A circle is given and it has a point inside the circle. How to construct a chord that passes through the point and equal to the radius of the circle with compass and straight edge?
I know how to construct with the intersecting chords theorem. Is there any other ways to costruct this?
Best Answer
Well, the side length of a regular hexagon equals the circumradius, so you may simply draw an inscribed regular hexagon and rotate it until one of its sides goes through the given point. Of course this can be done only if the distance of the given point from the center of the circle is $\geq \frac{\sqrt{3}}{2}R$.
Let $P$ be the given point and $O$ the center$^{(*)}$ of the given circle $\Gamma$. You may
$DP$ meets $\Gamma$ again at $E$ and $DE$ is a solution.
$(*)$ If $O$ is not previously drawn, you may simply construct it as the intersection of the axis of two different chords. Here an alternative construction which exploits the intersecting chords theorem:
If $VP$ meets $\Gamma$ again at $W$, $VW$ is a chord whose length equals $OQ$.
This construction is a bit more involved than the previous one, but both approaches can be easily modified in order to solve the problem of drawing a chord through $P$ of any allowed length.