# Proof: By Euclid

(related to Proposition: 3.02: Chord Lies Inside its Circle)

• For (if) not then, if possible, let it fall outside (the circle), like $AEB$ (in the figure).
• And let the center of the circle $ABC$ have been found [Prop. 3.1], and let it be (at point) $D$.
• And let $DA$ and $DB$ have been joined, and let $DFE$ have been drawn through.
• Therefore, since $DA$ is equal to $DB$, the angle $DAE$ (is) thus also equal to $DBE$ [Prop. 1.5].
• And since in triangle $DAE$ the one side, $AEB$, has been produced, angle $DEB$ (is) thus greater than $DAE$ [Prop. 1.16].
• And $DAE$ (is) equal to $DBE$ [Prop. 1.5].
• Thus, $DEB$ (is) greater than $DBE$.
• And the greater angle is subtended by the greater side [Prop. 1.19].
• Thus, $DB$ (is) greater than $DE$.
• And $DB$ (is) equal to $DF$.
• Thus, $DF$ (is) greater than $DE$, the lesser than the greater.
• The very thing is impossible.
• Thus, the straight line joining $A$ to $B$ will not fall outside the circle.
• So, similarly, we can show that neither (will it fall) on the circumference itself.
• Thus, (it will fall) inside (the circle).
• Thus, if two points are taken at random on the circumference of a circle then the straight line joining the points will fall inside the circle.
• (Which is) the very thing it was required to show.

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