FB$.
• Thus, in a circle, if any straight line through the center cuts in half any straight line not through the center then it also cuts it at right angles.
• And (conversely) if it cuts it at right angles then it also cuts it in half.
• (Which is) the very thing it was required to show.

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